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3 years ago

An object of mass 3 kg is accelerated at 4 m/s2. What force is acting on the object?

Physics

2 answers:

TEA [102]3 years ago

8 0

Answer: 12 N

Explanation:

maw [93]3 years ago

3 0

Answer:

12 N

Explanation:

Given,

Mass ( m ) = 3 kg

Acceleration ( a ) = 4 m/s^2

To find : Force ( F ) = ?

Formula : -

F = ma

= 3 x 4

= 12 N

Therefore, the force acting on the object is 12 N.

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Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As

Troyanec [42]

Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

Given

Mass = 45000kg

acceleration = 70m/s^2

Substitute

F = 45000 * 70

F = 3,150,000N

Hence the force required to be produced by the rocket engines is 3,150,000N

8 0

3 years ago

What is the ratio of the intensities and amplitudes of an earthquake P wave passing through the Earth and detected at two points

gavmur [86]

For both cases we will use the proportional values of the distance referring to the amplitude and intensity. Theoretically we know that the intensity is inversely proportional to the square of the distance, while the amplitude is inversely proportional to the distance, therefore,

PART A ) Intensity is inversely proportional to the square of the distance

$Intensity \propto \frac{1}{distance^2}$

Therefore the intensity of the two values would be

$\frac{I_{27}}{I_{13}} = \frac{(13km)^2}{(27km)^2}$

$\mathbf{\therefore \frac{I_{27}}{I_{13}} = 0.232 }$

PART B) Amplitude is inversely proportional to the distance

$Amplitude \propto \frac{1}{distance}$

$\frac{A_{27}}{A_{13}}= \frac{(13km)}{(27km)}$

$\mathbf{\therefore\frac{A_{27}}{A_{13}}= 0.4815}$

5 0

3 years ago

Ork is done if a boat has a force 60N and travels 600 meters

Travka [436]

Answer:

This is the answer.

Explanation:

is this right?

3 0

2 years ago

An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 a

Mrrafil [7]

Answer: $283.2\times 10^{-9}\ nC$

Explanation:

Given

Cross-sectional area $A=0.4\ cm^2$

Dielectric constant $k=4$

Dielectric strength $E=2\times 10^8\ V/m$

Distance between capacitors $d=5\ mm$

Maximum charge that can be stored before dielectric breakdown is given by

$\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC$

4 0

3 years ago

Read 2 more answers

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago