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ycow [4]
3 years ago
5

If your weight is 120 pounds and your mass is 54 kilograms how would those values change if you were on the moon

Physics
1 answer:
BlackZzzverrR [31]3 years ago
4 0
The gravitation acceleration on the moon is different than on Earth. It is 1.6 m/s^2. If you weigh 120 lbs, then you would multiply 120 pounds by the gravitational acceleration on the moon and then divide by the acceleration on Earth.

(120 lbs * 1.6) / 9.8 = 20 pounds.

The mass will always be the same no matter what planet you’re on, so it’s still 54 kg.
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Is a tape dispenser a pulley or a wheel and axis? 
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It is a wheel and axis
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3 years ago
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A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock
denis-greek [22]

Answer:

a) v=32.74\frac{m}{s}.

b) t=\frac{v-v_0}{g} =5.58s.

Explanation:

a) What is the speed of the rock just before it hits the street?

From Kinematics we have v^2= v_0^2+2a(y_f-y_i).

If we take the top of the roof as the position y_0=0m, then

y_f=-30m and we have (also, a=g=-9.8\frac{m}{s^2}) :

v^2= v_0^2+ 2ay_f ⇒ v= \sqrt{v_0^2+ 2ay_f}

⇒ v=32.74\frac{m}{s}.

b) How much time elapses from when the rock is thrown until it hits the street?

From Kinematics we have v(t)=v_0+at=v_0+gt

When the rock touches the ground:

v=-32.74\frac{m}{s}

With a minus sign to indicate the vector velocity points down.

t=\frac{v-v_0}{g} =5.58s

(remember that g=-9.8\frac{m}{s^2})

3 0
3 years ago
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Answer:

B Explain why the relationship between force and time is important to a helmet designer.

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5 0
3 years ago
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
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