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3 years ago

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The acceleration due to gravity on Mars is twice that on the Moon. If you hit a baseball on the Moon with the same effort (and t

herefore same speed and angle) as on Mars, how far yet wered rked out ofwould the ball would travel on the Moon compared to on Mars? Neglect air resistance on Mars. Flag estion Select one: a. twice as far as on Mars b. four times as far as on Mars c. the same distance as on Mars d, half as far as on Mars

1 answer:

Nimfa-mama [501]3 years ago

7 0

Answer:

option (d)

Explanation:

Acceleration due to gravity on mars = twice the acceleration due to gravity on moon

According to the conservation of energy, the kinetic energy at the time of hitting is totally converted into potential energy at height.

On the surface of moon, let the value of acceleration due to gravity is a and the ball rises upto height h.

So, kinetic energy = Potential energy

1/2 mv^2 = m x a x h ...... (1)

On the surface of mars, the value of acceleration due to gravity is 2a and the ball rises upto height h'.

So, kinetic energy = Potential energy

1/2 mv^2 = m x 2a x h' ...... (2)

By equating equation (1) and equation (2),we get

2h' = h

h' = h / 2

Thus, the ball rises upto half of the height as that of moon.

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Answer:

A. The mechanical energy transforms to thermal energy as the pendulum slows and eventually stops moving.

Explanation:

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3 years ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

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Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

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As we know the formula:

F = Eq

And we also know that :

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or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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4 years ago

A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plan

pshichka [43]

Answer:

a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

Wₓ = 1200 sin 30 = 600 N

W_y = 1200 cos 30 = 1039.23 N

Y axis

N- W_y = 0

N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

P -Wₓ -fr = 0

P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = 600+ 207.85

P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

P + fr -Wx = 0

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = Wₓ -fr

P = 600 - 207,846

P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

P - Wₓ + fr = 0

P = Wₓ - fr

fr = 0.15 1039.23

fr = 155.88 N

P = 600 - 155.88

P = 444.12 N

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3 years ago

A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to

dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So $1.41=2\times 3.14\sqrt{\frac{0.683}{K}}$

$0.2245=\sqrt{\frac{0.683}{K}}$

Squaring both side

$0.0504=\frac{0.4664}{K}$

$K=9.255N/m$

So spring constant of the spring will be equal to 9.255 N /m

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