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Ivahew [28]
3 years ago
6

Which describes the force between two magnets when their south poles are almost touching?

Physics
2 answers:
Oliga [24]3 years ago
7 0
<span>Which describes the force between two magnets when their south poles are almost touching?

Answer: C, the poles strongly repel each other.

HOPE THIS HELPS! ^_^</span>
almond37 [142]3 years ago
3 0
The poles strongly repel each other
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Most earthquakes occur along or near the edges of the earth's
Pepsi [2]
Most earthquakes occur along or near the edges of the earth's lithospheric<span> plate. </span>
8 0
4 years ago
A force of 3kN acts on a car to make it accelerate by 1.5m/s/s. What is the mass of the car?
Masja [62]

Answer:

2

Explanation:

To find force it's force = mass times acceleration so to find mass you would divide force by acceleration

5 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
4 years ago
65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!
otez555 [7]
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.



3 0
3 years ago
The direction of an electric field is the direction (5 points)
vlabodo [156]

The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field  Therefore the answer is b. a positive charge will move in the field.

5 0
3 years ago
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