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3 years ago

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A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in t

he magnetic field, the iron filings arranged themselves, creating lines of force. How do the energy and the lines of force change when a stronger bar magnet is used

1 answer:

Naya [18.7K]3 years ago

4 0

Answer: 1. The field energy will increase

2. The energy increases, and the lines of force are denser

3. It points toward the field of earths magnetic poles

4. 1 and 2 only

5. 2, 4, 1, 3

Explanation: just took it

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Explanation:Because luster is something you learned in science.

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3 years ago

Can someone pls help with the last one

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3 years ago

A girl sits on a tire swing and is pushed in a circular motion by her father. Her tangential speed is 2.8 m/s and the girl trave

Jet001 [13]

Formula: Ca=Vt^2/r

centripetal acceleration(Ca)= ?
tangential speed(Vt)= 2.8m/s
Radius(r)=2m

Substitute: Ca=2.8^2/2
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4 years ago

Read 2 more answers

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

The tire of your bicycle needs air so you attach a bicycle pump to it and begin to push down on the pump’s handle. If you exert

dusya [7]

Answer:

12.5 J

Explanation:

Force, F = 25 N

Distance, d = 0.5 m

The direction of force and the displacement is same.

Work is defined as the product of force in the direction of displacement and the displacement.

Work = Force x displacement x CosФ

Where, Ф be the angle between force and the displacement

Here, Ф = 0°

So, W = 25 x 0.5 x Cos0°

W = 12.5 J

6 0

4 years ago