You spot a plane that is 1.37 km north, 2.71 km east, and at an altitude 4.65 km above your position. (a) How far from you is the plane? (b) At what angle from due north (in the horizontal plane) are you looking? °E of N (c) Determine the plane's position vector (from your location) in terms of the unit vectors, letting î be toward the east direction, ĵ be toward the north direction, and k be in vertically upward. ( km)î + ( km)ĵ + ( km) k (d) At what elevation angle (above the horizontal plane of Earth) is the airplane?

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=[tex]T_b=101^{\circ}[/tex] C

Boiling point water=100 degree Celsius

[tex]K_f=1.86K/m[/tex]

[tex]K_b=0.512 K/m[/tex]

[tex]\Delta T_b=T-T_0[/tex]

Where [tex]T[/tex]=Boiling point of solution

[tex]T_0=[/tex]Boiling point of pure solvent

[tex]\Delta T_b=101-100=1^{\circ}[/tex]C

[tex]\Delta T_b=k_bm[/tex]

Using the formula

[tex]1=0.512\times m[/tex]

Molality,[tex]m=\frac{1}{0.512}[/tex] m

[tex]\Delta T_f=k_fm[/tex]

Using the formula

[tex]\Delta T_f=\frac{1}{0.512}\times 1.86[/tex]

[tex]\Delta T_f=3.63 C[/tex]

We know that

[tex]\Delta T_f=T_0-T_1[/tex]

Where [tex]T_0[/tex] =Freezing point of solvent

[tex]T_1=[/tex] Freezing point of solution

Using the formula

[tex]3.63=0-T_1[/tex]

Freezing point of water=0 degree Celsius

[tex]T_1=0-3.63=-3.63 C[/tex]

Hence, the freezing point of solution=-3.63 degree Celsius

Final answer:

The freezing point of an aqueous solution can be calculated using the freezing point depression equation ΔTf = Kf * m. In this case, the solution boils at 101°C, indicating the boiling point elevation constant (Kb) is provided. Assuming complete dissociation, we can calculate the freezing point depression to be -0.512°C.

Explanation:

The freezing point depression can be calculated by using the equation:

ΔTf = Kf * m

Where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution.

In this case, since we have the boiling point elevation constant (Kb), we need to use the equation:

ΔTf = Kb * m

Given that Kb is 0.512 K/m and assuming complete dissociation, a 1.0 m aqueous solution of a solute will contain 1.0 mol of particles per kilogram of water. Therefore, the freezing point depression (ΔTf) will be:

ΔTf = 0.512 K/m * 1.0 m = 0.512 K

Since the freezing point of pure water is 0°C, the freezing point of the aqueous solution will be:

0°C - 0.512 K = -0.512°C

Final answer:

Measuring 20 cycles of the pendulum motion to determine the period instead of just one cycle allows for a more accurate and representative measurement. By averaging out any errors or variations in the measurement process, the value for the period of the pendulum is more precise. Measuring multiple cycles also helps identify any potential changes in the period over time.

Explanation:

The reason we measure 20 cycles of the pendulum motion to determine the period, rather than just one cycle, is to obtain a more accurate and representative measurement of the period. By measuring multiple cycles, we can average out any errors or variations in the measurement process, resulting in a more precise value for the period of the pendulum.

For example, if we were to measure only one cycle of the pendulum, any small errors in the timing or counting of the cycles could significantly affect our measurement. However, by measuring 20 cycles and then dividing the total time by 20, we can minimize these errors and obtain a more reliable measurement of the period.

Furthermore, measuring multiple cycles allows us to observe any potential changes in the period over time. In certain situations, the period of a pendulum may vary slightly due to factors such as air resistance or changes in the length of the string. By measuring multiple cycles, we can identify and account for any such variations, providing a more comprehensive understanding of the pendulum's behavior.

Answer:

The mass of coal is 108 kg.

Explanation:

Given that,

Energy of coal = 1200 kg

Mass of aluminum = 1.0 kg

Energy required for single soda can = 15 g of Al

Energy required for 6 pack of cans = [tex]6\times15=90\ g\ of\ Al[/tex]

We need to calculate the mass of coal

Using formula of mass

[tex]\text{mass of coal}=\dfrac{\text{Energy of coal}\times\text{Energy required for 6 pack of cans}}{\text{Mass of aluminum}}[/tex]

Put the value into the formula

[tex]m=\dfrac{1200\times90\times10^{-3}}{1.0}[/tex]

Put the value into the formula

[tex]m=108\ Kg[/tex]

Hence, The mass of coal is 108 kg.

Answer:

42 minutes

Explanation:

First, let us find out the time required by the roommate, who is driving the truck, to reach the city.

We know that,

[tex]time=\frac{distance}{speed} =\frac{360}{50} hrs=7.2hrs[/tex]

Now, you are planning to rest a bit after traveling for an hour. So,

distance covered in that 1 hour = [tex]60mi/h\times1h=60 miles[/tex]

time required by you to cover the total distance = [tex]\frac{360mi}{60mi/h} =6hours[/tex]

If you wish reach at the destination half an hour (0.5 h) before your roommate, you can expend a total of [tex]7.2-0.5=6.7hours[/tex] throughout your journey.

Hence, you can rest for  [tex]6.7-6=0.7hours=(0.7\times60)minutes=42minutes[/tex]

After driving 60 miles in the first hour and intending to arrive half an hour before your roommate who travels at 50 mph, you can afford a break time of 30 minutes at the rest stop.

The subject of your question is

relative speed

and

time management

, which falls under Mathematics. Your overall travelling speed is 60 mph. After driving for an hour, you decide to take a break at a rest stop. In that one hour, you've driven 60 miles, so there are 300 miles left to the destination.  Your roommate, who doesn't stop for a break, continues to drive at 50 mph. You want to arrive half an hour before she does, which means you essentially have her travel time less 30 minutes to reach the destination. Because her speed in relation to the remaining distance is constant, her remaining travel time is 300 miles divided by 50 mph which equals to 6 hours. So, you actually have 6 - 0.5 = 5.5 hours to travel the remaining 300 miles including your rest stop. Considering your speed of 60 mph, it will take you 300/60 = 5 hours to reach the destination. Therefore, the length of the break you can take while still beating your roommate to the destination is 5.5 hours minus 5 hours, which is

30 minutes

.

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Answer with Explanation:

We are given that

r=3 m

Angular frequency=[tex]\omega[/tex]=560rev/min

A.1 revolution=[tex]2\pi[/tex] radian

560 revolutions=[tex]560\times 2\pi[/tex] rad

Angular frequency=[tex]2\times 3.14\times \frac{560}{60}[/tex]rad/s

1 min=60 s

[tex]\pi=3.14[/tex]

Angular frequency=[tex]\omega=58.6rad/s[/tex]

Linear speed=[tex]\omega r[/tex]

Using the formula

Linear speed=[tex]58.6\times 3=175.8m/s[/tex]

Hence, the linear speed of the blade tip=175.8m/s

B.Radial acceleration=[tex]a_{rad}=\frac{v^2}{r}[/tex]

By using the formula

Radial acceleration=[tex]a_{rad}=\frac{(175.8)^2}{3}= 10.301\times 10^3m/s^2[/tex]

Radial acceleration=[tex]\frac{10.301}{9.8}g\times 10^3=1.05\times 10^3 g[/tex]

Where [tex]g=9.8m/s^2[/tex]

Hence, the radial acceleration[tex]=1.05\times 10^3 g[/tex]

The efficiency of the machine is defined as

[tex]\eta = \frac{W_{out}}{W_{in}}[/tex]

Here

Work out is the work output and Work in is the work input

To find the Work in we have then

[tex]W_{in} = \frac{W_{out}}{\eta}[/tex]

[tex]W_{in} = \frac{mgh}{\eta}[/tex]

Replacing with our values

[tex]W_{in} = \frac{(58)(9.8)(3)}{73\%}[/tex]

[tex]W_{in} = 2335.89J[/tex]

The work done by the applied force is

W = Fd

Here,

F = Force

d = Distnace

Rearranging to find F,

[tex]F = \frac{W}{d}[/tex]

[tex]F = \frac{2335.89J }{18}[/tex]

F = 129.77N

Therefore the force exerted on the machine after rounding off to two significant figures is 130N

The force exerted on the pulley system when a rope is pulled 18.0 m in order to raise a 58 kg mass a height of 3.0 m with an efficiency of 73 percent is about 129.42 Newtons.

To solve this problem, we need to understand the concept of machine efficiency and work. The efficiency of a machine is the ratio of output work to input work.

In this case, the efficiency of the pulley system is given as 73%. Meaning output work is 73% of the input work. The force exerted on the machine is equivalent to the input work divided by the distance pulled.

The output work (W_out) can be determined using the formula W_out = mass * gravity * height = 58 kg * 9.8 m/s^2 * 3.0 m = 1701.6 Joules.

Then, we can find the input work (W_in) using the efficiency formula: W_in = W_out / efficiency = 1701.6 J / 0.73 = 2329.59 Joules.

Finally, we can find the force exerted on the pulley system (F_in) by dividing the input work by the distance pulled: F_in = W_in / distance = 2329.59 J / 18.0 m = 129.42 Newtons.

So, the force exerted on the machine would be approximately 129.42 Newtons.

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Answer:

The maximum volume flow rate is 0.03745m^3/s

Explanation:

Power input (Pi) = 9-hp = 9×746W = 6,714W

Elevation (h) = 15m

Efficiency (E) = 82% = 0.82

Density (D) of water = 1000kg/m^3

E = Po/Pi

Po (power output) = E×Pi = 0.82×6,714W = 5505.48W

Po = mgh/t

m/t (mass flow rate) = Po/gh = 5505.48/9.8×15 = 37.45kg/s

Volume flow rate = mass flow rate ÷ density = 37.45kg/s ÷ 1000kg/m^3 = 0.03745m^3/s

Answer:

Explanation:

Given

Astronaut in space gives an equal push to baseball and bowling ball.

Since the mass of the bowling ball is more than the baseball so inertia associated with a bowling ball is more as compared to baseball

Force applied on baseball and bowling ball is the same so the acceleration of baseball will be more because the mass of baseball is less.

[tex]Force=mass\times acceleration[/tex]

If both are traveling with the same speed then momentum associated with them is given by-product of mass and velocity

Since the mass of the bowling ball is more, therefore, the momentum of bowling ball is more as compared to baseball

Answer:

a)The keys were thrown with an initial velocity of 10.0 m/s.

b)The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).

Explanation:

Hi there!

The equations for the height and velocity of the keys are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h =height of the keys after a time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive)

v = velocity of the keys at time t.

a) We know that at t = 1.50 s, h = 4.00 m and let´s consider that the origin of the frame of reference is located at the point where the keys are thrown so that h0 = 0. Then, using the equation of height, we can obtain the initial velocity.

h = h0 + v0 · t + 1/2 · g · t²

4.00 m = v0 · 1.50 s - 1/2 · 9.81 m/s² · (1.50 s)²

4.00 m + 1/2 · 9.81 m/s² · (1.50 s)² = v0 · 1.50 s

v0 = ( 4.00 m + 1/2 · 9.81 m/s² · (1.50 s)²) / 1.50 s

v0 = 10.0 m/s

The keys were thrown with an initial velocity of 10.0 m/s.

b) Now, using the equation of velocity we can calculate the velocity at t = 1.50 s:

v = v0 + g · t

v = 10.0 m/s - 9.81 m/s² · 1.50 s

v = -4.72 m/s

The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).

A star that is cooler than the sun would appear red (a), because the color of a star is associated with its temperature, according to Wien's Law.

The color of a star is closely related to its temperature. In the case of a star that's cooler than the sun, it would appear (a) red. This is due to a principle in astrophysics known as Wien's Law, which states that the peak wavelength (color) of the light emitted by an object (such as a star) shifts to longer, redder wavelengths as the object gets cooler. So, cool stars like red giants appear red, while hotter stars appear blue or white.

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Answer:

a)[tex]q=14.5\ W/m^2[/tex]

b)Q= 58 W  

Explanation:

Given that

Thermal conductivity ,K = 0.029 W/m.k

The temperature difference ,ΔT= 10°C

The thickness ,L = 20 mm

We know that

[tex]Q=\dfrac{KA}{L}\times \Delta T[/tex]

Now by putting the values

[tex]Q=\dfrac{0.029\times 4}{0.02}\times 10\ W[/tex]

Q= 58 W

The heat flux through the sheet is given as

[tex]q=\dfrac{Q}{A}\ W/m^2[/tex]

[tex]q=\dfrac{58}{2\times 2}\ W/m^2[/tex]

[tex]q=14.5\ W/m^2[/tex]

a)[tex]q=14.5\ W/m^2[/tex]

b)Q= 58 W

Explanation:

Expression for energy balance is as follows.

        [tex]\Delta E_{system} = E_{in} - E_{out}[/tex]

or,          [tex]E_{in} = E_{out}[/tex]

Therefore,  

         [tex]m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})[/tex]

          [tex]h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}[/tex]

Hence, expression for exit velocity will be as follows.

           [tex]V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}[/tex]

                      = [tex]V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]

As [tex]C_{p}[/tex] for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.

       [tex]V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]                    

                  = [tex][(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}[/tex]

                 = 40.7 m/s

Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.

Answer:

time takes for the bullet to strike the ground is 0.5711 second

Explanation:

given data

height h = 1.60 m

initial speed u = 1058 m/s

solution

we get here time that is express here as

time = [tex]\sqrt{\frac{2h}{g}}[/tex]     ......................1

put here value and we will get here time that is

time = [tex]\sqrt{\frac{2*1.60}{9.81}}[/tex]

time = 0.5711 second

so time takes for the bullet to strike the ground is 0.5711 second

Final answer:

The time it takes for a bullet to strike the ground when fired horizontally from 1.60 m is calculated using the equation t = √(2h/g). Substitute 1.60 m for h and 9.81 m/s² for g to obtain t ≈ 0.571 seconds.

Explanation:

Calculating the Time for a Bullet to Strike the Ground

To calculate the time it takes for a bullet to strike the ground when fired from a horizontal position 1.60 meters above the ground with an initial speed of 1058 m/s, we can use the equations of motion for uniformly accelerated motion (in this case, the acceleration due to gravity). Since the bullet is fired horizontally, there is no initial vertical velocity component, so the vertical motion can be treated as a free-fall problem.

Using the formula for the time of free fall (t) from a height (h):
t = √(2h/g), where g is the acceleration due to gravity (approximately 9.81 m/s²).

Plugging in the given height of 1.60 m and solving for t gives us:
t = √(2 * 1.60 m / 9.81 m/s²) = √(0.3261 s²) ≈ 0.571 seconds.

Therefore, the bullet will take approximately 0.571 seconds to hit the ground.

Answer:

[tex]3.8469943828\times 10^{26}\ W[/tex]

Explanation:

[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ W/m^2K^4[/tex]

T = Surface temperature of the Sun = 5778 K

r = Radius of the Sun = 696000 km

From Stefan-Boltzmann law

[tex]F=\sigma T^4\\\Rightarrow F=5.67\times 10^{-8}\times 5778^4\\\Rightarrow F=63196526.546\ W/m^2K[/tex]

Power is given by

[tex]P=F4\pi r^2\\\Rightarrow P=63196526.546\times 4\pi\times (696000000)^2\\\Rightarrow P=3.8469943828\times 10^{26}\ W[/tex]

The power output of the Sun is [tex]3.8469943828\times 10^{26}\ W[/tex]

Explanation:

The given data is as follows.

     height (h) = 98.0 m,     speed (v) = 73.0 m/s,

Formula of height in vertical direction is as follows.

          h = [tex]\frac{gt^{2}}{2}[/tex],

or,      t = [tex]\sqrt{\frac{2h}{g}}[/tex]

Now, formula for the required distance (d) is as follows.

       d = vt

          = [tex]v \sqrt{\frac{2h}{g}}[/tex]  

        = [tex]73.0 m/s \sqrt{\frac{2 \times 98.0 m}{9.8 m/s^{2}}}[/tex]  

          = 326.5 m

Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

[tex]K_{rel}=(\gamma-1)mc^2[/tex]

[tex]\dfrac{K_{rel}}{mc^2}=\gamma-1[/tex]

Put the value into the formula

[tex]\gamma=\dfrac{105.7}{0.511}+1[/tex]

[tex]\gamma=208[/tex]

We need to calculate the electron’s velocity

Using formula of velocity

[tex]\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]

[tex]\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}[/tex]

[tex]\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1[/tex]

[tex]v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2[/tex]

Put the value into the formula

[tex]v^2=\dfrac{1-(208)^2}{-208^2}\times c^2[/tex]

[tex]v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}[/tex]

[tex]v=0.9999 c\ m/s[/tex]

Hence, The electron’s velocity is 0.9999 c m/s.

To solve this problem we will apply the principles of energy conservation. The kinetic energy in the object must be maintained and transformed into the potential electrostatic energy. Therefore mathematically

[tex]KE = PE[/tex]

[tex]\frac{1}{2} mv^2 = \frac{kq_1q_2}{r}[/tex]

Here,

m = mass (At this case of the proton)

v = Velocity

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge of each object

r= Distance between them

Rearranging to find the second charge we have that

[tex]q_2 = \frac{\frac{1}{2} mv^2 r}{kq_1}[/tex]

Replacing,

[tex]q_2 = \frac{\frac{1}{2}(1.67*10^{-27})(3*10^5)^2(7*10^{-2})}{(9*10^9)(1.6*10^{-19})}[/tex]

[tex]q_2 = 3.6531nC[/tex]

Therefore the charge on the sphere is 3.6531nC

Explanation:

The wavelength of green light is about 500 nanometers, or two thousandths of a millimeter. The typical wavelength of a microwave oven is about 12 centimeters, which is larger than a baseball.

The statement is falls, because the the wavelength of green light is about 500 nm or 500 × 10⁻⁹ m. But the size of an atom is  about 1.2  × 10⁻¹⁰ m. Hence atomic size not equals the wavelength of green light.

The wavelength of an electromagnetic wave is the distance between two consecutive crests or troughs.  The upward peak in the wave is called crests and the downward peaks are called troughs.

The wavelength of high energy waves will be shorter. Visible region is in between IR and UV rays in the electromagnetic spectrum. Green light is in the exact middle region of the VIBGYOR. Thus, it is having a wavelength of 500 -520 nm or  500 × 10⁻⁹ m.

The size of an atom is estimated in the range of 1.2  × 10⁻¹⁰ m and it varies from element to element. However the atomic size is not comparable with the wavelength of green light.

To find more on atomic size, refer here:

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To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

[tex]\Delta P = mgh[/tex]

[tex]\Delta P = (8.7)(9.8)(3)[/tex]

[tex]\Delta P = 255.78J[/tex]

Therefore the change in the internal energy of the system is 255.78

The correct equation to use to find the distance 's' when the iron block and magnet come to a rest is ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation represents the conversion of potential energy into work done against friction.

((mB)^2 + mM)gH = μ(mB + mM)gs.

This question represents a problem of mechanics in Physics, specifically involving both potential energy and work-energy theorem. The relevant equation to use in finding the distance 's' in this scenario is: ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation derives from setting the initial potential energy of the system equal to the final kinetic energy when it rests, including the energy dissipated due to friction.

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The correct equation to find 's' when an iron block slides down a hill and collides with a magnet before encountering a rough surface, is ((mB)^2 + mB + mM)gH = μ(mB + mM)gs. This equation is derived from equating the kinetic energy at the base of the hill with the work done by friction.

The correct equation to use to define s in this scenario would be: ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation stems from equating the energy at the top of the hill (kinetic + potential energy) with the work done against friction (which eventually stops the two masses). Let's understand it in a detailed step-by-step manner.

Step 1: At the base of the hill, the kinetic energy of the block is equal to the potential energy at the top of the hill, which can be represented as (mB + mM)gH = 0.5(mB + mM)v^2. From this, we can turn the speeds into velocities, which in turn gives us v = sqrt(2gH).

Step 2: When the two masses encounter the rough surface, they will lose their kinetic energy due to friction. The total kinetic energy lost, which is equal to work done by friction, can be represented as 0.5(mB + mM)v^2 = μ(mB + mM)gs. Substituting the value of v from step 1 into this equation gives us the final formula: ((mB)^2 + mB + mM)gH = μ(mB + mM)gs.

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The distance from the starting point is approximately 26.3 m and the compass direction is 25.3° west of north.

To find the distance between the starting point and the final position, we can use the Pythagorean theorem. We can calculate the distances in the north and west directions using the given lengths and angles. Using the given values, the distance from the starting point is approximately 26.3 m. The compass direction of the line connecting the starting point to the final position is 25.3° west of north.

Final answer:

To find the distance between the starting point and final position, use the Pythagorean theorem. The total distance is approximately 31.3 m. To determine the compass direction, use trigonometry. The compass direction is approximately 36.9° north of west.

Explanation:

To find the distance between your starting point and final position, you can use the Pythagorean theorem. First, calculate the horizontal distance by adding the horizontal components of the two legs of the walk: A = 18.0 m west and B = 0 m south. The horizontal distance is 18.0 m. Then, calculate the vertical distance by summing up the vertical components of the two legs of the walk: A = 0 m north and B = 25.0 m north. The vertical distance is 25.0 m. Apply the Pythagorean theorem to find the total distance by taking the square root of the sum of the squares of the horizontal and vertical distances. The total distance is approximately 31.3 m.

To determine the compass direction of a line connecting your starting point to your final position, you can use trigonometry. First, calculate the angle θ between the horizontal distance (18.0 m) and the hypotenuse (31.3 m). You can use the inverse tangent function: θ = arctan(vertical distance/horizontal distance). Calculate θ to be approximately 53.1°. Since you walked west and then north, the compass direction is 90° less than θ. Therefore, the compass direction is approximately 36.9° north of west.

Answer:

a)    vₓ = 6,457 m / s ,  v_{y}  = 0.518 m / s , b)    v = 6.478 m / s,      θ = 4.9°

Explanation:

a) This is a kinematic problem, let's use trigonometry to find the components of acceleration

           sin 31 = [tex]a_{y}[/tex] / a

           cos 31 = aₓ = a

           a_{y} = a sin31

           aₓ = a cos 31

Now let's use the kinematic equation for each axis

X axis

         vₓ = v₀ₓ + aₓ (t-t₀)

         vₓ = v₀ₓ + a cos 31 (t-t₀)

         vₓ = 2.6 + 0.45 cos 31  (20-10)  

         vₓ = 6,457 m / s

Y Axis

       v_{y} = v_{oy} + a_{y} t

       v_{y} = v_{oy} + a_{y}  sin31 (t-to)

       v_{y} = -1.8 + 0.45 sin31   (20-10)

       v_{y}  = 0.518 m / s

b) let's use Pythagoras' theorem to find the magnitude of velocity

       v = √ (vₓ² + v_{y}²)

       v = √ (6,457² + 0.518²)

       v = √ (41.96)

       v = 6.478 m / s

We use trigonometry for direction

       tan θ = v_{y} / vₓ

       θ = tan⁻¹  v_{y} / vₓ

       θ = tan⁻¹ 0.518 / 6.457

       θ = 4.9°

c) let's look for the vector at the initial time

        v₁ = √ (2.6² + 1.8²)

        v₁ = 3.16 m / s

        θ₁ = tan⁻¹ (-1.8 / 2.6)

        θ₁ = -34.7

We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.

          a = (v₂ -v₁) / (t₂-t₁)

Answer:

934701926.438 Hz

Explanation:

Mass of molecule

[tex]m=3\times 16\times 1.67\times 10^{-27}\ kg[/tex]

Moment of inertia is given by

[tex]I=\dfrac{1}{12}ml^2\\\Rightarrow I=\dfrac{1}{12}\times 3\times 16\times 1.67\times 10^{-27}\times (2.5\times 10^{-10})^2\\\Rightarrow I=4.175\times 10^{-46}\ kgm^2[/tex]

E = Electric field = 8000 N/C

p = Dipole moment = [tex]1.8\times 10^{-30}\ Cm[/tex]

l = Length of rod = [tex]2.5\times 10^{-10}\ m[/tex]

Frequency of oscillations is given by

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{pE}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{1.8\times 10^{-30}\times 8000}{4.175\times 10^{-46}}}\\\Rightarrow f=934701926.438\ Hz[/tex]

The frequency of oscillations is 934701926.438 Hz

We will make a graph to better understand the displacement of the individual. From the trigonometric properties we will find the required distances.

[tex]d_1 = \frac{450}{\sqrt{2}} = 318.198[/tex]

[tex]d_2 = \frac{400}{\sqrt{2}} = 282.843[/tex]

[tex]D = 500+d_1+d_2 = 1101.041[/tex]

Displacement ,

[tex]x = \sqrt{1101.041^2+(318.198-282.843)^2}}[/tex]

[tex]x = 1101.608m[/tex]

The angle would be

[tex]\theta = cos^{-1} (\frac{1101.041}{1101.608})[/tex]

[tex]\theta = E 1.838\° N[/tex]

Therefore the displacement was of 1101.608m to a angle of 1.838° from East to North.

The sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.

This is a problem of physics specifically in the field of mechanics, about displacement and vectors. To find the total displacement, the directions in which the sailor traveled need to be considered. A north direction has an angle of 90 degrees, northeast is 45 degrees, and northwest is 135 degrees.

Using the Pythagorean theorem and some trigonometry, we can calculate the displacement in the x and y directions. The total displacement in the x direction (East-West) is 550*cos(45) - 500*cos(45) = 35.35 meters (approximately). The total displacement in the y direction (North-South) is 600 + 550*sin(45) + 500*sin(45) = 1116.57 meters (approximately).

The total displacement would then be the square root of (35.35 ^2 + 1116.57 ^2) = 1118.16 meters (approximately).

To find the angle of the displacement, we can use the arctangent function (atan function in most calculators). The angle is atan (1116.57 / 35.35) = 88.13 degrees (approximately).

Therefore, the sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.

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To find out how far the firefighting crew should position their cannon from the building, we use the equations of projectile motion separately for the vertical and the horizontal components. Since there are two viable times that satisfy the conditions of motion, there are two potential positions of the cannon. This question combines the principles of physics, which are specifically related to projectile motion.

This question is about the physics of projectile motion, specifically under gravity. To find out how far from the building they should position their cannon, we have to deal with the vertical and horizontal components of the projectile motion separately.

First, we determine the time it takes for the water to reach the desired height. We get this by using the equation of motion: H = V*sin(theta)*t - 0.5*g*t^2, where H is the height above the ground level (10m), V is speed (25.0 m/s), sin(theta) is the vertical component of initial velocity, g is acceleration due to gravity (approximated as 10 m/s^2 for ease of calculation), and t is time. We can rearrange this equation to find t.

Next, we use the time obtained and the horizontal component of initial velocity (V*cos(theta)) to find the horizontal distance made by using the equation: s = V*cos(theta)*t.

The hint here is that there are two potential time values, a smaller one and a larger one, leading to two potential distances from the building that the cannon could be positioned. Both times satisfy the conditions of the motion.

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The firefighting crew should place their cannon either 8.0 meters or 56.1 meters from the building to reach a blaze 10.0 meters above ground.

To determine how far the firefighting crew should position their cannon, we need to analyze the projectile motion of the water. We'll break this down into x-direction (horizontal) and y-direction (vertical) components.

Step-by-Step Solution

1. Break down initial velocity into components

Initial speed, v₀ = 25.0 m/s

Angle, θ = 53.0°

v₀x = v₀ x cos(θ) = 25.0 x cos(53.0°) ≈ 15.0 m/s

v₀y = v₀ x sin(θ) = 25.0 x sin(53.0°) ≈ 20.0 m/s

2. Write the equations of motion

Vertical (y-direction): y = v₀y x t - 0.5 x g x t²

Given y = 10.0 m, v₀y = 20.0 m/s, g = 9.8 m/s²

Using the quadratic formula, we find the time (t) it takes for the water to reach 10.0 m:

10.0 = 20.0 x t - 0.5 x 9.8 x t²

4.9t² - 20.0t + 10.0 = 0

Solving for t, we get t ≈ 0.54 s or t ≈ 3.74 s

Both times are valid because water can take two paths: one on the way up (0.54 s) and one on the way down (3.74 s).

3. Calculate horizontal distance

Distance (x) = v₀x x t

For t ≈ 0.54 s: x = 15.0 x 0.54 ≈ 8.0 m

For t ≈ 3.74 s: x = 15.0 x 3.74 ≈ 56.1 m

Therefore, the cannon can be placed either 8.0 meters or 56.1 meters from the building.

Answer:

0.0167 m

Explanation:

f = Frequency = 18 GHz

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

When the speed of the wave is divided by the freqeuncy we get the wavelength

Wavelength is given by

[tex]\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{18\times 10^9}\\\Rightarrow \lambda=0.0167\ m[/tex]

The wavelength of the radio waves is 0.0167 m

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = [tex]20\times 1=20\ m[/tex]

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s[/tex]

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

The final speed of the motorist as she reaches the light is 5.72 m/s

To calculate the speed of the motorist as she reaches the light, we need to first find the deceleration of the motorist.

Formula:

Where:

Given:

Therefore,

Then,

Where:

we use the formula below to calculate the deceleration of the motorist.

Where:

Given:

Substitute these values into equation 2

Solve for a

Finally, we use the formula below to get the speed of the motorist as she reaches the light.

Given:

Substitute these values into equation 3

Hence, The final speed of the motorist as she reaches the light is 5.72 m/s

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Answer:

The force on the electron will become zero.

Explanation:

As electron has negative charge and proton has positive charge. The magnitude of charge on both particles is equal. Therefore, there will be a force of attraction between electron and proton. When another electron is brought near the proton the net charge in that area will become equal to zero. Therefore, first electron will not experience any force.

Final answer:

The original force between the first electron and proton remains unchanged when a second electron is placed next to the proton, but the first electron will experience an additional repulsive force from the second electron. This alters the net force acting on the first electron but does not directly change the force between it and the proton.

Explanation:

The question inquires about the effect on the force on an electron when a second electron is placed next to a proton, with all particles separated by a distance of 1 meter.

According to Coulomb's Law, which governs the electrostatic force between two charged particles, the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

When a second electron is placed next to the proton, this setup introduces additional forces: a repulsive force between the two electrons and an attractive force between the new electron and the proton.

However, the original question seems to focus on how the force on the first electron changes with this new arrangement. The presence of a second electron does not alter the magnitude of the force between the first electron and the proton directly since the forces here are evaluated pairwise, and the distance between them remains unchanged.

What changes, though, is the overall electrical environment, introducing a new set of forces that need to be considered for the complete system. The addition of the second electron introduces a repulsive force on the first electron, which is separate from but concurrent with the attractive force from the proton.

It is important to consider the net force acting on any charge in such scenarios, which would involve adding vectorially the attractive force from the proton and the repulsive force from the second electron.

Thus, while the force due to the proton-electron pair remains constant, the first electron experiences an additional repulsive force due to the second electron, affecting the net force on it but not the force between it and the proton directly.

Answer:

[tex]\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

Explanation:

Electric Potential of Point Charges

The electric potential from a point charge Q at a distance r from the charge is

[tex]\displaystyle V=k\frac{Q}{r}[/tex]

Where k is the Coulomb's constant. The total electric potential for a system of point charges is equal to the scalar sum of their individual potentials. The potential is not a vector, so there is no direction or vectors to deal with.

We are required to compute the total electric potential in the center of the square. We need to know the distance from each corner to the center. The diagonal of the square is

[tex]d=\sqrt2 a[/tex]

where a is the length of the side.

The distance from any corner to the center is half that diagonal, thus

[tex]\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}[/tex]

The total potential in the center is  

[tex]V_t=V_1+V_2+V_3+V_4[/tex]

Please note all the potentials must be calculated including the sign of the charges. Since all the charges are equal to Q, and the distances are the same, the total potential is 4 times the individual potential of each charge.

[tex]V_t=4\times V[/tex]

[tex]\displaystyle V=9\times 10^9 \frac{Q}{\frac{a}{\sqrt{2}}}[/tex]

Operating

[tex]\displaystyle V=9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

Thus:

[tex]\displaystyle V_t=4\times 9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

[tex]\boxed{\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}}[/tex]

Answer:

a)  y = A x² , b)   A = - ½ g / v₀², c)    v₀ = 15.46 m / s

Explanation:

For this problem of two-dimensional kinematics, we will use that the time to reach the wall is the same

X axis

         x = v₀ₓ t

         t = x / v₀ₓ

Y Axis  

         y = [tex]v_{oy}[/tex] t - ½ g t²

As it shoots horizontally the vertical speed is zero

        y = - ½ g t²

We replace

         y = - ½ g (x / v₀ₓ)²

The initial speed is all horizontal

        v₀ₓ = v₀

         y = - ½ g / v₀²    x²

        y = A x²

b) the expression for the constant is

           A = - ½ g / v₀²

c) we look for the initial speed

           v₀² = - ½ g x² / y

   As the object falls below the exit point its height is negative

          v₀ = √ (- ½  9.8   3²/ (-0.21))

          v₀ = 15.46 m / s