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4 years ago

If a 2 kg ball is traveling at a speed of 4 m/s, what is its kinetic energy?

Physics

2 answers:

Kay [80]4 years ago

8 0

Answer: 16J

Explanation:

K.E = $\frac{1}{2}$ m $v^{2}$

K.E = $\frac{1}{2}$ X 2 X 16

K.E = 16

Art [367]4 years ago

4 0

The kinetic energy of the ball is 16 J.

Answer: Option D

Explanation:

Kinetic energy is the form of energy exhibited by a body when it is in motion. The kinetic energy will be directly proportional to the mass of the body as well as to the square of the speed of the body at which it is moving. So the mathematical representation of kinetic energy of any object is

$\text {Kinetic energy}=\frac{1}{2} \times m \times(v)^{2}$

In the present case, the ball is the object with mass of around 2 kg and it is moving with the speed of about 4 m/s. So, the kinetic energy exhibited by the ball during its motion is

$\text {Kinetic energy of the ball}=\frac{1}{2} \times 2 \times(4)^{2}=16 \mathrm{J}$

So, the kinetic energy exhibited by the ball is 16 J.

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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

$E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

where

$\phi =$ Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

$\lambda =$ wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

$E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

Magnetic field Equation

$B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

Where B₀= E₀/c

$B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T$

$B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

6 0

3 years ago

A small block slides down an incline with a constant acceleration. The block is released from rest at the top of the incline. Af

zloy xaker [14]

Answer:

v = 3.24 m/s

Explanation:

Since we don't have time, we can use the formula;

(Final distance - initial distance)/time = (initial velocity + final velocity)/2

Thus;

(x_f - x_i)/t = ½(v_xi + v_xf)

We are given;

x_i = 0 m

x_f = 5 m

v_xi = 0 m/s

v_xf = 5 m/s

Thus, plugging in the relevant values;

(5 - 0)/t = (0 + 5)/2

5/t = 5/2

t = 2 s

Using Newton's first law of motion, we can find the acceleration.

v = u + at

Applying to this question;

5 = 0 + a(2)

5 = 2a

a = 5/2

a = 2.5 m/s²

To get the speed, in m/s, of the block when it had traveled only 2.1 m from the top, we will use the formula;

v² = u² + 2as

v² = 0² + 2(2.5 × 2.1)

v² = 10.5

v = √10.5

v = 3.24 m/s

3 0

4 years ago

Which of the following pieces of home-made equipment is appropriate for use?

denis-greek [22]

Answer:

your neighbors biycyle inner tubes

Explanation:

hope this helps :)

4 0

2 years ago

A tourist drops a rock from rest from a guard rail overlooking a valley. What is the velocity of the rock at 4.0 s? What is the

3241004551 [841]

At the end of 4.0 s the velocity of the rock is 39.24 m/s and its displacement from the point where it was dropped is 78.48 m.

The rock dropped from the guard rail is a freely falling body and is accelerated downwards under acceleration due to gravity.

To calculate the velocity after t=4.0 s, use the equation of motion,

$v=u+at$

Here, u is the rock's initial velocity, which is zero. It falls under acceleration due to gravity, hence a =g=9.81 m/s².

Therefore,

$v=u+at\\ v=gt\\ =(9.81m/s^2)(4.0s)\\ =39.24m/s$

If it falls through a distance s, then, use the equation,

$s=ut+\frac{1}2} gt^2$

Therefore,

$s=ut+\frac{1}2} gt^2\\ s=\frac{1}2} gt^2\\ =\frac{1}2}(9.81m/s^2)(4.0s)^2\\ =78.48 m$

Thus, after 4.0 s, the rock has a downward velocity of 39.24 m/s and it would have traveled a distance of 78.48m.

7 0

3 years ago

Your town is installing a fountain in the main square. If the water is to rise 28.0 m (91.8 feet) above the fountain, how much p

PilotLPTM [1.2K]

Explanation:

Given that,

The water is to rise to a height of 28 meters. Let P is pressure must the water have as it moves slowly toward the nozzle that sprays it up into the air.

We know that, the atmospheric pressure is : $P_a=10^5\ Pa$

P can be calculated as :

$P=P_a+d\times g\times h$

$P=10^5+1000\times 9.81\times 28$

P = 374680 Pa

So, the pressure of 374680 Pascal must the water have as it moves slowly toward the nozzle that sprays it up into the air. Hence, this is the required solution.

4 0

4 years ago