Ask question

Ask question

All categories

brilliants [131]

2 years ago

13

A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a friction

less table at 2m/s2, then what is the magnitude of the repulsive force?

1 answer:

makkiz [27]2 years ago

5 0

Answer:

0.1 N

Explanation:

using F = ma yields 0.1 N.

net

You might be interested in

Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver

zalisa [80]

Answer:

Total time taken=110 seconds

Total distance traveled=480m

Explanation:

First of all, we find the total time taken:

For that, we use the formula : Distance/Speed= Time

Time for part 1 : 200/5=40 seconds

Time for part 2 : 280/4=70seconds

Total time taken=110 seconds

Total distance traveled=480m

Average Speed= 480/110=4.36 m/s

Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)

Average Velocity=-80/110=-0.72 m/s

OR 0.72m/s towards west.

3 0

2 years ago

1. In a(n) ______________________ circuit, all parts are connected one after another along one path.

poizon [28]

Answer:

1. A <em>series circuit </em>is a closed circuit which has all loads connected in a row and there is only one path for the current to flow.

2. The <em>Ohm's Law </em>state that a current flow through a resistor is directly proportional to the voltage across it $R = \frac{V}{I}$

3. A <em>parallel circuit </em>is a closed circuit divided into branches that it has two o more paths for the current to flow and the loads are parallel to each other which mean the voltage across them is the same for all loads.

3 0

2 years ago

Read 2 more answers

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

8 0

2 years ago

Leto [7]

True
False
True
My answers

4 0

2 years ago

1) A train accelerates from 36 km/hr to 54 km/hr in 10<br> s. Find acceleration?

Crank

Answer: Given:

Initial velocity= 36km/h=36x5/18=10m/s

Final velocity =54km/h=54x5/18=15m/s

Time =10sec

Acceleration = v-u/ t

=15-10/10=5/10=1/2=0.5 m/s2

Distance =s=?

From second equation of motion:

S=ut +1/2 at^2

=10*10+1/2*0.5*10*10

=100+25

=125m

So distance travelled 125m

Hope it helps you

3 0

2 years ago