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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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John goes grocery shopping with his mother. His job is to push the cart. The cart is
lora16 [44]

Answer:

Beacause he has more grocceries and food heavy

Explanation:

7 0
3 years ago
1.imagine you were able to throw a ball in a frictionless environment such as outer space.once you let go of the ball,what will
Scorpion4ik [409]
Imagine you were able to throw a ball in a frictionless environment
such as outer space.  Once you let go of the ball, it will travel forever
in a straight line, and at a constant speed.  (At least until it bumps into
something.)

A car accelerates down the road.  The reaction to the tires pushing
on the road is the road pushing on the tires.

5 0
3 years ago
The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

6 0
3 years ago
Please don’t troll. I need someone to actually answer these questions.
mash [69]

Answer:

-3+3 i think this is the answer

Explanation:

i think you can ask someone else sorry

4 0
2 years ago
According to the picture, which is the least dense?
alexdok [17]

Answer:

a the chess peice

Explanation:

my head

4 0
3 years ago
Read 2 more answers
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