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3 years ago

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t

he moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 14 m/s at an angle 20 ∘ above the horizontal. For how much more time was the ball in flight?

1 answer:

castortr0y [4]3 years ago

5 0

Answer:

4.86 seconds

Explanation:

Velocity of projection, u = 14 m/s

angle of projection, θ = 20°

Formula for the time of flight

$T=\frac{2uSin\theta }{g}$

For earth

Te = (2 x 14 x Sin 20) / 9.8

Te = 0.98 s

For moon

g' = g/6 = 1.64 m/s^2

Tm = ( 2 x 14 x Sin 20) / 1.64

Tm = 5.84 seconds

Tm - Te = 5.84 - 0.98 = 4.86 s

So, it takes 4.86 s more time of flight on moon than the earth.

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A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into t

geniusboy [140]

Answer:

M = 1433.5 kg

Explanation:

This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,

B = ρ g V

with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition

Σ F = 0

B-W = 0

B = W

body weight

W = M g

the volume is

V = l to h

rho_liquid g (l to h) = M g

M = rho_liquid l a h

we calculate

M = 1000 4.7 6.10 0.05

M = 1433.5 kg

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3 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

3 years ago

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh

Svetradugi [14.3K]

Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

The magnitude of the stimulated acceleration due gravity is $a = 0.5 g$

The diameter of the spaceship is $d = 35m$

Generally the force acting on the spaceship is

$F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

$F = \frac{mv^2}{r}$

Thus

$ma = \frac{mv^2}{r}$

=> $\frac{v^2}{r} = a$

Generally the speed of this spaceship is mathematically represented as

$v = \frac{2 \pi}{T}$

=> $v^2 = [\frac{2\pi}{T}] ^2$

=> $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=> $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=> $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

substituting values

$T = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}$

$T = 16.8 s$

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If you, for example, poured it onto a wide cup with a volume equal to the total volume of the sand particles, the sand would not spread out to fill the container but would bunch up together in the middle.

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Answer:

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Explanation:

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