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2 years ago

1. Calculate the momentum of each car before the collision: SHOW YOUR WORK!

Physics

1 answer:

a_sh-v [17]2 years ago

3 0

Answer:

Momentum of red car = 5kgm/s

Momentum of blue car = 0kgm/s

Explanation:

Momentum = mass × velocity

For the red car

Mass = 1kg

Velocity = 5m/s

Momentum of the red car = 1kg × 5m/s

Momentum of the red car = 5kgm/s

For the blue car.

Mass = 1kg

Velocity = 0m/s(shows that the blue car is stationery)

Momentum = 1kg ×0m/s

Momentum of the blue car = 0kgm/s

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Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

2 years ago

The resistance of a wire depends upon the material's resistivity and the length and cross‑sectional area of the wire. What will

andriy [413]

Answer:

R' = 4R

The resistance will become 4 times the initial value.

Explanation:

The resistance of a wire at room temperature, is given by the following formula:

R = ρL/A ----------- equation 1

where,

R = Resistance of wire

ρ = resistivity of the material

L = Length of wire

A = Cross-sectional area of wire

Now, if the length (L) is multiplied by 4, then resistance will become:

R' = ρ(4L)/A

R' = 4 (ρL/A)

using equation 1:

<u>The resistance will become 4 times the initial value.</u>

6 0

3 years ago

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would

natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a) $v_{es} =v_{rms}$ = 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

$v_{rms} = \sqrt{ \frac{3RT}{M}}$

So, $v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}$

multiply each side by M_{o2}, so we have

$v_{rms,o2}^2 *M_{o2} =3RT_{o2}$

solving for temperature T_{o2}

$T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}$

In the question given, $v_{rms} =v_{es}$

$T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}$

T_{o2} =1.58*10^5 K

7 0

3 years ago

A hot air balloon exerts a force of 1200 N while

ValentinkaMS [17]

Answer:

Explanation:

For the explained scenario in the free body force diagram definitely the two forces 1200 N and 800 N should present as they are the acting forces

So A & D rules out.

Then you must think of B & C.

You also know that the weight of the load is always acting downwards as that force is generated by gravitational field of Earth. So 800 N should be downwards not upwards. That rules out B.

So answer is C

(Free body diagram is shown in the graph)

4 0

3 years ago

Why do we never notice quantization?

jarptica [38.1K]

Answer:

Explanation:

quantization of energy is only seen in atoms

7 0

3 years ago