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Travka [436]
3 years ago
5

Which is true concerning the acceleration due to Earth's gravity (ge) ? It decreases with increasing altitude. B. It is differe

nt for different objects in free fall. C. It is a fundamental quantity. D. It increases with increasing altitude. E. all of these
Physics
1 answer:
klasskru [66]3 years ago
6 0

Answer:

Option A decreases with increase in altitude

Explanation:

This can be explained as the value of gravitational acceleration, 'g' is not same everywhere.

It has its maximum value at poles of the Earth and minimum on its equator.

Thus a person will weigh more at poles than equator.

This variation is in accordance to:

g = \frac{GM_{E}}{radius^{2}}

Thus the gravitational acceleration changes as inverse square of the Radius of the Earth.

Thus as we move away from the Earth's center, gravitational acceleration, g decreases.

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A hockey player strikes a puck that is initially at rest. The force exerted by the stick on the puck is 975 N, and the stick is
Anna007 [38]

Explanation:

Given that,

The force exerted by the stick on the puck is 975 N

The stick is in contact with the puck for 0.0049 s

Initial speed of the puck, u = 0 (at rest)

(a) We need to find the impulse imparted by the stick to the puck.

Impulse = Force × time

J = 4.7775 kg-m/s

(b) Mass of the puck, m = 1.76 kg

We need to find the speed of the puck just after it leaves the hockey stick.

Let the speed be v.

As impulse is equal to the change in momentum.

J=m(v-u)\\\\4.7775=1.67(v-0)\\\\v=\dfrac{4.7775}{1.67}\\\\v=2.86\ m/s

So, when the puck leaves the hockey stick its speed is 2.86 m/s.

8 0
3 years ago
•Would a moving fan have energy? Why or why not.
Pepsi [2]
Moving fan has rotational kinetic energy
Non moving fan has no energy since it is in rest
7 0
4 years ago
The man is walking with speed v1 = 1.35 m/s to the right when he trips over a small floor discontinuity. Estimate his angular ve
soldier1979 [14.2K]

Answer:

The angular velocity, \omega = 1.05 rad/s clockwise

Explanation:

The mass of the man, m = 72 kg

The center of mass of the man will be at the middle of his body, particularly around his abdomen.

Height of the center of mass, h = 0.71 m

The speed of the man, v₁ = 1.35 m/s

Moment of inertia about the ankle joint, I = 66 kg/m²

Based on the principle of conservation of angular momentum (about the ankle joint):

Angular momentum before impact = Angular momentum after impact

Angular momentum before impact = -mv₁h

Angular momentum before impact = -(72 * 1.35 * 0.71)

Angular momentum before impact = -69.012 kg m²/s...............(1)

Angular momentum after impact = I \omega

Angular momentum after impact = 66 * \omega.................(2)

66 \omega = -69.012\\\omega = -69.012/66\\\omega = -1.05 rad/s

The angular velocity, \omega = 1.05 rad/s clockwise

8 0
3 years ago
Your forehead can withstand a force of about 6.0 kn before fracturing, while your cheekbone can withstand only about 1.3 kn. sup
Vinil7 [7]

a) The change in momentum of the ball is given by:

\Delta p = m \Delta v

where m=140 g=0.14 kg is the mass of the ball, and \Delta v =30 m/s is the change in velocity. Substituting, we find

\Delta p=(0.14 kg)(30 m/s)=4.2 kg m/s

The ball stops in t=1.5 ms=0.0015 s; the magnitude of the force that stops the baseball is given by the ratio between the change in momentum and the time taken:

F=\frac{\Delta p}{t}=\frac{4.2 kg m/s}{0.0015 s}=2800 N=2.8 kN


b) The force we found at point a) is the force that the head exerts on the ball to stop it. However, Newton's third law states that when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. If we apply this law to this case, we understand that the force exerted by the baseball on the head is equal to the force exerted by the head on the ball: therefore, the answer is still 2.8 kN.


c) The forehead is not in danger of a fracture, since it can withstand a maximum force of 6.0 kN, while the ball exerts a force of 2.8 kN. Instead, the cheek is in danger of fracture, because it can withstand only a maximum force of 1.3 kN.

6 0
3 years ago
If the archerfish spits its water 60 degrees from the horizontal aiming at an insect 1.4 m above
Arisa [49]

Answer:

6.1 m/s

Explanation:

Take up to be positive.  Given (in the y direction):

Δy = 1.4 m

vᵧ = 0 m/s

g = -10 m/s²

Find: v₀ᵧ

vᵧ² = v₀ᵧ² + 2aΔy

(0 m/s)² = v₀ᵧ² + 2 (-10 m/s²) (1.4 m)

v₀ᵧ = 5.29 m/s

The vertical component is 5.29 m/s.  So the total magnitude of the initial velocity is:

v₀ᵧ = v₀ sin θ

5.29 m/s = v₀ sin 60°

v₀ = 6.11 m/s

Rounding to two significant figures, the fish must spit the water at 6.1 m/s.

6 0
3 years ago
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