Answer:
Torque=6.261×10^-43Nm
Explanation:
Torque is given by÷
Torque=dqEsin(phi)
Where E is a constant=1.6022×10^-19
d= distance = 6.186×10^-30cm
Changing to metres=6.186×10^-28m
q= the charge=8500NC
Phi =42°
Torque= (6.186×10^-28)×8500×(1.6022×10^-19)sin42°
Torque= 8.425×10^-43 × 0.7431
Torque= 6.261 ×10^-43
The torque exerted by an 8500 N/C electric field exerts on the dipole is 6.261 ×10⁻⁴³ Nm.
What is torque?The force which causes the object to rotate about any axis is called perpendicular distance.
Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.
Given:
A single water molecule was oriented such that its dipole moment of magnitude 6.186 × 10 − 30 Cm is along a z-axis,
The electric field = 8500 N/C,
∅ = 42°
Calculate the torque as shown below,
Torque = d × q × E × sin(∅)
Where E is a constant = 1.6022×10⁻¹⁹
distance = 6.186×10⁻³⁰ cm = 6.186×10⁻²⁸ m
Torque = 6.186×10⁻²⁸ × 8500 × (1.6022 × 10⁻¹⁹) sin42°
Torque = 8.425×10⁻⁴³ × 0.7431
Torque = 6.261 ×10⁻⁴³
Thus, the torque is 6.261 ×10⁻⁴³ Nm.
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On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the circuit so that, maintaining the same speed, the intervals between them will decrease by 1 5
3 trams must be added
Explanation:
In this problem, there are 12 trams along the ring road, spaced at regular intervals.
Calling L the length of the ring road, this means that the space between two consecutive trams is
[tex]d=\frac{L}{12}[/tex] (1)
In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become
[tex]d'=(1-\frac{1}{5})d=\frac{4}{5}d[/tex]
And the number of trams will become
[tex]12+n[/tex]
So eq.(1) will become
[tex]\frac{4}{5}d=\frac{L}{n+12}[/tex] (2)
And substituting eq.(1) into eq.(2), we find:
[tex]\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3[/tex]
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Answer: 3
Explanation:
We need to do (1/(5-1))x12
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, downward. (a) How many electrons were added to or removed from the honeybee? (b) What is the ratio of the electric force on the bee to the bee’s weight (Fe/Fg)? (c) What electric field strength and direction would allow the bee to hang suspended in the air without effort?
The ratio of the electric force on the bee to the bee’s weight (Fe/Fg) is; 2.041 * 10⁻⁶
The electric field strength is; 4.9 * 10⁷ N/C
What is the electric field strength?
We are given;
Mass of bee; m = 0.12g = 0.00012 kg
Charge acquired by the bee q = +24 pC = 24 * 10⁻¹² C
Electric field; E = 100 N/C
B) The force F on a charge in electric field E is given by:
F = qE
F = 24 * 10⁻¹² * 100
F = 24 * 10⁻¹⁰ N
Weight is;
F_g = mg
F_g = 0.00012 * 9.8
F_g = 11.76 × 10⁻⁴ N
Ratio of force to weight is;
F_e/F_g = (24 * 10⁻¹⁰)/(11.76 × 10⁻⁴)
F_e/F_g = 2.041 * 10⁻⁶
C) To get the electric field strength, we will use the formula;
E' = mg/q
E' = (0.00012 * 9.8)/(24 * 10⁻¹²)
E' = 4.9 * 10⁷ N/C
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Final answer:
The honeybee gained 15,000 electrons. The ratio of the electric force on the bee to its weight is (100 N/C) x (24 x 10⁻¹² C) / [(0.12 g) x (9.8 m/s²)]. To hang suspended in the air without effort, the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).
Explanation:
(a) How many electrons were added to or removed from the honeybee?
To determine the number of electrons added to or removed from the honeybee, we can use the formula:
q = ne
Where q is the charge in Coulombs, n is the number of electrons, and e is the elementary charge, which is the charge of a single electron.
Given that the charge acquired by the honeybee is +24 pC (pC = picocoulombs = 10⁻¹² C), we can substitute the values into the formula:
24 x 10⁻¹² C = n x (1.6 x 10⁻¹⁹ C)
Solving for n, we find:
n = 15,000 electrons
Therefore, the honeybee gained 15,000 electrons.
(b) What is the ratio of the electric force on the bee to the bee’s weight (Fe/Fg)?
To calculate the ratio of the electric force on the bee to its weight, we can use the formula:
F = Eq
Where F is the force, E is the electric field strength, and q is the charge.
Given that the electric field near the surface of the Earth is 100 N/C downward, we can substitute the values into the formula:
F = (100 N/C) x (24 x 10⁻¹² C)
Using the weight formula Fg = mg, where m is the mass and g is the acceleration due to gravity, we can find the weight of the bee.
Fg = (0.12 g) x (9.8 m/s²)
Taking the ratio of Fe to Fg, we get:
Fe/Fg = (100 N/C) x (24 x 10⁻¹² C) / [(0.12 g) x (9.8 m/s²)]
(c) What electric field strength and direction would allow the bee to hang suspended in the air without effort?
The electric field needed for the bee to hang suspended in the air without any effort would need to balance the gravitational force acting on the bee. The gravitational force on the bee can be calculated using the formula Fg = mg, where m is the mass of the bee and g is the acceleration due to gravity. We can then set this force equal to the electric force using the formula F = Eq. Rearranging the formula, we get E = F/q.
Since we want the bee to hang suspended without any effort, the electric force needs to be equal in magnitude and opposite in direction to the gravitational force. This means the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).
Problem 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and that the moon requires 27.32 days to complete one full revolution about the earth.
The actual mass of the Earth (5.97 \times 10^24 kg).
To find the mass of the Earth using the given information about the moon's orbit and revolution period, we can use Kepler's third law, which states that the square of the orbital period of a planet or moon is proportional to the cube of its average distance from the planet or sun.
Let's call the mass of the Earth M and the mass of the moon m. The distance between the Earth and the moon is R = 238,910 mi.
According to Kepler's third law, we have:
(27.32 days)^2 = (238,910 mi)^3 / (M + m)
(38.79 days)^2 = (6,022,140,778 mi)^3 / M
(where 38.79 days is the sidereal period of Earth's rotation)
Dividing these two equations gives:
(38.79 days)^2 / (6,022,140,778 mi)^3 = (27.32 days)^2 / (238,910 mi)^3
Simplifying this expression:
M = (6,022,140,778 mi)^3 / [(38.79 days)^2 - (27.32 days)^2 / (1 + m/M)]^(1/3)
Since we know that m/M is very small (the mass of the moon is only about 1/81 of the mass of the Earth), we can approximate m/M as zero in this equation:
M = (6,022,140,778 mi)^3 / (38.79 days)^2 = 5.97 \times 10^24 kg
This is very close to the actual mass of the Earth (5.97 \times 10^24 kg).
At the instant that you fire a bullet horizontally from a rifle, you drop a bullet from the height of the gun barrel. If there is no air resistance, which bullet hits the level ground first? Explain.
Answer:
Both hit at the same time
Explanation:
If air resistant is ignored, then gravitational acceleration g is the only thing that affect the bullets vertical motion, no matter what their horizontal motion is. Since both bullets are starting from rest, vertically speaking their speeds are 0 initially, they are both subjected to the same acceleration g, then they travel at the same rate and would reach the ground at the same time.
What is the maximum number of 7.00 μF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC ?
Final answer:
To maintain the total charge below 953 µC for capacitors connected in parallel to a 3.00 V battery, a maximum of 45 capacitors, each with a capacitance of 7.00 µF, can be used. This is calculated by dividing the total charge by the charge per capacitor (953 µC / 21 µC per capacitor).
Explanation:
To find out the maximum number of 7.00 µF capacitors that can be connected in parallel while keeping the total charge under 953 µC, we use the formula for charge on a capacitor, Q=CV, where C is the capacitance and V is the voltage. Since the capacitors would be connected in parallel, the voltage across each capacitor would be the same as the voltage of the battery, which is 3.00 V.
The total charge Q for one capacitor is given by:
Q = C × V = 7.00 µF × 3.00 V = 21.00 µC per capacitor.
To find out how many such capacitors we could connect in parallel without exceeding the total charge of 953 µC, we divide the total permissible charge by the charge per capacitor:
Number of capacitors = Total charge / Charge per capacitor = 953 µC / 21.00 µC per capacitor ≈ 45.38.
Since we can't have a fraction of a capacitor, the maximum number you can use is 45 capacitors.
The maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored under 953 μC is 45 capacitors.
To determine the maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC, follow these steps:
First, recall the formula for the charge stored in a capacitor: Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage.Each 7.00 µF capacitor connected to a 3.00 V battery will store a charge of: Q = 7.00 µF × 3.00 V = 21.00 μCTo find the maximum number of capacitors (n) that can be connected while keeping the charge under 953 μC, set up the inequality. n × 21.00 μC < 953 μCSolving for n, we get: n < 953 μC / 21.00 μC ≈ 45.38Since the number of capacitors must be an integer, the maximum number of 7.00 µF capacitors that can be connected in parallel is 45 capacitors.
Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)
Answer:
In New York is at ocean level while Denver is at an altitude of 1 mile from ocean level. In this way, the breaking point of water is lower in Denver than in New York, that is, the water will boil at lower temperatures in Denver than in New York. If the breaking point of the water decreases, at this point it will put aside more effort to cook an egg. Now the time required to cook the egg is higher in Denver than in New York.
An egg cooks slower in Denver because water boils at a lower temperature due to the city's higher altitude and lower atmospheric pressure. This lower boiling point decreases the rate of the cooking chemical reactions.
Explanation:An egg cooks more slowly in boiling water in Denver than in New York City due to differences in atmospheric pressure and the effect this has on boiling point temperatures. Denver, known as the Mile-High City, is approximately one mile above sea level. This altitude results in a lower atmospheric pressure than New York City which is virtually at sea level. Because of this, water boils at a slightly lower temperature in Denver (approximately 202 degrees Fahrenheit) than it does in New York City (approximately 212 degrees Fahrenheit).
Since cooking is essentially a series of chemical reactions, and these reactions occur faster at higher temperatures, an egg will cook more slowly in Denver’s boiling water than in New York's. This is because the boiling water in Denver is at a lower temperature due to the effect of pressure on boiling point.
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If two such generic humans each carried 2.5 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 600 weight? *An average human weighs about 600 r= .............. km
Answer:
They would be [tex]r=9.7\,km [/tex] apart
Explanation:
Electric force between two charged objects is:
[tex] F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}[/tex] (1)
With q1, q2 the charges of the humans, r the distance between them and k the constant [tex] k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}[/tex], so if we want the electric force between them will be equal to their 600 N weight, we should make W=Fe=600 N on (1):
[tex]600=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} [/tex]
solving for r:
[tex]r^{2}=k\frac{\mid q_{1}q_{2}\mid}{600}[/tex]
[tex] r=\sqrt{k\frac{\mid q_{1}q_{2}\mid}{600}}[/tex]
[tex]r=\sqrt{(9.0\times10^{9})\frac{\mid(-2.5)(2.5)\mid}{600}} =9682 m[/tex]
[tex]r=9.7\,km [/tex]
A sperm whale can accelerate at about 0.120 m/s 2 0.120 m/s2 when swimming on the surface of the ocean. How far will a sperm whale travel if it starts at a speed of 1.10 m/s 1.10 m/s and accelerates to a speed of 2.05 m/s 2.05 m/s ? Assume the whale travels in a straight line.
Answer:
The distance traveled by the sperm whale is 12.46 meters.
Explanation:
Given that,
Acceleration of a sperm whale, [tex]a=0.12\ m/s^2[/tex]
Initial speed of the whale, u = 1.1 m/s
Final speed of the whale, v = 2.05 m/s
We need to find the distance traveled by the sperm whale. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
d is the distance covered
[tex]d=\dfrac{v^2-u^2}{2a}[/tex]
[tex]d=\dfrac{(2.05)^2-(1.1)^2}{2\times 0.12}[/tex]
d = 12.46 meters
So, the distance traveled by the sperm whale is 12.46 meters. Hence, this is the required solution.
The main waterline into a tall building has a pressure of 90 psia at 16 ft elevation below ground level. Howmuch extra pressure does a pump need to add to ensure a waterline pressure of 30 psia at the top floor 450 ft above ground?
Answer:
Explanation:
Given
initial Pressure [tex]P_1=90\ psia[/tex]
elevation [tex]z_1=16\ ft[/tex]
Final Pressure [tex]P_2=30\ psia[/tex]
elevation [tex]z_2=450\ ft[/tex]
Pressure after Pumping(pump inlet pressure ) is given by
[tex]P_{after\ pump}=P_{top}+\Delta P[/tex]
[tex]\Delta P=\rho gh[/tex]
where [tex]\rho [/tex]=density of water[tex](62.2\ lbm\ft^3)[/tex]
g=acceleration due to gravity[tex](32.2\ ft/s^2)[/tex]
h=elevation
[tex]\Delta P=62.2\times 32.2\times (16+450)\times \frac{1\ lbf}{32.174\ lbm\ ft}[/tex]
[tex]\Delta P=28,985\ lbf/ft^2[/tex]
[tex]\Delta P=201.3\ lbf/in.^2[/tex]
[tex]P_{after\ pump}=30+201.3=231.3\ lbf/in.^2[/tex]
Pressure required to be applied
[tex]\Delta P_{pump}=P_{after\ pump}-P_{bottom}[/tex]
[tex]\Delta P_{pump}=231.3-90=141.3\ psi[/tex]
Consider a bicycle that has wheels with a circumference of 1 m. What is the linear speed of the bicycle when the wheels rotate at 2 revolutions per second?
Answer:
V= 12.56 m/s
Explanation:
Given that
R= 1 m
The angular speed ,ω = 2 rev/s
We know that 1 rev = 2 π rad
Therefore
ω = 4 π rev/s
The linear speed V is given as
V= ω x R
V = 4 π x 1 m/s
V= 4 π m/s
Therefore the speed of the bicycle will be 4 π m/s
We know that , π = 3.14
V= 12.56 m/s
The linear speed will be 12.56 m/s
A ball is thrown at an angle of to the ground. If the ball lands 90 m away, what was the initial speed of the ball?
Answer:
The initial speed of the ball is 30 m/s.
Explanation:
It can be assumed that the ball is thrown at an angle of 45 degrees to the ground. The ball lands 90 m away. We need to find the initial speed of the ball. We know that the horizontal distance covered by the projectile is called its range. It is given by :
[tex]R=\dfrac{u^2\ sin2\theta}{g}[/tex]
u is the initial speed of the ball.
[tex]v=\sqrt{\dfrac{Rg}{sin2\theta}}[/tex]
[tex]v=\sqrt{\dfrac{90\times 9.8}{sin2(45)}}[/tex]
v = 29.69 m/s
or
v = 30 m/s
So, the initial speed of the ball is 30 m/s. Hence, this is the required solution.
The exact initial speed requires the angle [tex]\( \theta \)[/tex]; assuming [tex]\( \theta = 45^\circ \)[/tex], the initial speed is 29.71 m/s.
To solve this problem, we need to use the kinematic equations for projectile motion. The horizontal distance (range) \( R \) that the ball travels can be found using the following equation:
[tex]\[ R = \frac{{v_0^2 \sin(2\theta)}}{g} \][/tex]
Given that the ball lands 90 m away and assuming that the angle of projection [tex]\( \theta \)[/tex] is known (but not provided in the question), we can rearrange the equation to solve for [tex]\( v_0 \)[/tex]:
[tex]\[ v_0^2 = \frac{R \cdot g}{\sin(2\theta)} \][/tex]
[tex]\[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \][/tex]
However, since the angle [tex]\( \theta \)[/tex] is not given, we cannot calculate the exact initial speed [tex]\( v_0 \)[/tex] without additional information. The problem as stated is incomplete because it requires the value of [tex]\( \theta \)[/tex] to find [tex]\( v_0 \)[/tex].
If we assume that the angle [tex]\( \theta \)[/tex] is such that [tex]\( \sin(2\theta) \)[/tex] is maximized (which occurs at [tex]\( \theta = 45^\circ \)[/tex], then [tex]\( \sin(2 \cdot 45^\circ) = \sin(90^\circ) = 1 \)[/tex], and the equation simplifies to:
[tex]\[ v_0 = \sqrt{R \cdot g} \][/tex]
[tex]\[ v_0 = \sqrt{90 \, \text{m} \cdot 9.81 \, \text{m/s}^2} \][/tex]
[tex]\[ v_0 = \sqrt{882.9 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v_0 =29.71 \, \text{m/s} \][/tex]
Therefore, under the assumption that [tex]\( \theta = 45^\circ \)[/tex], the initial speed of the ball would be [tex]\( 29.71 \, \text{m/s} \)[/tex].
In conclusion, without the value of [tex]\( \theta \)[/tex], we cannot determine the exact initial speed of the ball. However, if we assume the most favorable condition for maximum range, where [tex]\( \theta = 45^\circ \)[/tex], the initial speed [tex]\( v_0 \)[/tex] would be [tex]\( 29.71 \, \text{m/s} \)[/tex].
If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?
Answer:
[tex]heigth=2.86m[/tex]
Explanation:
Given data
time=0.19 s
distance=1.6 m
To find
height
Solution
First we need to find average velocity
[tex]V_{avg}=\frac{distance}{time}\\V_{avg}=\frac{1.6m}{0.19s}\\V_{avg}=8.42m/s[/tex]
Also we know that average velocity
[tex]V_{avg}=(V_{i}+V_{f})/2\\[/tex]
Where
Vi is top of window speed
Vf is bottom of window speed
Also we now that
[tex]V_{f}=V_{i}+gt\\V_{f}=V_{i}+(9.8)(0.19)\\V_{f}=V_{i}+1.862[/tex]
Substitute value of Vf in average velocity
So
[tex]V_{avg}=(V_{i}+V_{f})/2\\where\\V_{f}=V_{i}+1.862\\and\\V_{avg}=8.42m/s\\So\\8.42m/s=(V_{i}+V_{i}+1.862)/2\\2V_{i}+1.862=16.84\\V_{i}=(16.84-1.862)/2\\V_{i}=7.489m/s\\[/tex]
Vi is speed of balloon at top of the window
Now we need to find time
So
[tex]V_{i}=gt\\t=V_{i}/g\\t=7.489/9.8\\t=0.764s[/tex]
So the distance can be found as
[tex]distance=(1/2)gt^{2}\\ distance=(1/2)(9.8)(0.764)^{2}\\ distance=2.86m[/tex]
To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight. Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can solve for the initial height to be approximately 0.01485 m or 14.85 cm above the top of the window.
Explanation:To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can plug in these values into the equation to solve for the initial height:
1.6 = 0.5*9.8*(0.19)^2
Simplifying the equation, we find that the balloon was dropped from a height of approximately 0.01485 m or 14.85 cm above the top of the window.
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A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa). How many electrons are transferred? Avogadro’s number is 6.022 × 1023 /mol, and the elemental charge is 1.602 × 10−19 C.
Answer:
Number of electrons, [tex]n=9.98\times 10^{19}[/tex]
Explanation:
A strong lightning bolt transfers an electric charge of about 16 C to Earth, q = 16 C
We need to find the number of electrons that transferred. Let there are n electrons transferred. It is given by using quantization of electric charge as :
q = ne
[tex]n=\dfrac{q}{e}[/tex]
e is elemental charge
[tex]n=\dfrac{16}{1.602\times 10^{-19}}[/tex]
[tex]n=9.98\times 10^{19}[/tex]
So, there are [tex]9.98\times 10^{19}[/tex] electrons that gets transferred. Hence, this is the required solution.
A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).
A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa).
What is the electric charge?Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.
We want to calculate the number of electrons that have a charge of 16 C. We have to consider the following relationships:
The charge of 1 electron is 1.602 × 10⁻¹⁹ C.There are 6.022 × 10²³ electrons in 1 mole (Avogadro's number).16 C × 1 electron/1.602 × 10⁻¹⁹ C = 1.0 × 10²⁰ electron
1.0 × 10²⁰ electron × 1 mol/6.022 × 10²³ electron = 1.7 × 10⁻⁴ mol
A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).
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Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper jumps back and forth between the cars (he has strong legs!) with a constant horizontal velocity of 15 m/s relative to the ground. The insect jumps the instant he lands, so he spends no time resting on either car. What total distance does the grasshopper travel before the cars hit?
Answer:
Total distance does the grasshopper travel before the cars hit is 150 m
Explanation:
Each car moves x=100 m before they collide. Both the cars moving in constant velocity. time taken t by each car is
[tex]t=\frac{x}{v}[/tex]
where x is the distance traveled with velocity v
[tex]t=\frac{100}{10}\\t=10 sec[/tex]
The insect is moving through this time period with a constant velocity of 15 m/s
The distance traveled by grasshopper is
[tex]distance=V_{gh} \times t\\distance=15 \times 10\\distance=150 m[/tex]
The grasshopper travels a total distance of 2600 m before the cars collide.
Explanation:We can solve this problem by calculating the time it takes for the cars to collide. Since the cars are moving towards each other at a combined speed of 20 m/s (10 m/s + 10 m/s), and they start 200 m apart, it will take them 200 m / 20 m/s = 10 seconds to collide.
During these 10 seconds, the grasshopper keeps jumping back and forth with a velocity of 15 m/s relative to the ground. To find the total distance the grasshopper travels, we need to calculate the number of jumps the grasshopper can make in 10 seconds. Since the grasshopper jumps the instant it lands, the number of jumps is equal to the number of times the grasshopper crosses from one car to the other. Given that the grasshopper has a velocity of 15 m/s, we divide the total distance the grasshopper travels by this velocity to find the number of jumps: 200 m / (15 m/s) = 13.33 jumps.
The grasshopper is not able to make a fraction of a jump, so we take only the whole number of jumps that the grasshopper can make: 13 jumps. Therefore, the total distance the grasshopper travels before the cars hit is 13 jumps · 200 m per jump = 2600 m.
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A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0°, and 60.0°.
The height and distance for 30 degrees are 14.4 m and 100 m, for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.
In projectile motion, consider the motion along vertical and horizontal separately.
Given:
Initial velocity, [tex]u =33.6\ m/s\\[/tex]
For angle [tex]30^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(30.0)\\=29.14m/s\\u_y =33.6\times sin30.0\\ =16.80m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{16.80^2}{2\times9.8}\\=14.4\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.14}{9.8}\\=100\ m[/tex]
For angle [tex]45^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(45.0)\\=23.75m/s\\u_y =33.6\times sin45\\ =23.75m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{23.75^2}{2\times9.8}\\=28.77\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times23.75\times23.75}{9.8}\\=115.11\ m[/tex]
For angle [tex]60^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(60)\\=16.8m/s\\u_y =33.6\times sin60\\ =29.09m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{29.09^2}{2\times9.8}\\=43.17\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.09}{9.8}\\=100\ m[/tex]
Therefore, the distance and height for 30 degrees are 14.4 m and 100 m for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.
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To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. The distance traveled can be calculated using the formula d = v²sin(2θ) / g.
To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s²). Once we have the height, we can calculate the distance traveled using the formula d = v²sin(2θ) / g.
Height (h) = (33.6²sin²30.0°) / (2 * 9.8) = 19.22 meters
Distance (d) = 33.6²sin(2 * 30.0°) / 9.8 = 152.19 meters
Height (h) = (33.6²sin²45.0°) / (2 * 9.8) = 38.45 meters
Distance (d) = 33.6²sin(2 * 45.0°) / 9.8 = 203.43 meters
Height (h) = (33.6²sin²60.0°) / (2 * 9.8) = 57.67 meters
Distance (d) = 33.6²sin(2 * 60.0°) / 9.8 = 228.05 meters
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Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it 2.20 m above the ground. Ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
a. 6.69m/s
b. y=4.48m
c. t=1.43secs
Explanation:
Data given, acceleration,a=35m/s^2
distance covered,d=64cm=0.64m,
a. to determine the speed, we use the equation of motion
initial velocity,u=0m/s
if we substitute values we arrive at
[tex]v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\[/tex]
b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2
and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.
Hence we can write the equation above again
[tex]v^{2}=u^{2}-2a(y-2.2)\\[/tex]
if we substitute values we have
[tex]v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m[/tex]
c. the time it takes to arrive at 1.83m is obtain by using the equation below
[tex]1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37[/tex]
if we insert the values, we solve for t , hence t=1.43secs
(a) The speed of the shot when Sam releases it 6.75 m/s.
(b) The height risen by the shot above the ground is 4.52 m.
(c) The time taken for the shot to return to 1.8 m above the ground is 1.44 s.
The given parameters;
constant acceleration, a = 35 m/s²height above the ground, h₀ = 64 cm = 2.2 mheight traveled, Δh = 64 cm = 0.64 mThe speed of the shot when Sam releases it is calculated as;
[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2a(\Delta h)\\\\v = \sqrt{2a(\Delta h)} \\\\v = \sqrt{2\times 35(0.65)} \\\\v = 6.75 \ m/s[/tex]
The height risen by the shot is calculated as follows;
[tex]v^2 = u^2 + 2gh\\\\at \ maximum \ height , v = 0\\\\0 = (6.75)^2 + 2(-9.8)h\\\\19.6h = 45.56 \\\\h = \frac{45.56}{19.6} \\\\h =2.32 \ m[/tex]
The total height above the ground = 2.20 m + 2.32 m = 4.52 m.
The time taken for the shot to return to 1.8 m above the ground is calculated as follows;
the time taken to reach the maximum height is calculated as;
[tex]h = vt - \frac{1}{2} gt^2\\\\2.32 = 6.75t - (0.5\times 9.8)t^2\\\\2.32 = 6.75t -4.9t^2\\\\4.9t^2 -6.75t + 2.32=0\\\\a = 4.9, \ b = -6.75, \ c = 2.32\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-6.75) \ +/- \ \ \sqrt{(-6.75)^2 - 4(4.9\times 2.32)} }{2(4.9)} \\\\t = 0.72 \ s\ \ or \ \ 0.66 \ s[/tex]
[tex]t \approx 0.7 \ s[/tex]
height traveled downwards from the maximum height reached = 4.52 m - 1.8 m = 2.72 m
[tex]h = vt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(2.72)}{9.8} } \\\\t = 0.74 \ s[/tex]
The total time spent in air;
[tex]t = 0.7 \ s \ + \ 0.74 \ s\\\\t = 1.44 \ s[/tex]
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If you find two stars with the same Right Ascension, are they necessarily close together in the sky? Why or why not?
In space, spatial coordinates can be roughly divided into measures of Right ascension and declination. The declination is measured in degrees while the ascent is measured in hours, minutes, seconds. When you have objects in space such as those of the characteristics presented we will have to they are not necessarily close together in the sky because we can find two stars on the same right ascension but on different declination lines (Which means they can be very far apart from each other)
A plane flies 1.4hours at 110mph on a bearing of 40degrees.It then turns and flies 1.7hours at the same speed on a bearing of 130degrees.How far is the plane from its starting point?
Answer:
Distance from starting point= 242.249871 miles
Explanation:
Distance covered on bearing of 40 degree=a= [tex]1.4*110\ mph[/tex]
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=[tex]1.7*110\ mph[/tex]
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=[tex]130-40[/tex]
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=[tex]\sqrt{a^2+b^2}[/tex]
Distance from starting point=[tex]\sqrt{154^2+187^2}[/tex]
Distance from starting point= 242.249871 miles
Answer:
Distance from starting point= 242.249871 miles
Distance covered on bearing of 40 degree=a=
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=
Distance from starting point=
Distance from starting point= 242.249871 miles
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Two 1.4 g spheres are charged equally and placed 1.6 cm apart. When released, they begin to accelerate at 110 m/s2 . Part A What is the magnitude of the charge on each sphere? Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]q = 2.17\times 10^{-15}~{\rm C}[/tex]
Explanation:
The relation between the acceleration and the electrical force between the spheres can be found by Newton's Second Law.
[tex]\vec{F} = m\vec{a}\\\vec{F} = (1.4\times 10^{-3})(110) = 0.154~N[/tex]
This is equal to the Coulomb's Force.
[tex]F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{2q}{(1.6 \times 10^{-2})^2} = 0.154\\q = 2.17\times 10^{-15}~C[/tex]
Find the magnitude of the net electric force exerted on this charge. Express your answer in terms of some or all of the variables q, R, and appropriate constants.
Answer:
Th steps is as shown in the attachment
Explanation:
from the diagram, its indicates that twelve identical charges are distributed evenly on the circumference of the circle. assuming one of gthe charge is shifted to the centre of he circle alomng the x axis, as such the charge is unbalanced and there is need ot balanced all the identical charges for the net force to be equal to zero.
The mathematical interpretation is as shown in the attachment.
A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is fixed in place at the bottom of the incline. If the inclined plane makes an angle θ of 35.0 ∘ with the horizontal, what is the magnitude of the acceleration of the box when it is 61.0 cm from the bottom of the incline?
The magnitude of the acceleration of the box is [tex]a=6.41\ m/s^2.[/tex]
First, we need to calculate the force between the two charges using Coulomb's law:
[tex]\[ F = \frac{qQ}{4\pi\epsilon_0r^2} \][/tex]
Given that [tex]\( q = +2.50 \, \mu C = 2.50 \times 10^{-6} \, C \), \( Q = +75.0 \, \mu C = 75.0 \times 10^{-6} \, C \)[/tex], and [tex]\( r = 61.0 \, cm = 0.610 \, m \)[/tex], we can plug these values into the equation:
[tex]\[ F = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.610 \, m)^2} \][/tex]
Solving this, we get the force [tex]\( F \)[/tex] in newtons.
Next, we need to find the component of this force parallel to the incline, which is given by:
[tex]\[ F_{\parallel} = F \sin(\theta) \][/tex]
However, since the box is on an incline, we use [tex]\( \tan(\theta) \)[/tex] instead of [tex]\( \sin(\theta) \)[/tex] to find the acceleration:
[tex]\[ F_{\parallel} = F \tan(\theta) \][/tex]
Given [tex]\( \theta = 35.0^{\circ} \)[/tex], we can calculate [tex]\( F_{\parallel} \)[/tex].
Finally, the acceleration \( a \) of the box is given by Newton's second law:
[tex]\[ a = \frac{F_{\parallel}}{m} \][/tex]
where[tex]\( m = 495 \ , g = 0.495 \, kg \)[/tex].
Combining all the steps, we have:
[tex]\[ a = \frac{qQ}{4\pi\epsilon_0mr^2} \tan(\theta) \][/tex]
Plugging in the values, we get:
[tex]\[ a = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.495 \, kg)(0.610 \, m)^2} \tan(35.0^{\circ}) \][/tex]
Calculating the value of [tex]\( a \)[/tex], we find the magnitude of the acceleration of the box.
[tex]a=6.41\ m/s^2[/tex]
A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured ������� – ��������� with respect to the positive � − ����. What are the magnitude and angle (measured ������� – ��������� with respect to the positive � − ����) of the equilibrant vector? Please show all steps in your calculations
Answer:
223 degree
Explanation:
We are given that
Magnitude of resultant vector= 8 units
Resultant vector makes an angle with positive -x in counter clockwise direction
[tex]\theta=43^{\circ}[/tex]
We have to find the magnitude and angle of the equilibrium vector.
We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.
Therefore, magnitude of equilibrium vector=8 units
x-component of a vector=[tex]v_x=vcos\theta[/tex]
Where v=Magnitude of vector
Using the formula
x-component of resultant vector=[tex]v_x=8cos43=5.85[/tex]
y-component of resultant vector=[tex]v_y=vsin\theta=8sin43=5.46[/tex]
x-component of equilibrium vector=[tex]v_x=-5.85[/tex]
y-component of equilibrium vector=[tex]-v_y=-5.46[/tex]
Because equilibrium vector lies in III quadrant
[tex]\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}[/tex]
The angle [tex]\theta'[/tex]lies in III quadrant
In III quadrant ,angle =[tex]\theta'+180^{\circ}[/tex]
Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree
5. An automobile cooling system holds 16 L of water. How much heat does it absorb if its temperature rises from 20o C to 90o C?
Answer:
[tex]Q=4704000J\\Q=4.7MJ[/tex]
Explanation:
Given data
Heat capacity c=4.2 J/gC
Water weigh m=1000g
Temperature Risen T=20°C to 90°C
To find
Heat Q
Solution
The formula to find Heat is:
Q=mcΔT
Where
m is mass
c is heat capacity
ΔT is change in temperature
So
[tex]Q=4.2(16*1000)*(90-70)\\Q=4704000J\\Q=4.7MJ[/tex]
The automobile cooling system would absorb approximately 4,699,840 Joules of heat when its temperature rises from 20°C to 90°C, according to the formula for heat absorption Q = mcΔT.
Explanation:To answer this question, we apply the formula for heat absorption: Q = mcΔT.
Here, 'Q' is the total heat absorbed, 'm' is the mass of the substance (water in this case), 'c' is the specific heat capacity of the substance, and 'ΔT' is the change in temperature. The mass can be defined as the volume of water multiplied by the density of water (1 g/cm^3). So for 16L, m = 16000g. The specific heat capacity of water is approximately 4.186 J/g °C, and the change in temperature, ΔT, is 70°C (90°C - 20°C).
Plugging in these values, we get Q = 16000g× 4.186 J/g °C × 70°C = 4,699,840 Joules. Therefore, it's estimated that the automobile cooling system would absorb around 4,699,840 Joules of heat.
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An experiment is performed in the lab, where the mass and the volume of an object are measured to determine its density. Two completely different valid methods are used. Each experimental method to measure the density is performed, two considerable sets of data are taken on each and the results are compared. The results of the density measurement by each method should be:______.
a. dependent on the mass and the volume of the object
b. completely different, because it was measured different ways.
c. correct on the most accurate method and wrong in the other
d. the same within the uncertainty of each measurement method
Answer:
d. the same within the uncertainty of each measurement method
Explanation:
The density of an object and in general any physical property, has the same value regardless of the method used to measure it, either directly or indirectly. Since two completely different valid methods are used, the results must be the same, taking into account the level of precision of each of the methods.
What is the resolution of an analog-to-digital converter with a word length of 12 bits and an analogue signal input range of 100V? Show work.
The resolution of an ADC with a word length of 12 bits and an input range of 100V is approximately 0.0244V. Calculating this gives us a resolution of approximately 0.0244V.
Explanation:The resolution of an analog-to-digital converter (ADC) is determined by the number of bits used to represent the digital output.
In this case, the ADC has a word length of 12 bits.
The resolution can be calculated using the formula:
Resolution = Full Scale Range / (2^Word Length)
In this case, the Full Scale Range is 100V. Plugging in the values:
Resolution = 100V / (2^12)
Calculating this gives us a resolution of approximately 0.0244V.
a skydiver jumps out of a plane waering a suit and decleration due to air i s 0.45 m/s^2 for each 1 m/s, skydyvers initial value problems that models the skydiver velocity is v(t). temrinal speed assuming that accelration due to gravity is 9.8m/s
Answer:
[tex]v = 21.77\ m/s[/tex]
Explanation:
Given,
Air resistance, = 0.45 m/s² for each 1 m/s
Air resistance for velocity v = 0.45 v.
Terminal velocity = ?
acceleration due to gravity= g = 9.8 m/s²
now,
a = 0.45 v
we know,
[tex]a =\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=0.45 v[/tex]
[tex]\dfrac{dv}{v} = 0.45 dt[/tex]
integrating both side
[tex]\int \dfrac{dv}{v} = 0.45\int dt[/tex]
[tex] ln(v) = 0.45 t[/tex]
[tex]t = e^{0.45 t}[/tex]
When a body is moving at terminal velocity, Force due to gravity is equal to force due to air resistance.
m a = m g
0.45 v = 9.8
[tex]v = \dfrac{9.8}{0.45}[/tex]
[tex]v = 21.77\ m/s[/tex]
Hence, the terminal velocity of the skydiver is equal to 21.77 m/s.
A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each filled balloon is a sphere 30.0 cm in diameter at 27.0oC and absolute pressure of 1.20 atm? Assume all the helium is transferred to the balloons.
Answer:
884 balloons
Explanation:
Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.
So if pressures reduces from 100 to 1.2, the new volume would be
[tex]V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2[/tex]
The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is
[tex]V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3[/tex]
The number of balloons that 12.5 m3 can fill in is
[tex]V_2/V_b = 12.5 / 0.014 = 884[/tex]
Final answer:
Based on the volume of the tank, temperature, and pressure, we can calculate the number of moles of helium. By dividing this number by the number of moles of helium in one balloon, we can determine how many balloons can be blown up.
Explanation:
To find out how many balloons can be blown up with the given amount of helium, we need to calculate the total volume of helium in the tank and divide it by the volume of each balloon.
Given:
The volume of the tank is 0.150 m³
The temperature of the helium in the tank is 27.0°C
The pressure of the helium in the tank is 100 atm
The volume of each balloon is 30.0 cm in diameter, which is equivalent to a radius of 15.0 cm or 0.15 m
The temperature of the filled balloon is 27.0°C
The absolute pressure of the filled balloon is 1.20 atm
First, we need to convert the volume of the tank to liters:
0.150 m³ * 1000 L/m³ = 150 L
Next, we need to calculate the number of moles of helium in the tank using the ideal gas law:
P * V = n * R * T
n = (P * V) / (R * T)
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Using the given values:
Pressure (P) = 100 atm
Volume (V) = 150 L
Ideal gas constant (R) = 0.0821 L·atm/K·mol
Temperature (T) = 27.0°C + 273.15 = 300.15 K
Calculating the number of moles:
n = (100 atm * 150 L) / (0.0821 L·atm/K·mol * 300.15 K) = 6.19 moles
Now we can calculate the number of balloons that can be blown up:
Number of balloons = (Number of moles of helium) / (Number of moles of helium in one balloon)
The number of moles of helium in one balloon can be found using the ideal gas law:
P * V = n * R * T
n = (P * V) / (R * T)
Using the given values:
Pressure (P) = 1.20 atm
Volume (V) = (4/3) * π * (0.15 m)³ = 0.141 m³
Ideal gas constant (R) = 0.0821 L·atm/K·mol
Temperature (T) = 27.0°C + 273.15 = 300.15 K
Calculating the number of moles:
n = (1.20 atm * 0.141 m3) / (0.0821 L·atm/K·mol * 300.15 K) = 0.006 moles
Finally, calculating the number of balloons:
Number of balloons = 6.19 moles / 0.006 moles = 1031
Therefore, 1031 balloons can be blown up with the given amount of helium.
A 1500 kg car traveling at 90 km/hr East collides with a 3000 kg truck traveling at 60 km/hr South. They stick together and move which direction?
Answer:
They both move towards South direction.
Explanation:
NB: The car traveling due east must be backing west and the car traveling due south must be backing north.
From their momentum in calculation below:
Momentum of the car traveling east is MU= 1500x90x1000/3600= 37,500kgm/s.
Momentum of car traveling due South =MU = 3000x60x1000/3600= 50,000. So, from here we see the the momentum of the car moving south has the high momemtum so they stick together and move towards the direction of south.
You are going to an outdoor concert, and you'll be standing near a speaker that emits 60 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Answer:
[tex]r=44m[/tex]
Explanation:
β is calculated as:
[tex]\beta =(10dB)log_{10}(I/I_{o} )\\ I=I_{o}10^{\frac{\beta }{10dB} }\\ I=(1.0*10^{-12}W/m^{2} )10^{\frac{\(94dB }{10dB} }\\I=2.51mW/m^{2}[/tex]
The distance r is defined as the radius of spherical wave.solve for r
We have
[tex]I=\frac{P_{source} }{4\pi r^{2} }\\ r=\sqrt{\frac{P_{source}}{4\pi I} }\\ r=\sqrt{\frac{60W}{4\pi (2.51mW/m^{2} )} }\\r=44m[/tex]
sky divers jump out of planes at an altitude of 4000m. How much timewill passuntill they deploy their parachutes at an altitutde of 760m?
Answer:
Explanation:
Given
Altitude from which Diver steps out [tex]H=4000\ m[/tex]
altitude at which diver opens the Parachutes [tex]h=760\ m[/tex]
total height covered by diver[tex]=H-h=4000-760=3240\ m[/tex]
Initial velocity of diver is zero
using equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where s=displacement
u=initial velocity
a=acceleration
t=time
[tex]3240=0+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=661.224[/tex]
[tex]t=25.71\ s[/tex]