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Alex Ar [27]
3 years ago
8

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car?

Physics
1 answer:
aksik [14]3 years ago
3 0

ANSWER


C. F=4.5 \times10^3. newtons


EXPLANATION


According to Newton's second law,



F_{net}=ma, where



m=1.00\times 10^3kg is the mass measured in kilograms.


and


a=4.5ms^{2} is the acceleration in metres per second square.



We substitute these values to obtain,


F=1.00\times10^3 \times 4.5.



We rearrange to get,


F=1.00\times4.5 \times10^3.


We multiply out the first two numbers and leave our answer in standard form to get,



F=4.5 \times10^3 N.



The correct answer is C


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A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0
3 years ago
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

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The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

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μFs = F

where;

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μ = F/Fs

μ = F/(mgcosθ)

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μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

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