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2 years ago

5

Suppose the moon were held in its orbit not by gravity but by a massless cable attached to the center of the earth. What would b

e the tension in the cable?

1 answer:

Mars2501 [29]2 years ago

5 0

Answer:

$1.98\cdot 10^{20} N$

Explanation:

The tension in the cable would be then equal to the gravitational force that keeps the Moon in circular orbit around the Earth. So, we just need to calculate the magnitude of this force, which is given by:

$F=G\frac{mM}{r^2}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=7.35\cdot 10^{22} kg$ is the mass of the Moon

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$r=3.84\cdot 10^8 m$ is the distance between the Earth and the Moon

Substituting these numbers into the formula, we find:

$F=(6.67\cdot 10^{-11} ) \frac{(7.35 \cdot 10^{22})(5.97\cdot 10^{24})}{(3.84\cdot 10^8)^2}=1.98\cdot 10^{20} N$

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A Porsche 944 Turbo has a rated engine power of 217hp . 30% of the power is lost in the drive train, and 70% reaches the wheels.

scZoUnD [109]

Explanation:

(a) It is given that two-third of weight is over the drive wheels. So, mathematically, w = $\frac{2}{3}mg$ .

Hence, maximum force is expressed as follows.

$F_{max} = \mu_{s} \times w$

$m \times a_{max} = \mu_{s} (\frac{2}{3} mg)$

Hence, the maximum acceleration is calculated as follows.

$a_{max} = \frac{2}{3} \mu_{s} \times g$

= $\frac{2}{3} \times 1.00 \times 9.8 m/s^{2}$

= 6.53 $m/s^{2}$

Hence, the maximum acceleration of the Porsche on a concrete surface where μs = 1 is 6.53 $m/s^{2}$ .

(b) Since, 30% of the power is lost in the drive train. So, the new power is 70% of $P_{max}$ .

That is, new power = $0.7 \times P_{max}$

Now, the expression for power in terms of force and velocity is as follows.

P = $F_{max} \nu$

$0.7 P_{max} = ma_{max} \nu$

Therefore, speed of the Porsche at maximum power output is as follows.

$\nu = 0.7 \times \frac{P_{max}}{ma_{max}}$

= $0.7 \times \frac{217 hp \times \frac{746 W}{1 hp}}{1500 kg \times 6.53 m/s^{2}}$

= 11.568 m/s

= 11.57 m/s

Therefore, speed of the Porsche at maximum power output is 11.57 m/s.

(c) The time taken will be calculated as follows.

time = $\frac{\text{velocity}}{\text{acceleration}}$

= $\frac{11.57 m/s}{6.53 m/s^{2}}$

= 1.77 s

Therefore, the Porsche takes 1.77 sec until it reaches the maximum power output.

6 0

2 years ago

Which equation is most likely used to determine the acceleration from a velocity vs:time graph?

tresset_1 [31]

Acceleration, a = (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.

7 0

2 years ago

Read 2 more answers

A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to

Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

$y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}$

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:

$x_{A}=V_{x}t$

$x_{A}=(100m/s)(3.19)$

$x_{A}=319m$

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

$x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}$

So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0

2 years ago

Li is riding her bicycle at 8.0 m/s. She slows down to 4.0 m/s. Her change in velocity is m/s. If Li takes 2 seconds to make thi

forsale [732]

You will have to use this formula:
$v = vo + a \times t$

Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs

Then:

-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2

Ps: It's value is negative because the she was in retrograde motion.

Answer: Her acceleration is -2 m/s^2.

4 0

2 years ago

Read 2 more answers

The standing vertical jump is a good test of an athlete's strength and fitness. The athlete goes into a deep crouch, then extend

SashulF [63]

Answer:

<em>The athlete will rise 1.10 meters off the ground</em>

Explanation:

<u>Vertical Motion</u>

If an object is launched vertically upwards at an initial speed vo, then it will reach a maximum height given by

$\displaystyle y_m=\frac{v_o^2}{2g}$

The athlete can exert a net force upwards equal to twice his weight. It makes him accelerate upwards at

$\displaystyle a=\frac{F_n}{m}=\frac{2W}{m}=2g$

The speed at the end of his push can be computed by

$v^2=2ay$

Replacing the value of a obtained above:

$v^2=4gy$

where y is the length of this crouch

$v^2=4\cdot 9.8\cdot 0.55$

$v=4.64\ m/s$

This is the initial speed of this vertical launch, thus

$\displaystyle y_m=\frac{4.64^2}{2\cdot 9.8}$

$y_m=1.10\ m$

5 0

2 years ago