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TiliK225 [7]
3 years ago
14

A 900 N crate is being pulled across a level floor by a force F of 340 N at an angle of 23° above the horizontal. The coefficien

t of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate.
Physics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

0.958891203 m/s²

Explanation:

N = Weight of crate = 900 N

\mu = Coefficient of friction = 0.25

Force of friction acting on the force applied

f=\mu N\\\Rightarrow f=0.25\times 900\\\Rightarrow f=225\ N

Force used to pull the crate

F=340cos 23\\\Rightarrow F=312.97165\ N

The net force is

F_n=F-f\\\Rightarrow F_n=312.97167-225\\\Rightarrow F_n=87.97167\ N

Acceleration is given by

a=\frac{F_n}{m}\\\Rightarrow a=\dfrac{87.97167}{\dfrac{900}{9.81}}\\\Rightarrow a=0.958891203\ m/s^2

The magnitude of the acceleration of the crate is 0.958891203 m/s²

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nalin [4]

Answer:

13 W/m^2

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

I \propto \frac{1}{r^2}

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} (1)

where in this problem, we have:

I_1 = 1300 W/m^2 apparent brightness at a distance r_1, where

r_1 = 150 million km

We want to estimate the apparent brightness at r_2, where r_2 is ten times r_1, so

r_2 = 10 r_1

Re-arranging eq.(1), we find I_2:

I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2

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3 years ago
When you jump, you exert a pushing force against the ground. Gravity pulls you back down. Why can a person jump higher on the mo
Dennis_Churaev [7]

Answer:

This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher

7 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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A type of energy embodied in oscillating electric and magnetic fields is called
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Answer: Electromagnetic radiation

Explanation:

Electromagnetic radiation is a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

To understand it better:

This radiation is spread thanks to the electromagnetic fields produced by moving electric charges and their sources can be natural or man-made.

It should be noted that the energy of electromagnetic radiation can vary and depending on its frequency it can be useful for various situations.

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