If argon could exist as a solid, what would best represent the speed through solid argon?

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod

cluponka [151]

Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

Explanation:

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

$F=\dfrac{mv^2}{r}$

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

$F'=\dfrac{2m\times (4v)^2}{4r}\\\\F'=\dfrac{2m\times 16v^2}{4r}\\\\F'=8\dfrac{mv^2}{r}\\\\F'=8F$

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

8 0

3 years ago

The motion of a car on a position time graph is represented with a horizontal tine What does this indicate about the car's motio

GalinKa [24]

Answer:

Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.

The motion of a car on a position-time graph that is represented with a horizontal line indicates that the car has stopped moving.

A straight line with a positive slope indicates that the car is moving at a constant velocity, and thus the slope is constant. On the other hand, a curve with a changing slope, shows that the velocity is changing.

8 0

3 years ago

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force $F_p$ by the worker must be equal to the friction force $F_f$ on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

$F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N$

b. The work is done on the crate by this force is the product of its force $F_p$ and the distance traveled s = 4.35

$W_p = F_ps = 79.1*4.35 = 344 J$

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

$W_f = F_fs = -79.1*4.35 = -344 J$

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

$W_p + W_f = 344 - 344 = 0 J$

8 0

3 years ago

Read 2 more answers

Now, find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 11. Then div

Hunter-Best [27]

Answer

1.0/5

4

IlaMends

Ambitious

2.1K answers

12.9M people helped

Explanation:

When pH of the solution is 11.

..(1)

At pH = 11, the concentration of ions is .

When the pH of the solution is 6.

..(2)

At pH = 6, the concentration of ions is .

On dividing (1) by (2).

The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is .

Difference between the ions at both pH:

This means that Hydrogen ions in a solution at pH = 7 has ions fewer than in a solution at a pH = 6

8 0

3 years ago

Read 2 more answers

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, $A_p = 180 cm^2$

- The charge on each plate, $q = 17 * 10^-^6 C$

- Permittivity of free space, $e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}$

- The radius for the flux region, $r = 3.3 cm$

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

$A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2$

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

σ = $\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\$

σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

$E+ = E- = \frac{sigma}{2*e_o} \\\\$

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

$E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\$

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

Φ = E_net * Ar * cos ( θ )

Φ = (106214689.26553) * (0.00342) * cos ( 5 )

Φ = 361872 N.m^2 / C

5 0

3 years ago