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Lady_Fox [76]
3 years ago
6

ASAP WILL GIVE BRAINLIEST!!!

Physics
1 answer:
Oksana_A [137]3 years ago
5 0

Answer:

<em>The change in momentum of the car is 3575 Kg.m/s</em>

Explanation:

<u>Impulse and Momentum</u>

The impulse (J) experienced by the object equals the change in momentum of the object (Δp).

The formula that represents the above statement is:

J = Δp

The impulse is calculated as

J = F.t

Where F is the applied force and t is the time.

The car hits a wall with a force of F=6500 N and stops in 0.55 s. Thus, the impulse is:

J = 6500 * 0.55

J = 3575 Kg.m/s

The change in momentum of the car is:

\Delta p= J = 3575\ Kg.m/s

The change in momentum of the car is 3575 Kg.m/s

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If an object accelerating at −1.5m/s^2 takes 1.2s to reach 5.0m/s, what was its initial speed?
USPshnik [31]

-1.5 m/s^2 x 1.2 seconds = -1.8 m/s

It is a negative value which means the object slowed down. The object would have originally been going that amount more.


5.0 + 1.8 = 6.8 m/s


answer: 6.8 m/s

3 0
2 years ago
How do consumers sometimes interact with producers?
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In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra
shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0
4 years ago
A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?
poizon [28]

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

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Answer:

a). Single replacement.

Explanation:

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