Answer:
The radius of the curve that Car 2 travels on is 380 meters.
Explanation:
Speed of car 1, [tex]v_1=65\ km/h[/tex]
Radius of the circular arc, [tex]r_1=95\ m[/tex]
Car 2 has twice the speed of Car 1, [tex]v_2=130\ km/h[/tex]
We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
According to given condition,
[tex]\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}[/tex]
[tex]\dfrac{65^2}{95}=\dfrac{130^2}{r_2}[/tex]
On solving we get :
[tex]r_2=380\ m[/tex]
So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.
Final answer:
For Car 2 to have the same centripetal acceleration while traveling at twice the speed of Car 1, it must travel on a curve with a radius that is four times larger, which would be 380 meters.
Explanation:
The centripetal acceleration (ac) of a car going around a level curve at a constant speed is given by the formula ac = v2 / r, where v is the speed of the car and r is the radius of the circular path. If Car 1 is traveling at 65 km/h and has a centripetal acceleration on a curve with a radius of 95 m, and Car 2 is traveling at twice the speed of Car 1, we are asked to find the required radius of the curve for Car 2 to maintain the same centripetal acceleration.
To keep the same centripetal acceleration for Car 2, which is moving at twice the speed, the radius r2 must be increased proportionally to the square of the speed ratio. Since Car 2 is traveling at twice the speed, the radius must be increased by a factor of 22, or 4 times larger than that for Car 1. Therefore, if Car 1's radius is 95 m, Car 2's radius needs to be 95 m * 4, which is 380 m.
How strong an electric field is needed to accelerate electrons in an X-ray tube from rest to one-tenth the speed of light in a distance of 5.1 cm?
Answer:
E= 50.1*10³ N/C
Explanation:
Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:
[tex]vf^{2} -vo^{2} = 2*a*x[/tex]
We know that v₀ = 0 (it starts from rest), that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:
[tex]a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2[/tex]
According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.
Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:
F = q*E (1)
In this case, q=e= 1.6*10⁻19 C
But this force, can be expressed in this way, according Newton's 2nd Law:
F = m*a (2) ,
where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.
From (1) and (2), we can solve for E, as follows:
[tex]E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C[/tex]
⇒ E = 50.1*10³ N/C
A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door. Where is the pivot point and what is the maximum torque?
Answer:
Explanation:
Given
Force applied [tex]F=80\ N[/tex]
Door is [tex]d=3\ m[/tex] wide
for gate to revolve Properly Pivot Point must be at center i.e. 1.5 from either end
Torque applied is [tex]T=force\times distance[/tex]
Maximum torque
[tex]T_{max}=F\times \frac{d}{2}[/tex]
[tex]T_{max}=80\times \frac{3}{2}[/tex]
[tex]T_{max}=120\ N-m[/tex]
Final answer:
The pivot point of a revolving door is at its central axis, and the maximum torque exerted by the businesswoman on the revolving door is 120 Nm, calculated by multiplying the force (80 N) by the perpendicular distance from the pivot point to the point of application (1.5 m).
Explanation:
The question is asking about the concept of torque, which is a measure of the turning force on an object. In the scenario described, a businesswoman is applying a force of 80 N to a revolving door. The maximum torque can be calculated by multiplying the force applied by the perpendicular distance from the pivot point to the point of application. The pivot point of a revolving door is its central axis, which is the center of the door.
To calculate the maximum torque, we use the formula:
Torque = Force * Distance.
Here, Torque = 80 N * 1.5 m (since the force is applied at the edge of the 3 m wide door, the distance from the pivot point is half the width).
Thus, Torque = 120 Nm
This is the maximum torque exerted by the woman on the door relative to its central.
A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 4 s. The additional power needed to achieve this acceleration is (a) 56 kW (b) 222 kW (c) 2.5 kW (d) 62 kW (e) 90 kW
To solve this problem we will apply the concepts related to power as a function of the change of energy with respect to time. But we will consider the energy in the body equivalent to kinetic energy. The change in said energy will be the difference between the two velocity data given by half of the mass. We will first convert the given units into an international system like this
Initial Velocity,
[tex]V_i = 60km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]V_i = 16.6667m/s[/tex]
Final Velocity,
[tex]V_f = 100km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]V_f = 27.7778m/s[/tex]
Now Power is defined as the change of Energy over the time,
[tex]P = \frac{E}{t}[/tex]
But Energy is equal to Kinetic Energy,
[tex]P = \frac{\frac{1}{2} m\Delta v^2}{t}[/tex]
[tex]P = \frac{\frac{1}{2} m(v_f^2-v_i^2)}{t}[/tex]
Replacing,
[tex]P = \frac{\frac{1}{2} (900)(27.7778^2-16.6667^2)}{4}[/tex]
[tex]P = 56kW[/tex]
Therefore the correct answer is A.
Final answer:
The additional power needed to achieve the acceleration is 62 kW.So,option (d) 62 kW is correct.
Explanation:
To calculate the additional power needed to achieve acceleration, we can use the formula for power: Power = Force x Velocity. We know the mass of the car (900 kg), the initial velocity (60 km/h), and the final velocity (100 km/h). We can convert the velocities to m/s and calculate the force required to accelerate the car. Then, we can multiply the force by the change in velocity to find the additional power needed.
In this case, the additional power needed is approximately 62 kW. Therefore, option (d) 62 kW is the correct answer.
Does the speedometer of a car measure speed or velocity? Explain.
The speedometer of a car measures speed, which is a scalar quantity indicating how fast the car is moving without regard to direction. Velocity, on the other hand, includes both speed and direction, which the speedometer does not display. The odometer measures total distance traveled, not displacement, and dividing distance by time gives average speed, not velocity.
Explanation:The speedometer of a car measures speed, not velocity. Speed is a scalar quantity, which means it only describes how fast an object is moving regardless of its direction. On the other hand, velocity is a vector quantity that describes both the speed and the direction of an object's movement. For example, if a car is moving at 60 miles per hour (mph), the speedometer shows this speed, but it does not indicate whether the car is traveling north, south, east, or west – that would be necessary information to determine the car's velocity.
A car's odometer, in contrast, measures the total distance traveled by the car. This distance is a scalar quantity as well, which means it does not account for the direction of travel, only the cumulative distance covered. When you divide the total distance traveled, as shown on the odometer, by the total time taken for the trip, you are calculating the average speed of the car, not the magnitude of average velocity. These two quantities – average speed and the magnitude of average velocity – are the same when the car moves in a straight line without changing its direction.
A person is placed in a large, hollow, metallic sphere that is insulated from ground. (a) If a large charge is placed on the sphere, will the person be harmed upon touching the inside of the sphere? Yes No Correct: Your answer is correct. (b) Explain what will happen if the person also has an initial charge whose sign is opposite that of the charge on the sphere.
Answer:
Explanation:
It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.
If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.
If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.
If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere
A person inside a charged, hollow, metallic sphere, known as a Faraday cage, would not be harmed upon touching the interior, regardless of their own charge, due to the neutralization of the electric field inside the conductor.
A person placed inside a large, hollow, metallic sphere that is insulated from the ground and charged will not be harmed upon touching the inside of the sphere. The phenomenon that explains this is known as the Faraday cage effect. According to this principle, an external static electric field will cause the charges within a conductor to rearrange themselves in such a way that the electric field inside the conductor cancels out.
Now, if the person inside the sphere also has a charge of the opposite sign to that of the sphere, an interesting interaction occurs. However, due to the conductor's nature, the electric field inside the metallic sphere remains null. The charges within the conductor would still redistribute to neutralize the electric field within the conductor. Therefore, if a person inside touched the inner surface of the sphere, they would not be directly harmed by the electric field, as it is neutralized within the conductor. The charges from the person would likely redistribute on the sphere's inner surface to maintain electrostatic equilibrium without causing harm.
Two charges q1 and q2 are separated by a distance d and exert a force F on each other. What is the new force F ′ , if charge 1 is increased to q′1 = 5q1, charge 2 is decreased to q′2 = q2/2 , and the distance is decreased to d′ = d 2 ?
Answer:
The new force is 10 times the force F
Explanation:
Electric force between charged particles q1 and q2 at distance d is:
[tex]F=k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex] (1)
A new force between two different particles at a different distance is:
[tex]F'=k\frac{\mid q_{1}'q_{2}'\mid}{d'^{2}}=k\frac{\mid 5q_{1}\frac{q_{2}}{2}\mid}{(\frac{d}{2})^{2}}=\frac{5}{\frac{2}{4}}k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex]
[tex]F'=10k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex]
Note that on the right side of the equation the term [tex]k\frac{\mid q_{1}q_{2}\mid}{d^{2}}=F [/tex] on (1), so:
[tex]F'=10F [/tex]
A wave on a string has a wavelength of 0.90 m at a frequency of 600 Hz. If a new wave at a frequency of 300 Hz is established in this same string under the same tension, what is the new wavelength? Group of answer choices
Answer:
λ₂ = 1.8 m
Explanation:
given,
wavelength of the string 1 = 0.90 m
frequency of the string 1 = 600 Hz
wavelength of string 2 = ?
frequency of the string 2 = 300 Hz
we now,
[tex]f\ \alpha\ \dfrac{1}{\lambda}[/tex]
now,
[tex]\dfrac{f_1}{f_2}=\dfrac{\lambda_2}{\lambda_1}[/tex]
[tex]\dfrac{600}{300}=\dfrac{\lambda_2}{0.9}[/tex]
λ₂ = 2 x 0.9
λ₂ = 1.8 m
Hence, the wavelength of the second string is equal to λ₂ = 1.8 m
Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after accelerating through potential difference 4V?
Final answer:
When a charge is accelerated through a potential difference, its speed can be calculated using the equation v = √(2qV/m). In the given example, an electron is accelerated through a potential difference of 4V, resulting in a speed of approximately 290 mV.
Explanation:
When a charge is accelerated through a potential difference, it gains kinetic energy. The relationship between the potential difference and the speed of the charge can be calculated using the equation:
v = √(2qV/m)
Where v is the final speed of the charge, q is the charge of the particle, V is the potential difference, and m is the mass of the particle.
In the given example, an electron with a charge of -1.60 × 10-19 C and a mass of 9.11 × 10-31 kg is accelerated to a speed of 10 × 104 m/s through a potential difference. To calculate the potential difference, we can rearrange the equation to:
V = m(v²)/(2q)
Substituting the values:
V = (9.11 × 10-31 kg)(10 × 104 m/s)²/ (2)(-1.60 × 10-19 C)
V = 2.91 × 10-2 V
So, the potential difference is approximately 29 mV.
A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint between the two particles.
The electric potential at the midpoint between a proton and an electron is zero.
Explanation:The electric potential at the midpoint between a proton and an electron can be calculated using the formula:
V = k * (q1 / r1 + q2 / r2)
where V is the electric potential, k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r1 and r2 are the distances between the particles and the midpoint. In this case, the charges of the proton and the electron are equal in magnitude but opposite in sign, so q1 = -q2. The distances from the particles to the midpoint are equal because they are fixed in space, so r1 = r2. Plugging in the values, we get:
V = k * (-q / r1 + q / r1) = 0
Therefore, the electric potential at the midpoint between the two particles is zero.
A snorkeler with a lung capacity of 4.3 L inhales a lungful of air at the surface, where the pressure is 1.0 atm. The snorkeler then descends to a depth of 49 m , where the pressure increases to 5.9 atm. What is the volume of the snorkeler's lungs at this depth? (Assume constant temperature.)
To find the volume of the snorkeler's lungs at a depth of 49m, use Boyle's Law by calculating the volume at the new pressure using the initial volume and pressure.
The volume of the snorkeler's lungs at a depth of 49 m can be calculated using Boyle's Law. Boyle's Law states that pressure and volume are inversely proportional when temperature is constant. To find the volume at depth, you can use P1V1 = P2V2, where P1V1 is the initial condition and P2V2 is the final condition.
Using the given data:
P1 = 1.0 atm, V1 = 4.3 L (lung capacity at the surface)P2 = 5.9 atm (pressure at 49 m depth), V2 = unknown (volume at 49 m depth)By rearranging the formula:
V2 = (P1 * V1) / P2 = (1.0 atm * 4.3 L) / 5.9 atm = 0.72 L
Therefore, at a depth of 49 m, the volume of the snorkeler's lungs would be approximately 0.72 liters.
Spring A has a spring constant of 100 N/m and spring B has a spring constant of 200 N/m, and both have the same displacement of 0.20 m. The spring potential energy of spring A is _____ that of spring B.
Answer:
Explanation:
Given
Spring constant of spring A is [tex]k_A=100\ N/m[/tex]
Spring constant of spring B is [tex]k_B=200\ N/m[/tex]
If displacement in both the springs is [tex]x=0.2\ m[/tex]
Potential Energy stored in the spring is given by
[tex]U=\frac{1}{2}kx^2[/tex]
where k=spring constant
x=compression or extension
[tex]U_A=\frac{1}{2}\times 100\times (0.2)^2----1[/tex]
[tex]U_B=\frac{1}{2}\times 200\times (0.2)^2----2[/tex]
Divide 1 and 2
[tex]\frac{U_B}{U_A}=\frac{200}{100}=2[/tex]
[tex]U_A=\frac{U_B}{2}[/tex]
So Potential Energy Stored in Spring A is half of Spring B
The spring potential energy of spring A is half that of spring B, if the spring constant of spring A is 100 N/m and that of B is 200 N/m.
Spring constant of spring A, k₁ = 100 N/m
Spring constant of spring B, k₂= 200 N/m
Displacement of both the spring, x = 0.20 m
The potential energy stored in a spring is given by the equation U = 0.5kx²
For spring A, the potential energy is U₁ = 0.5 × 100×(0.20)² = 2 J.
For spring B, the potential energy is U₂ = 0.5 × 200×(0.20)² = 4 J
Therefore, the potential energy stored in spring A is half that of spring B.
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You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/s. (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling?
Final answer:
The speed of the putty just before it strikes the ceiling is approximately 0.342 m/s, and it takes approximately 0.97 seconds to reach the ceiling.
Explanation:
To calculate the speed of the putty just before it strikes the ceiling, we can use the equation for vertical motion:
v = u + gt
Where:
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity (-9.8 m/s^2)
t is the time
In this case, the object is thrown upwards, so the final velocity when it reaches the ceiling will be zero. The initial velocity is 9.50 m/s and the acceleration due to gravity is -9.8 m/s^2. Substitute these values into the equation and solve for t:
0 = 9.50 - 9.8t
t = 0.97 seconds
Therefore, the speed of the putty just before it strikes the ceiling is 9.50 - 9.8(0.97) = -0.342 m/s. Since speed cannot be negative, the actual speed is 0.342 m/s.
To calculate the time it takes for the putty to reach the ceiling, we can use the same equation and solve for t:
0 = 9.50 - 9.8t
t = 0.97 seconds
So, it takes the putty approximately 0.97 seconds to reach the ceiling.
Final answer:
The speed of the putty just before it strikes the ceiling is 9.50 m/s, the same as its initial speed due to conservation of energy. The time it takes for the putty to reach the ceiling can be calculated using kinematic equations.
Explanation:
The problem presented concerns the motion of a glob of putty thrown upward and involves calculations based on the principles of kinematics in physics.
Answer to Part (a): What is the speed of the putty just before it strikes the ceiling?
The speed of the putty just before it strikes the ceiling can be found by using the kinematic equation for velocity under constant acceleration (gravity). Since the putty is thrown upwards, it will slow down under the influence of gravity until it reaches its maximum height. At this point, it will start to fall back down, accelerating under gravity until it hits the ceiling. Assuming no energy loss, the speed of the putty just before it strikes the ceiling will be the same as its initial speed when it leaves the hand, which is 9.50 m/s.
Answer to Part (b): How much time from when it leaves your hand does it take the putty to reach the ceiling?
The time it takes for the putty to reach the ceiling can be calculated using the kinematic equation s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is the acceleration due to gravity (approximately -9.81 m/s² when upwards is considered positive), and t is the time. Solving for t gives us the time taken for the putty to reach the height of 3.60 m.
Here we'll see that an emf can be induced by the motion of the conductor in a static magnetic field. The U-shaped conductor in (Figure 1) lies perpendicular to a uniform magnetic field B⃗ with magnitude B=0.75T , directed into the page. We lay a metal rod with length L=0.10m across the two arms of the conductor, forming a conducting loop, and move the rod to the right with constant speed v=2.5m/s . What is the magnitude of the resulting emf?
With the given magnetic field and rod length, what must the speed be if the induced emf has magnitude 0.73 V ?
Answer:
a) ΔV = 0.1875 V , b) v = 9.73 m / s
Explanation:
For this exercise we can use Necton's second law, as the speed is constant the forces on the driver are equal
[tex]F_{E} - F_{B}[/tex] = 0
[tex]F_{E}[/tex] = F_{B}
q E = q v B
E = v B
The electrical force induced in the conductor is
ΔV = E l
ΔV = v B l
Let's calculate
ΔV = 2.5 0.75 0.10
ΔV = 0.1875 V
b) If ΔV = 0.73 what speed should it have
v = DV / B l
v = 0.73 / 0.75 0.1
v = 9.73 m / s
The pressure drop needed to force water through a horizontal 1-in.-diameter pipe is 0.55 psi for every 8-ft length of pipe. (a) Determine the shear stress on the pipe wall. Determine the shear stress at distances (b) 0.3 and (c) 0.5 in. away from the pipe wall.
Answer:
(a). The shear stress on the pipe wall is 0.2062 lb/ft²
(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²
(c). The shear stress at the distance 0.5 in away from the pipe wall is zero.
Explanation:
Given that,
Diameter = 1 in
Pressure = 0.55 psi
Length = 8 ft
We need to calculate the radius of the pipe
Using formula of radius
[tex]r=\dfrac{D}{2}[/tex]
Put the value into the formula
[tex]r=\dfrac{1}{2}[/tex]
[tex]r=0.5\ in[/tex]
(a). We need to calculate the shear stress on the pipe wall
Using formula of shear stress
[tex]\dfrac{\Delta p}{L}=\dfrac{2\tau}{r}[/tex]
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0.5}{2\times8\times12}[/tex]
[tex]\tau=0.2062\ lb/ft^2[/tex]
(b). We need to calculate the shear stress at the distance 0.3 in
Using formula of shear stress
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0.3}{2\times8\times12}[/tex]
[tex]\tau=0.12375\ lb/ft^2[/tex]
(c). We need to calculate the shear stress at the distance 0.5 in away from the pipe wall
r = 0.5-0.5 = 0
Using formula of shear stress
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0}{2\times8\times12}[/tex]
[tex]\tau=0[/tex]
Hence, (a). The shear stress on the pipe wall is 0.2062 lb/ft²
(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²
(c). The shear stress at the distance 0.5 in away from the pipe wall is zero.
A thin, square, conducting plate 40.0 cm on a side lies in the xy plane. A total charge of 4.70 10-8 C is placed on the plate. You may assume the charge density is uniform. (a) Find the charge density on each face of the plate. 1.47e-07 What is the definition of surface charge density? C/m2 (b) Find the electric field just above the plate. magnitude N/C direction (c) Find the electric field just below the plate. magnitude N/C direction
Answer:
(a) Surface charge density is the charge per unit area.
[tex]\sigma = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]
(b) [tex]\vec{E} = (+\^z)~8.34\times 10^3~{\rm N/C}[/tex]
(c) [tex]\vec{E} = (-\^z)~8.34\times 10^3~{\rm N/C}[/tex]
Explanation:
(a) Surface charge density is the charge per unit area. The area of the square plate can be calculated by its side length.
[tex]A = l^2 = (0.4)^2 = 0.16 ~{\rm m^2}[/tex]
Half of the total charge is distributed on one side and the other is distributed on the other side.
Therefore, surface charge density on each face of the plate is
[tex]\sigma = Q/A = \frac{2.35 \times 10^{-8}}{0.16} = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]
(b) To find the electric field just above the plate, Gauss' Law can be used. Normally, Gauss' Law can only be used in infinite sheet (considering the flat surfaces), but just above the surface can be considered that the distance from the surface is much much smaller than the length of the plate (x << l).
In order to apply Gauss' Law, we have to draw an imaginary cylinder with radius r. The cylinder has to stay perpendicular to the plane.
[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r^2 = \frac{\pi r^2 \sigma}{\epsilon_0}\\E = \frac{\sigma}{2\epsilon_0}\\E = \frac{1.46 \times 10^{-7}}{2\times 8.8\times 10^{-12}} = 8.34\times 10^3~{\rm N/C}[/tex]
The direction of the electric field is in the upwards direction.
(c) The magnitude of the electric field is the same as that of upper side. Only the direction is reversed, downward direction.
A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 13°, and that the angle of depression to the bottom of the tower is 4°. How far is the person from the monument? (Round your answer to three decimal places.)
Answer:
665 ft
Explanation:
Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.
The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is
[tex]dtan13^0 = 0.231d[/tex]
Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees
[tex]dtan4^0 = 0.07d [/tex]
Since the 2 sides length above make up the 200 foot monument, their total length is
0.231d + 0.07d = 200
0.301 d = 200
d = 200 / 0.301 = 665 ft
To find the distance to the 200-foot tall monument from a specified point, we use tangent trigonometric functions with the given angles of elevation and depression. The total horizontal distance is calculated by separately finding the distances to the top and the bottom of the monument and combining them.
Explanation:The question involves solving a problem using trigonometry to find the distance to a monument given the angles of elevation and depression from a certain point. Since the height of the monument is given as 200 feet, and we have the angles of elevation (13°) to the top of the monument and the angle of depression (4°) to the bottom, we can use trigonometric functions to calculate the distances to the top and the bottom of the monument separately and then sum them to find the total distance from the observer to the monument.
To find the horizontal distances (D) from the observer to the top and bottom of the monument, we can use the tangent function (tan) from trigonometry. The tangent of the angle of elevation (13°) is equal to the height of the monument divided by the distance to the top (Dtop), and the tangent of the angle of depression (4°) is equal to the height from the window to the base of the monument divided by the distance to the bottom (Dbottom).
Assuming the window is at the same height as the base of the monument (which is not given explicitly, but implied by the angle of depression), we get two equations:
tan(13°) = 200 / Dtoptan(4°) = 0 / DbottomWe can then solve for Dtop and Dbottom using these equations and add the two distances to find the total horizontal distance to the monument.
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The total mass of all the planets is much less than the mass of the Sun. (T/F)
Answer:
True
Explanation:
The total mass of all the planets is much less than the mass of the Sun is a True statement.
Sun contributes to most of the mass of the solar system. Sun contributes to 99.8% mass of the solar system. Only 0.2% of the mass of the solar system is given by the planets. Hence, the above statement is absolutely correct.
Charge of uniform surface density (0.20 nC/m2 ) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z = 2.0 m
Answer:
E= 11.3 N/C.
Explanation:
The electric filed at any point on the z-axis is given by the formula
[tex]E= \frac{\sigma}{2\epsilon_0}[/tex]
here, sigma is the charge density and ε_o is the permitivity of free space.
therefore,
[tex]E= \frac{0.2\times10^{-9}}{2\times8.85\times10^{-12}}[/tex]
solving it we get
E= 11.3 N/C.
Hence, the required Electric field is E= 11.3 N/C.
Two electrons are separated by a distance of 1.00 nm and held fixed in place. A third electron, initially very far away, moves toward the other two electrons and stops at the point exactly midway between them. Calculate the speed of the third electron when it was very far away from the other electrons.
Answer:
The speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s
Explanation:
qV = 0.5Mv²
where;
V is the potential difference, measured in Volts
q is the charge of the electron in Coulomb's = 1.6 × 10⁻¹⁹ C
Mass is the mass of the electron in kg = 9 × 10⁻³¹ kg
v is the velocity of the electron in m/s
Applying coulomb's law, we determine the Potential difference V
V = kq/r
V = (8.99X10⁹ * 1.6 × 10⁻¹⁹)/(1X10⁻⁹)
V = 14.384 X 10¹⁹ V
The speed of the electron can be determined as follows;
v² = (2qV)/M
v = √(2qV)/M)
v = √(2*1.6 × 10⁻¹⁹* 14.384 X 10¹⁹)/(9 × 10⁻³¹)
v = √(5.1143 X 10³¹) = 7 X 10¹⁵m/s
Therefore, the speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s
What is the relationship between the applied force of a hanging mass on a spring and the spring force of the spring?
Answer:
elastic force and weight are related to the acceleration of the System.
Explanation:
The relationship between these two forces can be found with Newton's second law.
[tex]F_{e}[/tex] - W = m a
K x - m g = m a
We see that elastic force and weight are related to the acceleration of the System.
If a harmonic movement is desired, an extra force that increases the elastic force is applied, but to begin the movement this force is eliminated, in general , if the relationship between this external and elastic force is desired, the only requirement is that it be small for harmonic movement to occur
The relationship between the applied force of a hanging mass and the spring force of the spring is governed by Hooke's law, where the force is proportional to displacement. At equilibrium, the spring force balances the gravitational force, and any displacement results in harmonic oscillation.
Explanation:The relationship between the applied force of a hanging mass on a spring and the spring force is described by Hooke's law. When a mass is attached to a vertical spring, it stretches until the spring force equals the force of gravity on the mass. This point is called the equilibrium position. The spring force at this point is k times the displacement from the spring's unstretched length, where k is the spring constant. If the mass is displaced from this equilibrium position, a restoring force acts to return the mass to equilibrium, leading to harmonic oscillation.
In the context of springs, two scenarios arise. When the spring is relaxed, there is no force on the block, indicating an equilibrium situation. Upon stretching or compressing the spring, Hooke's law predicts the force on the block as being proportional to the displacement from its equilibrium position, x, with force F being -Kx. Here, -K represents the negative spring constant, signifying a restoring force opposite to the direction of displacement.
For a massive continuous spring with negligible gravitational effect compared to the spring force (KL » mg), when the top of the spring is driven up and down, the position of the bottom can be described as a function of time. This shows that both vertical and horizontal spring-mass systems follow similar dynamics and obey the laws governing harmonic motion.
Compared with ultraviolet radiation, infrared radiation has greater (a) wavelength; (b) amplitude; (c) frequency; (d) energy.
Answer:
(a) WaveLength
Explanation:
Compared with ultraviolet radiation, infrared radiation has greater wavelength. Option A.
UV vs IRCompared with ultraviolet radiation, infrared radiation has greater wavelength.
Wavelength refers to the distance between two consecutive peaks or troughs of a wave. In the electromagnetic spectrum, infrared radiation has longer wavelengths than ultraviolet radiation.
Ultraviolet radiation has shorter wavelengths and higher energy, while infrared radiation has longer wavelengths and lower energy.
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A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.
What is the electric potential at the center of the sphere?
Answer:
The electric potential at the center of the sphere is 36 V.
Explanation:
Given that,
Radius R= 0.600 m
Distance D = 1.20 m
Electric potential = 18.0 V
We need to calculate the electric potential
Using formula of electric potential
[tex]V=\dfrac{kq}{r}[/tex]
Put the value into the formula
[tex]18.0=\dfrac{9\times10^{9}\times q}{1.20}[/tex]
[tex]q=\dfrac{18.0\times1.20}{9\times10^{9}}[/tex]
[tex]q=2.4\times10^{-9}\ C[/tex]
We need to calculate the electric potential at the center of the sphere
Using formula of potential
[tex]V=\dfrac{kq}{r}[/tex]
Put the value into the formula
[tex]V=\dfrac{9\times10^{9}\times2.4\times10^{-9}}{0.600}[/tex]
[tex]V=36\ V[/tex]
Hence, The electric potential at the center of the sphere is 36 V.
The electric potential at the center of the sphere is 36 V.
The given parameters;
radius of the, R = 0.6 mdistance of the charge, r = 1.2 m potential difference, V = 18 VThe magnitude of the charge is calculated as follows;
[tex]V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\q = \frac{18 \times 1.2}{9\times 10^9} \\\\q = 2.4 \times 10^{-9} \ C[/tex]
The electric potential at the center of the sphere;
[tex]V = \frac{kq}{r} \\\\V = \frac{9\times 10^9 \times 2.4 \times 10^{-9}}{0.6} \\\\V = 36 \ V[/tex]
Thus, the electric potential at the center of the sphere is 36 V.
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Which of the following statements are true?
a. Electric field lines and equipotential surfaces are always mutually perpendicular.
b. When all charges are at rest, the surface of a conductor is always an equipotential surface.
c. An equipotential surface is a three-dimensional surface on which the electric potential is the same at every point.
d. The potential energy of a test charge increases as it moves along an equipotential surface.
e. The potential energy of a test charge decreases as it moves along an equipotential surface.
Answer:
a,b and c are true.
Explanation:
Following are true statements
a. Electric field lines and Equipotential surfaces are always mutually perpendicular is a true statement.
b. When all charges are at rest, the surface of a conductor is always an equipotential surface.
c. An equipotential surface is a three-dimensional surface on which the electric potential is the same at every point.
Following are False statements
d. The potential energy of a test charge increases as it moves along an equipotential surface.
e. The potential energy of a test charge decreases as it moves along an equipotential surface.
Reason: A t any point in an equipotential surface, the potential is same throughout. There is no increase or decrease in potential energy as the test charge moves in an equipotential environment.
Statements a, b, and c are correct: Electric field lines and equipotential surfaces are always mutually perpendicular; when all charges are at rest, the surface of a conductor is always an equipotential surface; and an equipotential surface is a three-dimensional surface on which the electric potential is the same at every point. Statements d and e are not correct because the potential energy of a test charge does not change as it moves along an equipotential surface.
Explanation:The following statements are true about electrical fields and potential:
a. Electric field lines and equipotential surfaces are always mutually perpendicular. This statement is correct. An electric field line shows the direction of the force a positive test charge would experience. An equipotential line or surface is one where the potential is the same at any point on the line or surface. As such, they will always be perpendicular to each other.b. When all charges are at rest, the surface of a conductor is always an equipotential surface. This statement is correct. In static conditions, the surface of a conductor is at a uniform potential because charges flow until they reach an equilibrium.c. An equipotential surface is a three-dimensional surface on which the electric potential is the same at every point. This statement is right. That's why we named it equipotential (equal potential).d. The potential energy of a test charge increases as it moves along an equipotential surface. This statement is not correct. Since it's an equipotential surface, the potential energy stays the same.e. The potential energy of a test charge decreases as it moves along an equipotential surface. This statement is also not correct for the same reason stated above.Learn more about Electric Field and Potential here:https://brainly.com/question/31256352
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Two electrons, each with mass mmm and charge qqq, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed vvv toward electron B in the positive x direction, and electron B has initial speed 3v3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.
What is the minimum separation r_min that the electrons reach?
Final answer:
The minimum separation r_min between two electrons moving towards each other and stopping due to repulsion is found by conserving mechanical energy, transforming the initial kinetic energy into electrostatic potential energy at the point of closest approach.
Explanation:
The minimum separation r_min that two electrons reach when moving directly toward each other and slowing down due to electric repulsion is determined by the conservation of energy. The total mechanical energy (kinetic plus potential) of the system must be conserved. At the maximum approach, both electrons will momentarily be at rest before repelling each other again, so all their kinetic energy is converted into electrical potential energy. The initial kinetic energy of both electrons, which is due to their motion, is transformed into electrostatic potential energy at the point of minimum separation.
The concept involves the calculation of this potential energy and setting it equal to the initial kinetic energy to find r_min. One might use the relationship between kinetic energy (1/2)m(v^2), electrostatic potential energy (k(q^2)/r), and the conservation of energy to solve for the unknown r_min. It's important to note that in such a scenario, relativistic effects are not considered as long as the velocity of the electrons is significantly less than the speed of light (c).
An empty glass soda bottle is to be use as a musical instrument. In order to be tuned properly, the fundamental frequency of the bottle be 440.0 Hz.
(a) If the bottle is 25.0 cm tall, how high should it be filled with water to produce the desired frequency?
(b) What is the frequency of the nest higher harmonic of this bottle?
Explanation:
For a pipe with one end open ,we have the formula for fundamental frequency as
[tex]f_1= \frac{v}{4L}[/tex]
v= velocity of sound in air =340 m/s
L= length of pipe
hence, [tex]L= \frac{v}{4f_1}[/tex]
given f_1 = 440 Hz
Substituting , L = 340/(4×440) = 0.193 m
a) Let the bottle be filled to height , h.
Given that height of bottle ,H = 25 cm = 0.25 m
Also we found out that length of pipe L = 0.193 m
So, we have h = H - L
= 0.25 - 0.193 = 0.057 m = 5.7 cm.
Answer:
(a). The bottle filled at the height is 5.5 cm
(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.
Explanation:
Given that,
Frequency = 440.0 Hz
Height of bottle = 25.0 cm
Suppose,
Let the bottle be filled to height h.
For a pipe with one end open,
We need to calculate the length of the pipe
Using formula of fundamental frequency
[tex]F=\dfrac{v}{4L}[/tex]
[tex]L=\dfrac{v}{4f}[/tex]
Where, L = length
v = speed of sound
Put the value into the formula
[tex]L=\dfrac{343}{4\times440}[/tex]
[tex]L=0.195\ m[/tex]
(a). We need to calculate the height
Using formula of height
[tex]h=H-l[/tex]
Where, H = height of bottle
l = length of pipe
Put the value into the formula
[tex]h=25.0\times10^{-2}-0.195[/tex]
[tex]h=0.055\ m[/tex]
[tex]h=5.5\ cm[/tex]
(b). We need to calculate the frequency of the nest higher harmonic of this bottle
Using formula of frequency
[tex]f_{next}=nf_{1}[/tex]
Where, [tex]f_{1}[/tex]=fundamental frequency
Put the value into the formula
[tex]f_{2}=2\times440[/tex]
[tex]f_{2}=880\ Hz[/tex]
Hence, (a). The bottle filled at the height is 5.5 cm
(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.
A bicycle racer sprints near the end of a race to clinch a victory. The racer has an initial velocity of 12.1 m/s and accelerates at the rate of 0.350 m/s2 for 6.07 s. What is his final velocity?
Answer:
v= 14.22 m/s
Explanation:
Given that
u = 12.1 m/s
a=0.35 m/s².
t= 6.07 s
We know v = u +at
v=final velocity
u=initial velocity
t=time
a=acceleration
Now by putting the values in the above equation
v= 12.1 + 0.35 x 6.07 m/s
v= 14.22 m/s
Therefore the final velocity will be 14.22 m/s.
The final velocity of a bicycle racer with an initial velocity of 12.1 m/s and an acceleration of 0.350 m/s² over 6.07 s is approximately 14.22 m/s.
Explanation:To compute the final velocity of a bicycle racer sprinting towards the end of a race, we can employ the kinematic equation for velocity under constant acceleration:
v = u + at
Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given that the racer has an initial velocity u of 12.1 m/s, accelerates at a rate a of 0.350 m/s² for a duration t of 6.07 s, we substitute these values into the equation:
v = 12.1 m/s + (0.350 m/s² × 6.07 s)
Calculation:
v = 12.1 m/s + (0.350 m/s² × 6.07 s)
v = 12.1 m/s + 2.1245 m/s
v = 14.2245 m/s
The final velocity of the racer is approximately 14.22 m/s.
If a guitar string has a fundamental frequency of 500 Hz, what is the frequency of its second overtone?
At the time t = 0, the boy throws a coconut upward (assume the coconut is directly below the monkey) at a speed v0. At the same instant, the monkey releases his grip, falling downward to catch the coconut. Assume the initial speed of the monkey is 0, and the cliff is high enough so that the monkey is able to catch the coconut before hitting the ground. The time taken for the monkey to reach the coconut is
Answer:
The time taken for the monkey to reach the coconut, t is = H/v₀
Explanation:
Let the coordinate of the boy's hand be y = 0. The height of the tree above the boy's hand be H.
Coordinate of where the monkey meets coconut = y
Using the equations of motion,
For the monkey, initial velocity = 0m/s, time to reach coconut = t secs and the height at which coconut is reached = H-y
For the coconut, g = -10 m/s², initial velocity = v₀, time to reach monkey = t secs and height at which coconut meets monkey = y
For monkey, H - y = ut + 0.5gt², but u = 0,
H - y = 0.5gt²..... eqn 1
For coconut, y = v₀t - 0.5gt² ....... eqn 2
Substituting for y in eqn 1
H - y = H - (v₀t - 0.5gt²) = 0.5gt²
At the point where the monkey meets coconut, t=t
H - v₀t + 0.5gt² = 0.5gt²
v₀t = H
t = H/v₀
Solved!
A ball is thrown horizontally off a cliff. If the initial speed of the ball is (15.0 + A) m/s and the cliff is (25.0 + B) m high, how far from the base of the cliff will the ball land in the water below? Calculate the answer in meters (m) and round to three significant figures.A=4B=54
Answer:
76.3 m
Explanation:
We are given that
Initial speed of the ball,u=(15+A)m/s
Height of cliff,h=(25.9+B) m
We have to find the distance from the base of the cliff the ball will land in the water below.
A=4 and B=54
Distance=[tex]u\sqrt{\frac{2h}{g}}[/tex]
Using the formula and substitute the values
[tex]D=(15+4)\sqrt{\frac{2(25+54)}{9.8}}[/tex]
Because [tex]g=9.8m/s^2[/tex]
[tex]D=19\sqrt{\frac{158}{9.8}}[/tex]
D=76.3
Hence, the distance from the base of the cliff the ball will land in the water below=76.3 m
Calculate the acceleration of a 1400-kg car that stops from 39 km/h "on a dime" (on a distance of 1.7 cm).
Answer:
[tex]a=-3449.67\frac{m}{s^2}[/tex]
Explanation:
The car is under an uniforly accelerated motion. So, we use the kinematic equations. We calculate the acceleration from the following equation:
[tex]v_f^2=v_0^2+2ax[/tex]
We convert the initial speed to [tex]\frac{m}{s}[/tex]
[tex]39\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=10.83\frac{m}{s}[/tex]
The car stops, so its final speed is zero. Solving for a:
[tex]a=\frac{v_0^2}{2x}\\a=-\frac{(10.83\frac{m}{s})^2}{2(1.7*10^{-2}m)}\\a=-3449.67\frac{m}{s^2}[/tex]
To calculate the acceleration of the car, use the formula a = (vf - vi) / t. Convert the initial velocity and distance to the appropriate units before substituting them into the formula.
To calculate the acceleration of the car, we can use the formula:
a = (vf - vi) / t
Where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken. In this case, the initial velocity is 39 km/h, which is converted to m/s by vi = 39 km/h * (1000 m/1 km) * (1 h/3600 s). The final velocity is 0 m/s since the car stops. The time taken can be found by t = d / vi, where d is the distance and vi is the initial velocity. The distance is given as 1.7 cm, which is converted to m by d = 1.7 cm * (1 m/100 cm). Substituting these values into the formula, we get the acceleration of the car.
So the acceleration of the car is approximately -8.45 m/s^2.
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