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Elodia [21]
3 years ago
6

You are pushing a box up a ramp with 70 N of force. If the ramp is made steeper and shorter, what needs to happen to the box for

it to continue moving?
Physics
1 answer:
Elenna [48]3 years ago
6 0

Answer:

A larger force than 70 N will be required for the box to continue moving

Explanation:

A ramp is an inclined plane surface that is tilted to form a slope on its opposite sides

A ramp provides mechanical advantage or force amplification, by allowing less force to lift heavier load from having to move through a longer distance to reach a particular elevation when the slope of the ramp is gentle

Therefore, when the slope is steeper, and shorter, more force than 70 N will be required for the box to continue moving.

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What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s
Nikolay [14]

Answer: 20 m/s

Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:

Ekinetic=(1/2)*(m*v^2)

then E=0.5*30Kg*(20 m/s)^2=15*400=6000J

8 0
3 years ago
What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure
mezya [45]

Answer:

this is a no brainer

Explanation:

As air pressure in an area increases, the density of the gas particles in that area increases.

8 0
3 years ago
A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c
Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

I= \Delta p = m \Delta v = m (v-u)

where

m is the mass of the object

\Delta v is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s

(b) 155 N

The impulse can also be rewritten as

I=F \Delta t

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

\Delta t is the duration of the collision

In this situation, we have

\Delta t = 0.06 s

So we can re-arrange the equation to find the magnitude of the average force:

F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N

6 0
3 years ago
Newton
vodomira [7]

Given parameters:

Mass on earth  = 50kg

Unknown:

Mass on planet Xenon = ?

Weight on planet Xenon = ?

Mass is the amount of matter contained in a particular substance.

Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.

         Weight  = mass x acceleration due to gravity

Planet Xenon has half the gravitational force of Earth.

 This translated gives \frac{9.8}{2}   = 4.9m/s²

Now, mass is always the same every where if the amount of matter in a substance does not change.

In this problem, mass = 50kg on planet xenon.

Weight =  mass x acceleration due to gravity  = 50 x 4.9  = 245N

The weight on Xenon is 245N and the mass is 50kg

4 0
4 years ago
The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi
melomori [17]

Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC    = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC        = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

5 0
3 years ago
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