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2 years ago

Why doesn't a transformer work with direct current? why is ac required?

1 answer:

4 0

The electric field depends not on the magnetic field, but change in magnetic field. if the magnetic field is constant. electric field will be zero.

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Why does a 10 gram piece if copper take up more space than a 10 gram piece of gold

algol [13]

Answer:

because the mass of the copper is higher than the mass of the gold.

Explanation:

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3 years ago

Is it true or false that air pollution only occurs as a result of human activity

gregori [183]

Hi alexander it is very true air pollution is only caused by human activity

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3 years ago

Read 2 more answers

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

$r_{cm}=[12.73,12.73]cm$

Explanation:

The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

Any differential of mass can be calculated as:

$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

$X_{cm}=2*a/\pi=12.73cm$

$Y_{cm}=2*a/\pi=12.73cm$

Therefore, the position of the center of mass is:

$r_{cm}=[12.73,12.73]cm$

5 0

2 years ago

Pushing a heavy sofa across a carpeted room is an example of

Diano4ka-milaya [45]

Force is the answer so

4 0

3 years ago

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

Scilla [17]

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

now we will have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}$

$KE = mv^2$

now by work energy theorem we can say

$W = KE_f - KE_i$

$842 J = mv_f^2 - mv_i^2$

$842 = 3(v_f^2) - 3\times 11^2$

now solve for final speed

$v_f = 20 m/s$

3 0

2 years ago