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2 years ago

The amount of gravity between 1kg of lead and Earth is ___ the amount of gravity between 1kg of marshmallows and Earth.

1 answer:

3 0

The amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.

Answer: Option C

<u>Explanation:</u>

According to universal law of gravity, the gravitational force is directly proportionate to the product of masses of the interacting objects and inversely proportionate to its distance squared between them.

Since the mass of Earth is very high compared to the mass of the other objects, be it lead or marshmallows, and the distance will be mostly same. Thus, there will be no change in the amount of gravity acting between them. Thus, the amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.

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A 2kg watermelon is dropped from a 4m tall roof a) use the appropriate kinematic equations to determine the instantaneous veloci

Ivenika [448]

Answer:

8.85m/s

Explanation:

The potential energy the watermelon held before dropping is Ep=mgh=2*9.8*4=78.4J.

When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.

We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s

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2 years ago

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

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2 years ago

Recall the thermal energy model you used earlier to simulate friction with the chemistry and physics books. How does heat travel

dsp73

Answer:

due to condection of heat from pan in thermal energy according to 9th

Explanation:

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2 years ago

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A bungee jumper starts with 1000 J in their GPE store. After they jump they fall and are brought to a stop with the bungee cord.

pochemuha

Answer:

energy is equal to 1000 J

Explanation:

When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J

This is a case of energy transformation, but the total value of mechanical energy does not change

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2 years ago

okay so there is a wheel and it shows it when its high, in the middle, and below. my question is when is its potenial energy the

earnstyle [38]

At the highest point

cheers

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3 years ago