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3 years ago

6

PLEASE HELP!!! I have an electromagnet. If I detach the two wires from the battery and reattach them to the opposite terminals,

how would that change the current and magnetic field?

1 answer:

lakkis [162]3 years ago

4 0

Explanation:

hope it helps thanks

pls mark me as brainliest

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HELP ASAP ILL GIVE BRAINLIST

OlgaM077 [116]

Answer:

Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps

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2 years ago

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A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu

ira [324]

Answer:

$-20.0 kg m^2/s$

Explanation:

The angular momentum of an object in rotation is given by

$L=I \omega$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, initially we have

$I=2 kg m^2$ is the moment of inertia of the wheel

$\omega_i = 6.0 rad/s$ is the initial angular speed

So the initial angular momentum is

$L_i = I\omega_i = (2)(6.0)=12 kg m^2/s$

Later, a counterclockwise torque of

$\tau=-5.0 Nm$ is applied

So the angular acceleration of the wheel is:

$\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2$ in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

$\omega_f = \omega_i + \alpha t$

where

t = 4.0 is the time interval

Solving,

$\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s$

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

$L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s$

So, the change in angular momentum is:

$\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2$

7 0

3 years ago

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

Harrizon [31]

Answer:

It will take you 30.8 s to travel the 120 m of the ramp.

Explanation:

Hi there!

The equation for the position of an object moving in a straight line is:

x = x0 + v * t

Where:

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, we will consider the start of the ramp as the origin of our reference system so that x0 = 0.

Now, let´s calculate the speed of the person walking on the ground:

x = v * t

120 m = v * 72 s

v = 120 m / 72 s

v = 1.7 m/s

If you walk on the ramp with that speed, your total speed will be your walking speed plus the speed of the ramp because both are in the same direction. Then, using the equation for the position:

x = v * t

In this case, v = speed of the ramp + walking speed

v = 2.2 m/s + 1.7 m/s = 3.9 m/s

120 m = 3.9 m/s * t

t = 120 m / 3.9 m/s = 30.8 s

It will take you 30.8 s to travel the 120 m

8 0

3 years ago

It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of

nadya68 [22]

Answer:

44 N/m

Explanation:

The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

The work needed to stretch a spring by <em>e</em> is given by

$W = \frac{1}{2} ke^2$

where <em>k</em> is spring constant.

$k = \dfrac{2W}{e^2}$

Using the appropriate values,

$k = \dfrac{2\times 49\text{ J}}{1.5^2\text{ m}^2} = 43.55\ldots\text{ N/m} \approx 44\text{ N/m}$

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3 years ago

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Svetllana [295]

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3 years ago