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3 years ago

Why is the earth's magnetic field important for us?

1 answer:

8 0

It protects us from the magnetic/electrical radiation that comes from the sun. High radiation periods coincide with solar storms.

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2. A woman prevents a 3kg brick from falling by pressing it against a vertical wall. The coefficient of friction

Anettt [7]

<h3>Answer:</h3>

49 N

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of the brick as 3 kg
The coefficient of friction as 0.6

We are required to determine the force that must be applied by the woman so the brick does not fall.

We need to importantly note that;
For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.

That is; Friction force = Mg

But; Friction force = μ F

Therefore;

μ F = mg

0.6 F = 3 × 9.8

0.6 F = 29.4

F = 49 N

Therefore, she must use a force of 49 N

6 0

3 years ago

A 1500 kg car has the same momentum as a 2500 kg truck moving at 25 m/s. How fast is the car moving in m/s

Alenkinab [10]

Answer:

41.6m/s

Explanation:

P=mv

2500kg × 25m/s = 62500kgm/s

62500kgm/s ÷ 1500 kg = 41.6m/s

5 0

3 years ago

A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.

nikklg [1K]

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R$

where $m=4.10\,\rm kg$ , $v=2.85\frac{\rm m}{\rm s}$ , and $R$ is the radius of the circular path.

As shown in the diagram, we can see that

$\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)$

where $r=1.69\,\rm m$ , so that

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}$

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

$F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}$

Solve for $\theta$ :

$\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0$

Complete the square:

$\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}$