what is the gravitational potential energy of a 150-kg object that is suspended 5-m above the earth's surface

I WILL MARK BRAINLIEST IF CORRECT!!!

julsineya [31]

0.05*0.2=0.05*0.15+0.015*m2',

m2'=1/6 m/s,

m2'=0.17 m/s

N I C E - D A Y!

6 0

3 years ago

5. Three children are at work. One is pushing a chair, another is pushing the sofa and the third one is pushing an empty trolley

katen-ka-za [31]

Answer:

sofa

Explanation:

because the sofa is heavy and the amount of friction is high due to the amount of opposite friction is pushing it back

4 0

3 years ago

Please help me with this and if I have good answer u will be marked as brilliant !!!!

AysviL [449]

Number one- numbers of items sold. Number two- Thursday and Friday. Number three- 1,200. Number four-150

3 0

3 years ago

Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of

mart [117]

Answer:

The value is $B = 3.33 *10^{-6} \ T$

Explanation:

From the question we are told that

The distance of separation is $d = 0.6 \ m$

The current on the one wire is $I_1 = 9 \ A$

The current on the second wire is $I_2 = 4 \ A$

Generally the magnitude of the field exerted between the current carrying wire is

$B = B_1 - B_2$

Here $B_1$ is the magnetic field due to the first wire which is mathematically represented as

$B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}$

Here $d_1$ is the distance to the half way point of the separation and the value is

$d_1 = 0.3 \ m$

$B_2$ is the magnetic field due to the first wire which is mathematically represented as

$B_2 = \frac{\mu_o * I_2 }{2 \pi * d_2}$

Here $d_2$ is the distance to the half way point of the separation and the value is

$d_2 = 0.3 \ m$

This means that $d_1 = d_2 = a = 0.3$

So

$B = \frac{\mu_o * I_1 }{2 \pi * d_1} - \frac{\mu_o * I_2 }{2 \pi * d_2}$

=> $B = \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }$

=> $B = \frac{ 4\pi * 10^{-7} * (9- 4)}{2 * 3.142 *0.3 }$

=> $B = 3.33 *10^{-6} \ T$

5 0

3 years ago

A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the

blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0

3 years ago