The type of function that describes the amplitude of damped oscillatory motion is _. The type of function that describes the amplitude of damped oscillatory motion is _. quadratic sinusoidal inverse exponential linear

Answers

Answer 1

Answer:

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × $e^{(-bt)/(2m)}$ × cos(ωt + ∅ ) ..................................... ( 1 )

here function A × $e^{(-bt)/(2m)}$ is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

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Each plate of a parallel‑plate capacitor is a square of side 0.0479 m, and the plates are separated by 0.479 × 10 − 3 m. The capacitor is charged and stores 8.11 × 10 − 9 J of energy. Find the electric field strength E inside the capacitor.

Answers

Explanation:

It is known that the relation between electric field and potential is as follows.

E = $(V)/(d)$

And, formula to calculate the capacitance is as follows.

C = $(\epsilon_(o) A)/(d)$

= $(8.854 * 10^(-12) * (0.479 m)^(2))/(0.479 * 10^(-3))$

= $4.24 * 10^(-9)$ F

Hence, energy stored in a capacitor is as follows.

W = $(1)/(2)CV^(2)$

V = $\sqrt{(2W)/(C)}$

E = $\sqrt{(2W)/(d^(2)C)}$

= $(2 * 8.11 * 10^(-9) J)/((0.479 * 10^(-3))^(2) * 4.24 * 10^(-9))$

= $16.687 * 10^(3) N/C$

Thus, we can conclude that electric field strength E inside the capacitor is $16.687 * 10^(3) N/C$ .

An electron traveling horizontally to the right enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the electric field?

Answers

Answer:

Upward

Explanation:

For charged particles immersed in an electric field:

- if the particle is positively charged, the direction of the force is the same as the direction of the electric field

- if the particle is negatively charged, the direction of the force is opposite to the direction of the electric field

In this problem, we have an electron - so a negatively charged particle - so the direction of the force is opposite to that of the electric field.

Since the electric field is directed downward, therefore, the electric force on the electron will be upward.

A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

Answers

Answer:

(a) Fw = 101.01 N

(b) W = 282.82 J

(d) N = 368.61 N

(e) Net force = 0 N

Explanation:

(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:

$F_g-F_w=0$ (1)

Fg: gravitational force over the object

Fw: worker's force

However, in an incline you have that the gravitational force on the object, due to its weight, is given by:

$F_g=Wsin\theta=Mg sin\theta$ (2)

M: mass of the ice block = 39 kg

g: gravitational constant = 9.8m/s^2

θ: angle of the incline

You calculate the angle by using the information about the distance of the incline and its height, as follow:

$sin\theta=(0.74m)/(2.8m)=0.264\n\n\theta=sin^(-1)(0.264)=15.32\°$

Finally, you solve the equation (1) for Fw and replace the values of all parameters:

$F_w=F_g=Mgsin\theta\n\nF_w=(39kg)(9.8m/s^2)sin(15.32\°)=101.01N$

The worker's force is 101.01N

(b) The work done by the worker is given by:

$W=F_wd=(101.01N)(2.8m)=282.82J$

(c) The gravitational force on the block is, without taking into account the rotated system for the incline, only the weight of the ice block:

$F_g=Mg=(39kg)(9.8m/s^2)=382.2N$

The gravitational force is 382.2N

(d) The normal force is:

$N=Mgcos\theta=(39kg)(9.8m/s^2)cos(15.32\°)=368.61N$

(e) The speed of the block when it slides down the incle is constant, then, by the Newton second law you can conclude that the net force is zero.

What happens to a black body radiator as it increases in temperature? A. it gives off a range of electromagnetic radiation of shorter wavelengths.
B. It gives off only one wavelength of electromagnetic radiation
C. It releases only ultraviolet waves of electromagnetic radiation
D. It becomes hotter but gives off less electromagnetic radiation

Answers

The black body radiator as it increases in temperature gives off a range of electromagnetic radiation of shorter wavelengths so, the option A is correct.

What is radiation?

Radiation is the movement of atomic and subatomic particles as well as waves, such as those that define X-rays, heat rays, and light rays. Radiation of both types, from cosmic and earthly sources, is constantly being thrown at all matter.

The characteristics and behavior of radiation, as well as the matter it interacts with, are outlined in this article, which also explains how energy is transferred from radiation to its surroundings.

The effects of such an energy transfer to living matter, including the typical effects on numerous biological processes, are given a great deal of attention (e.g., photosynthesis in plants and vision in animals).

Thus, the black body radiator gives off a range of electromagnetic radiation of shorter wavelengths.

To know more about radiation:

brainly.com/question/29333363

#SPJ2

Answer: A

Explanation:

Answer is a hope this helps guys!

_ is the name given to the heat energy received from the sun

Answers

Answer:

The think the answer is solar radiation.

Explanation:

here, we gain the heat from the sun through a radiation. When it travels from the sun the harmful radiation are absorbed by ozone layer and heat enegry is provided to the surface of the Earth.

hopeit helps..

A 4.0-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70 kg construction worker stands at the far end of the beam.What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?

Answers

Answer:

T=12544 N*m

Explanation:

Given

L=4.0m

ms=500kg

mw=70kg

Torque is the force in a distance the relation is proportional so the torque of weight first is:

Ts = Fs*d

Ts = ms*g*L

Ts = 500kg*9.8m/s^2*2m

Ts = 9800 N*m

now torque of the worker

Tw = Fw*d

Tw = 70kg*9.8m/s^2*4m

Tw = 2744 N*m

Torque net is

Tnet = Tw+Ts

Tnet= 2744 + 9800 =12544 N*m

Final answer:

The total torque about the bolt due to the worker and the weight of the beam is 12544 Nm. This is found by adding the torque due to the beam and the worker which can be calculated using their weights and their distance from the pivot point (bolt).

Explanation:

The key to solving this question is understanding torque, which in physics represents the rotational effect of a force. Torque is calculated using the formula τ = r x F, where τ is the torque, r is the distance from the pivot point, and F is the force applied.

In this case, there are two forces to consider: the weight of the beam and the weight of the worker. Both of these can be calculated using the formula for weight (F = m*g), where m is mass and g is gravitational acceleration, which is approximately 9.8 m/s^2 on Earth. The weight of the beam is therefore 500 kg * 9.8 m/s^2 = 4900 N, and the weight of the worker is 70 kg * 9.8 m/s^2 = 686 N.

The distance from the pivot (bolt) for the beam's weight is considered to be the midpoint of the beam, so it is 4.0 m / 2 = 2.0 m. For the worker, r equals the full length of the beam, which is 4.0 m. The total torque can be calculated by adding the torque due to the beam and the worker. Therefore, the total torque τ = (2.0 m * 4900 N) + (4.0 m * 686 N) = 9800 Nm + 2744 Nm = 12544 Nm.