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umka21 [38]
3 years ago
6

With a bar magnet where are the lines of force closest together

Physics
2 answers:
attashe74 [19]3 years ago
8 0
At the tip of either of the magnets poles
MA_775_DIABLO [31]3 years ago
3 0

Answer: at the tips of the magnet

Explanation: charges are more at the tip of a substance so likewise for a magnet

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A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
Bumek [7]

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

v = d/t .....1

For each of the cases, equation 1 would be used to calculate the velocity.

a) v1 = d1/t1

d1= 16m

t1 = 2.9s

v1 = 16m/2.9s

v1 = 5.52m/s

b) v2 = d2/t2

d2 = -2.5m

t2 = 1.65s

v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

v3 = 4.62m/s

d) vt = dt/tt

dt = 16m - 2.5m + 24m = 37.5m

tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

5 0
4 years ago
If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula
snow_lady [41]

Here we know that

\omega_i = - 8.4 rad/s

\alpha = - 2.8 rad/s^2

t = 1.5 s

now from kinematics we have

\omega_f = \omega_i + \alpha t

now from above all values we have

\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)

\omega_f = -8.4 + (-4.2)

\omega_f = -12.6 rad/s

so final angular speed is -12.6 rad/s

6 0
3 years ago
How do objects with the same charger interact
maxonik [38]

The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.

7 0
3 years ago
Read 2 more answers
If you given volume. for example 200 cm³ how can you change it to area m²​
PIT_PIT [208]

Answer:

move the decimal 6 places to the left.

Explanation:

um I assume you meant to say area m^3

7 0
3 years ago
A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc
julsineya [31]

Answer:

15

Explanation:

mass, M = 5Kg

horizontal force, F_h = 40N

acceleration, a =5 m/s^2

frictional force, F_f =?

net force = ma

net force = F_h - F_f = 40N - F_f

40  - F_f = 5 x 5

- F_f = 25 - 40

multiply both side by -1

F_f = 40 - 25 = 15

the frictional force is 15N

4 0
3 years ago
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