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3 years ago

If an object is accelerating, can the net force acting on it be zero?

Physics

2 answers:

kaheart [24]3 years ago

8 0

No, because if the net force were 0, it would either be at rest or moving in constant velocity.

Ulleksa [173]3 years ago

5 0

Answer:

No, the net force cannot be zero.

Explanation:

From newton's second law,

ΣFnet = ma

Since a is accelerating I.e it is change velocity with respect to time taken, since acceleration can only occur when the force acting on the body is not balanced.

Then Fnet cannot be zero.

Now, if Fnet is zero

Fnet = ma=0

It shows that, the acceleration is zero. So since the body is accelerating, then the net force can not be zero

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A)If the rms value of the electric field in an electromagnetic wave is doubled, by what factor does the rms value of the magneti

Ghella [55]

A) The magnetic field doubles as well

The relationship between rms value of the electric field and rms value of the magnetic field for an electromagnetic wave is the following:

$E_{rms}=cB_{rms}$

where

E_rms is the magnitude of the electric field

c is the speed of light

B_rms is the magnitude of the magnetic field

From the equation, we see that the electric field and the magnetic field are directly proportional: therefore, if the rms value of the electric field is doubled, then the rms value of the magnetic field will double as well.

B) The intensity will quadruple

The intensity of an electromagnetic wave is given by

$I=\frac{1}{2}c\epsilon_0 E_0^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

$E_0$ is the peak intensity of the electric field

The rms value of the electric field is related to the peak value by

$E_{rms}=\frac{E_0}{\sqrt{2}}$

So we can rewrite the equation for the intensity as

$I=c\epsilon_0 E_{rms}^2$

we see that the intensity is proportional to the square of the rms value of the electric field: therefore, if the rms value of the electric field is doubled, the average intensity of the wave will quadruple.

7 0

3 years ago

What is the power generation in a machine that produces 76 j in 3.7s

ICE Princess25 [194]

Power is the ratio between energy and time:
$P= \frac{E}{t}$
In our problem we have E=76 J and t=3.7 s. Therefore, the power is
$P= \frac{76 J}{3.7 s} =20.5 W$

3 0

3 years ago

Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a dista

Gennadij [26K]

Answer:

The speed with which the skier was going is approximately 2.9906 m/s

Explanation:

The given parameters are;

The distance the skier slides before coming to rest, s = 12.4 m

The coefficient of friction between the skier and the snow, $\mu _k$ = 0.0368

Therefore, for conservation of energy, we have;

Initial kinetic energy = Work done by the kinetic friction force

Initial kinetic energy = 1/2·m·v²

The work done by the kinetic friction force = $\mu _k$ ×m×g×s

Where;

m = The mass of the skier

v = The speed with which the skier was going

g = The acceleration due to gravity = 9.8 m/s²

s = The distance the skier slides before coming to rest = 12.4 m

$\mu _k$ = The kinematic friction = 0.0368

Therefore, for conservation of energy, we have;

1/2·m·v² = $\mu _k$ ×m×g×s

1/2·v² = $\mu _k$ ×g×s

v² = 2× $\mu _k$ ×g×s = 2 × 0.0368 × 9.8 × 12.4 = 8.943872

v = √(8.943872) ≈ 2.99063070271 ≈ 2.9906 m/s

The speed with which the skier was going = v ≈ 2.9906 m/s.

3 0

3 years ago

True or False: There are many different types of leukocytes. True False

Fantom [35]

the answer is a.) true

6 0

3 years ago

A pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s a

arsen [322]

Answer:

The apparent weight is 5 times greater than the original weight at the bottom.

Explanation:

Given:

Mass of the pilot, m = 84 kg

Velocity of the jet, v = 345 m/s

Radius of the loop, R = 3.033 km = 3.033 * 10^3 m

We have to find the apparent weight that the pilot feels.

Let the apparent weight be "N" .

Apparent weight :

It is based on where is the position of the pilot in the loop-the-loop.
The apparent weight is the highest at the bottom of the loop-the-loop.
Because the weight acts down and the normal force acts towards the center of the circle.

From the FBD shown we can say that :

apparent weight (N)

⇒ $N=mg+\frac{mv^2}{R}$