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Sonbull [250]
3 years ago
9

Austin and his friends are pushing a heavy box up a ramp. How can they increase the rate at which the box is moving?

Physics
1 answer:
Zarrin [17]3 years ago
6 0

If they were pulling the box instead of pushing it up .

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PART A:
olga55 [171]

Answer:

4 m/s^{2}

Explanation:

From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.

The Net force = Mass * Acceleration.

Since Net Force = 20 Newton, Mass = 5kg, therefore;

20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;

Acceleration = \frac{20}{5} which gives 4.

Note: RHS means Right Hand Side.

         LHS means Left Hand Side.

 

7 0
2 years ago
Which statement is NOT true?
Nuetrik [128]

Answer:

light doesn't need a medium through which to travel because the speed of light is experimentally constant

4 0
2 years ago
A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Virty [35]

Hello!

A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result? A) law of differential mass B) law of conservation of momentum C) law of unequal forces D) law of accelerated collision

We have the following data¹:

ΔP (momentum before impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 5 m/s

mB (mass) = 5 kg

vB (velocity) = 0 m/s

Solving:

ΔP = mA*vA + mB*vB

ΔP = 10 kg*5 m/s + 5 kg*0 m/s

ΔP = 50 kg*m/s + 0 kg*m/s

Δp = 50 kg*m/s ← (momentum before impact)

We have the following data²:

ΔP (momentum after impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 0 m/s

mB (mass) = 5 kg

vB (velocity) = 10 m/s

Solving:

Δp = mA*vA + mB*vB

Δp = 10 kg*0 m/s + 5 kg*10 m/s

Δp = 0 kg*m/s + 50 kg*m/s

Δp = 50 kg*m/s ← (momentum after impact)

*** Then, which principle explains the result ?

Law of conservation of momentum, <u>since the total momentum of body A and B before impact is equal to the total momentum of body A and B after impact.</u>

Note:  Bodies of different masses and velocities may have the same kinetic energy, if proportionality between the units is maintained it can occur that they have the same kinetic energy.

Answer:

B) law of conservation of momentum

_______________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

5 0
2 years ago
A student mixes two clear liquids together. After a few minutes, a white powdery solid can be seen settling on the bottom of the
SIZIF [17.4K]

I think the answer is A, please lmk if that is incorrect!

4 0
3 years ago
Read 2 more answers
This procedure has been used to "weigh" astronauts in space: A 42.5 kg chair is attached to a spring and allowed to oscillate. W
Bingel [31]

Answer:

The mass of the astronaut is approximately 119.74 kg

Explanation:

Assuming this problem as a Simple Harmonic Motion of a mass-spring system, the period (T) of the oscillations for a mass (m) and spring constant (k) is:

T=2\pi\sqrt{\frac{m}{k}} (1)

First, we have to calculate the spring constant using equation (1) and the data provided for the oscillations without the astronaut:

<em>(it’s important to note that one complete vibration is the period of the movement)</em>

1.3=2\pi\sqrt{\frac{42.5}{k}}\Longrightarrow k=42.5(\frac{2\pi}{1.3})^{2}

k\approx992.8\,\frac{N}{m}

Now with the value of k, we can use again (1) to find the mass of the astronaut (Ma) that makes the period to be 2.54 seconds

2.54=2\pi\sqrt{\frac{42.5+M_{a}}{992.8}}\Longrightarrow M_{a}=992.8(\frac{2.54}{2\pi})^{2}-42.5

\mathbf{M_{a}\approx119.74\,kg}

3 0
3 years ago
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