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4 years ago

The electromagnet in a galavanometer.

Physics

1 answer:

hoa [83]4 years ago

8 0

<span>The electromagnet in a Galvanometer moves a pointer
along a numbered scale in response to a current. </span>

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Please indicate how long each bar is in centimeters and millimeters.

Maru [420]

Answer:

1) 74 cm or 740 mm

2) 33 cm or 330 mm

3) 76 cm or 760 mm

4) 17 cm or 170 mm

brainliest !?!?

6 0

3 years ago

How much force must be applied for a 150.W motor to keep a package moving at 3.00m/s?

leva [86]

The force must be applied for motor to keep the package moving is 50 N.

<h3>What is force?</h3>

Force is the action of push or pull the object in order to make it move or stop.

Power is related to force and velocity as

P = F x v

150 W = F x 3 m/s

F =50 N

Thus, the force must be applied for motor to keep the package moving is 50 N.

Learn more about force.

brainly.com/question/13191643

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2 years ago

You have 2 minutes to get to PE from science class before you get a tardy. If PE is 100m away and you walk at a speed of 1.1m/s

mart [117]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

If are spaced closely together on the map,there is a drastic temperature change over the distance

AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

6 0

3 years ago

Read 2 more answers

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

7 0

4 years ago