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matrenka [14]
2 years ago
14

A box is against a wall. A person pushes on the box, but the box does not move. Is this

Physics
2 answers:
Serhud [2]2 years ago
7 0

Answer:

it is a force

Explanation:

a force is a push or a pull

Levart [38]2 years ago
4 0

Answer:

Yes, this is an example of a force because a force is a push or pull

Explanation:

Since the person is adding force to push the box.

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Which is the best example of the transfer of heat through radiation
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Nitially the flame produces radiation<span> which heats the tin can. The tin can then</span>transfers heat<span> to the water </span>through<span> conduction. The hot water then rises to the top, in the convection process. </span>
7 0
3 years ago
A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf
makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

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6 0
3 years ago
Which are functions of proteins?
NikAS [45]
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5 0
3 years ago
Read 2 more answers
An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can a
kipiarov [429]

Answer:

The longest wavelength of light is 209 nm.

Explanation:

Given that,

Spring constant = 74 N/m

Mass of electron m= 9.11\times10^{-31}\ kg

Speed of light c= 3\times10^{8}\ m/s

We need to calculate the frequency

Using formula of frequency

f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Where, k= spring constant

m = mass of the particle

Put the value into the formula

f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}

f=1.434\times10^{15}\ Hz

We need to calculate the longest wavelength that the electron  can absorb

\lambda=\dfrac{c}{f}

Where, c = speed of light

f = frequency

Put the value into the formula

\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}

\lambda=2.092\times10^{-7}\ m

\lambda=209\ nm

Hence, The longest wavelength of light is 209 nm.

6 0
3 years ago
A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0
3 years ago
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