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3 years ago

What is the area of space around a magnet called?

Physics

1 answer:

erma4kov [3.2K]3 years ago

8 0

Answer:

These acts of attraction and repulsion are called “magnetism”, and the magnetic space around a magnet is called the “magnetic field”

Explanation:

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A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

3 years ago

Which car is harder to stop and why?

kondor19780726 [428]

Answer:

The momentum of bath cars is 40000 Ns which make the difficulty to stop each car in aspect of fprce is the same.

Explanation:

Momentum (P) =mass(m) × velocity (v)

For car A,

P = m × v = 1000 × 40 = 40000 Ns

For car B,

P = m × v = 4000 × 10 = 40000 Ns

Force (F) = Momentum change(ΔΡ)/ time taken(t)

F = ΔΡ/t

When stopping the car the momentum changes from 40000 Ns to 0

So momentum change in both cars is the same. So to stop the two cars in a given time (t) you need the same force, which means you will feel same difficulty.

8 0

3 years ago

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As a ray of light becomes brighter, its corresponding light wave, shown here, will increase in amplitude, making it look MOST li

grandymaker [24]

Answer:

the answer is C

Explanation:

i did the usatestprep

6 0

3 years ago

Read 2 more answers

Should the hypothesis always be correct for a conclusion

dusya [7]

No it shouldn't, a hypothesis doesn't need to be correct but must have an idea for why x variable effects y variable and have good reasoning. In the conclusion you should state if it's correct or not and explain why it's correct/incorrect and what answer you've determined from data.

4 0

3 years ago