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riadik2000 [5.3K]
3 years ago
12

A driver hits his brakes and stops a car that was traveling at 96 km/hr in 3.5 s. what was the force applied to the car if its m

ass is 2100 kg?
Physics
1 answer:
vagabundo [1.1K]3 years ago
6 0

This cab be solve using Newtons second law of motion :

F = ma

Where F is the force applied

A is the accelaration = v/s

V is the velosity

S is the time

M is the mass

F = (2100 kg)*( 96km/hr)( 1000m/ 1km)( 1 hr/3600 s) / 3.5 s

F = 16000 N is the force applied

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A longitudinal wave transports energy through the medium without permanently transporting matter.


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Force is applied to an object and the object is moved over a distance in the same direction of the applied force" is the definit
just olya [345]

Answer: The correct answer is "work".

Explanation:

The expression for the work done in terms of force and displacement is as follows;

W= Fd

Here, W is the work done, F is the applied force and d is the displacement.

If the force is applied to an object and the object is displaced to some distance then the work is done.

It is given in the problem that force is applied to an object and the object is moved over a distance in the same direction of the applied force". The given statement is the definition of work. The work done can be zero, negative and positive. In the given case, the positive work is said to be done as the direction of the applied force and the distance moved by the object are same.

Therefore, the correct option is (d).

7 0
3 years ago
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Explain two ways in which changes on the earth’s surface are connected to changes below the earth’s surface.
FinnZ [79.3K]

Answer:

The Earth's surfaces changes in slow processes, erosion and weathering, and some changes are due to rapid processes, such as: landslides, volcanic eruptions, tsunamis and earthquakes.

Explanation:

4 0
4 years ago
A merry-go-round accelerating uniformly from rest achieves itsoperating speed of 2.5rpm in five revolution.
g100num [7]

Answer:

(A) The magnitude of its angular acceleration 1.09 x 10⁻³ rad/s²

(B) Time of motion 240.2 s

Explanation:

Given;

final angular speed, ωf = 2.5 RPM

angular distance, θ = 5 rev = 5 x 2π = 10π rad

initial angular speed, ωi = 0

final angular speed in rad/s;

\omega_f = \frac{2.5 \ rev}{min} \times \ \frac{2\pi}{1 \ rev} \times \ \frac{1 \min}{60 s} = 0.2618 \ rad/s \\

(A) the magnitude of its angular acceleration(rad/s^2);

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\(0.2618)^2 = 0 + (2\times 10\pi)\alpha\\\\0.0685 = 20\pi \alpha\\\\\alpha = \frac{0.0685}{20\pi} \\\\\alpha = 1.09\times 10^{-3} \ rad/s^2

(B) Time of motion;

\omega_f = \omega_i + \alpha t\\\\0.2618 = 0 + 1.09\times 10^{-3} t\\\\t = \frac{0.2618}{1.09\times 10^{-3}} \\\\t = 240.2 \ s

7 0
3 years ago
WILL GIVE BRAINLIEST
Vladimir [108]

Answer:

2\:\mathrm{m/s^2}

From Newton's 2nd Law, we have \Sigma F=ma. Substituting given values, we have:

40=20a,\\a=\frac{40}{20}=\boxed{2\:\mathrm{m/s^2}}

7 0
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