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Tasya [4]
1 year ago
11

Auto companies frequently test the safety of automobiles by putting themthrough crash tests to observe the integrity of the pass

enger compartment.If a 1000-kg car is sent toward a cement wall with a speed of 14 m/s and theimpact brings it to a stop in 8.00 x 10-2 s, with what average force is it,brought to rest?
Physics
1 answer:
e-lub [12.9K]1 year ago
4 0
Answer:

The average force that brings the car to rest is 175000N

Explanation:

The mass of the car, m = 1000 kg

Initial speed, u = 14 m/s

Final speed, v = 0 m/s (Since the car comes to a stop)

The time taken, t = 8 x 10^-2 s

The average force is calculated as:

\begin{gathered} F=\frac{m(u-v)}{t} \\ F=\frac{1000(14-0)}{8\times10^{-2}} \\ F=\frac{14000}{0.08} \\ F=175000N \end{gathered}

The average force that brings the car to rest is 175000N

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Part C: Quantitative Problems when vf is not 0
Alina [70]

Answer:

(a)

\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s

(b)

1120 N

Explanation:

Change in velocity, \triangle v is given by subtracting the initial velocity from the final velocity and expressed as \triangle v= v_f -v_i

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

\triangle v=0-8=-8\ m/s

To find m\triangle v then we substitute 7 kg for m and -8 m/s for \triangle v therefore \triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}

Substituting t with 0.05 s then a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

3 0
3 years ago
First to answer will be the brainliest i need the answer ASAP
slavikrds [6]
The answer is chemical.
5 0
3 years ago
In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5220 km away, and released. It found its wa
Korolek [52]

Answer:

4.6834625323 m/s

0 m/s

Explanation:

s = Displacement

t = Time

Velocity is given by

v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{5220000}{12.9\times 24\times 60\times 60}\\\Rightarrow v=4.6834625323\ m/s

The bird's average velocity for the return flight is 4.6834625323 m/s

In the whole episode the bird went 5220 km away from its nest and came back. This means the displacement is zero.

Hence, the average velocity for the whole episode is 0 m/s

7 0
2 years ago
Read 2 more answers
Arocket launches at an angle of 33.6 degrees from the horizontal at a
babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, \theta=33.6

Initial velocity of the rocket is, u=58.5

A vector at an angle \theta with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle \theta with the horizontal, then the horizontal and vertical components are given as:

A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

u_y=u\sin \theta

Plug in the given values and solve for u_y. This gives,

u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})

Therefore, the Y component of initial velocity is 32.37.

4 0
2 years ago
The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction bet
kirill [66]
The answer would be 54 m/s as the maximum speed
8 0
3 years ago
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