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2 years ago

If the net force on an object stays the same but the mass of the object is doubled, what will happen to the

Physics

1 answer:

Sergio [31]2 years ago

5 0

Answer:

The acceleration of the object decreases I think

Explanation:

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WORD BANK:

Mkey [24]

Answer:

1. Vector, base

2. Vector, derived

3. Vector, ?

4. Scalar, derived

5. scalar, base

4 0

2 years ago

What is the difference between resilience and modulus of resilience

pshichka [43]

Resilience is the amount of energy that can be put into a volume of material and still be stored elastically. ie When the energy goes away, the material regains its undeformed shape. The Mod of R is the amount that can be stored by a unit volume of that material. The Mod of R is heavily related to Youngs Modulus.

6 0

3 years ago

6. Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. Which light bulb has the greater filament resi

Nonamiya [84]

$\\ \bull\tt\longrightarrow P=\dfrac{V^2}{R}$

P is power
R is resistance

$\\ \bull\tt\longrightarrow R=\dfrac{V^2}{P}$

Hence

$\\ \bull\tt\longrightarrow R\propto V$

$\\ \bull\tt\longrightarrow R\propto \dfrac{1}{P}$

Therefore if power is low then resistance will be high.

The first bulb has less power hence it has greater filament resistance.

5 0

3 years ago

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m bef

Dimas [21]

Supposing there's no air resistance, horizontal velocity is constant, which makes it very easy to solve for the amount of time that the rock was in the air.

Initial horizontal velocity is: 
cos(30 degrees) * 12m/s = 10.3923m/s

15.5m / 10.3923m/s = 1.49s

So the rock was in the air for 1.49 seconds.

Now that we know that, we can use the following kinematics equation:

d = v i * t + 1/2 * a * t^2

Where d is the difference in y position, t is the time that the rock was in the air, and a is the vertical acceleration: -9.80m/s^2.

Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s

So:

d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89

d = -1.95

This means that the initial y position is 1.95 m higher than where the rock lands.

5 0

3 years ago