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3 years ago

If you are in a rear-wheel skid, you must release your brakes and turn your steering wheel __________

1 answer:

3 0

I think the correct answer would be A. If you are in a rear-wheel skid, you must release your brakes and turn your steering wheel into the direction of the skid. A skid in vehicles is a situation wherein one or more wheels are slipping in relation to the road and affecting the whole handling of the vehicle. To do a rear wheel skid, you should be releasing the break with caution and move or steer the wheel into the direction you would want to move.

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In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue

Andru [333]

Answer:

a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

p₀ = m v₁₀ + 0

p₀ = 0.6 2

p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

p₀ₓ = 1.2 kg m / s

p_{fx} = m v1f cos 20 + m v2f cos θ

p₀ = p_f

1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

1.2482 = v_{2f} cos θ

Y axis

p_{oy} = 0

p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

0.2736 = v_{2f} sin θ

we write our system of equations

0.2736 = v_{2f} sin θ

1.2482 = v_{2f} cos θ

divide to solve

0.219 = tan θ

θ = tan⁻¹ 0.21919

θ = 12.36

let's look for speed

0.2736 = v_{2f} sin θ

v_{2f} = 0.2736 / sin 12.36

v_{2f} = 1.278 m / s

4 0

3 years ago

A 4n force is applied to an object with a mass of 2kg what is the accelertion of the object

Pavel [41]

As per Newton's law we know that

F = ma

here we know that

F = 4 N

m = 2 kg

so from this equation we will have

4 = 2 a

a = 4/2 = 2 m/s^2

so the acceleration will be 2 m/s^2

5 0

2 years ago

Please help me with this question

Oksanka [162]

Car at rest:

velocity= 0m/s

Acceleration:

0.2m/s²

Since total time:

3 min = 180s

Formula of acceleration:

acceleration = [final velocity - initial velocity] ÷ [total time]

Velocity at end:

0.2m/s² = [final velocity - 0m/s] ÷ [180s]

0.2m/s² × 180s = [final velocity]

[final velocity] = 36m/s

Distance travelled:

Velocity = displacement(distance) ÷ time

36m/s = displacement(distance) ÷ 180s

displacement(distance) = 36m/s × 180s

displacement(distance) = 6480m

Hey I'm sorry but i do not understand why the answer on your worksheet for distance travelled is 3240m... its half of what my answer is...

6 0

2 years ago

after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

2 years ago

I need help on 2/3 please asap thx ♥️

ch4aika [34]

questions 2

F=m

therefore

a=F/m

a=408/68

=6m/s^2.

5 0

3 years ago