All categories

3 years ago

When is work performed?

1 answer:

7 0

Answer: In summary, work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement.

Explanation:

You might be interested in

What area of science compares and describes quantities and movements of matter and energy?

qaws [65]

Answer:

the answer is physics

4 0

4 years ago

A traveling electromagnetic wave in a vacuum has an electric field amplitude of 93.3 V/m. Calculate the intensity S of this wave

7nadin3 [17]

Answer:

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Explanation:

The expression for the intensity of the electromagnetic wave is,

$I = \frac{1}{2} C{ {\varepsilon _0}E_m^2$

Here, $\varepsilon _0$ is the permittivity of the free space,

$E_m$ is the electric field amplitude and

c is the speed of the light.

substitute

⁸m/s for c

8.85×10 −12 C² /N⋅m² for ${\varepsilon _0}$

and 93.3 V/m for ${E_{\rm{m}$

$I = \frac{1}{2} \times (3\times10^8)\times(8.85\times10^-^1^2)(93.3)\\\\I = 11.56W/m^2$

The expression for the energy is,

E = I×A×t

Here, I is the intensity of the electromagnetic wave,

A is the area, and

t is the time.

Substitute

11.56W/m² for I

0.0287m ² for A

13.7s for t

$E = (11.56)\times(0,0287)\times(13.7)\\E = 4.55J$

The energy flowing through the given area is 4.55 J

5 0

4 years ago

A certain CD has a playing time of 74.0 minutes. When the music starts, the CD is rotating at an angular speed of 480 revolution

Rashid [163]

Answer: $0.00636\ rad/s^2$

Explanation:

Given

CD has a playing time of $t=74\ min\ or\ 74\times 60\ s$

Initial angular speed of CD is $480\ rpm$

Final angular speed of DC is $210\ rpm$

Angular speed, when rpm is given

$\omega =\dfrac{2\pi N}{60}$

$\omega_i=\dfrac{2\pi \times 480}{60}\\\\\Rightarrow \omega_i=16\pi \ rad/s$

Final speed

$\Rightarrow \omega_f=\dfrac{2\pi \times 210}{60}\\\\\Rightarrow \omega_f=7\pi \ rad/s$

Using equation of angular motion

$\Rightarrow \omega_f=\omega_i+\alpha t$

Insert the values

$\Rightarrow 7\pi =16\pi +\alpha \times 74\times 60\\\Rightarrow -9\pi =\alpha \cdot (4440)\\\\\Rightarrow \alpha=-\dfrac{9\pi}{4440}\\\\\Rightarrow \alpha=-0.00636\ rad/s^2$

Magnitude of angular acceleration is $0.00636\ rad/s^2$

3 0

3 years ago

A 2800 kg truck moving at 12 m/s to the right hits a stopped 1100 kg car. What is the combined velocity the moment they stick to

leva [86]

Answer:

The combined velocity is 8.61 m/s.

Explanation:

Given that,

The mass of a truck, m = 2800 kg

Initial speed of truck, u = 12 m/s

The mass of a car, m' = 1100 kg

Initial speed of the car, u' = 0

We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.

$m_1v_1+m_2v_2=(m_1+m_2)V\\\\V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\V=\dfrac{2800\times 12+0}{2800+1100}\\\\V=8.61\ m/s$

So, the combined velocity is 8.61 m/s.

5 0

3 years ago

An electron is released from rest in a unifor electric field and accelerates to the north at a rate of 145 m/s^2. What is the ma

julsineya [31]

Answer:

E = 8.26*10⁻¹⁰ N/C, due south.

Explanation:

Assuming no other forces acting on the electron than the electrostatic force due to the electric field, we can apply Newton's 2nd law as follows:

$F = -eE =ma (1)$

Solving for E, we can find its magnitude as follows:

$E =\frac{m*a}{e} = \frac{9.1e-31 kg*145m/s2}{1.6e-19C} = 8.26e-10 N/C (1)$

The direction of the electric field is by definition the one that would take a positive test charge, so if the electron is accelerated to the north, the electric field would exactly oppose to this direction, so it is directed due south.

7 0

3 years ago