Can an object have both kinetic energy and gravitational potential energy? Explain.

A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of

Damm [24]

You have to reduce 2.00 an5.00 I order to use the×that=0.800

3 0

3 years ago

A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g

dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

$f = \frac{4.04}{60} =0.067 rev/s$

here we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(0.067)$

$\omega = 0.42 rad/s$

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0

3 years ago

A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that

NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

$\tau=NIAB\sin\theta$

B is magnetic field

$B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T$

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

6 0

3 years ago

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

cluponka [151]

Answer:

The thickness is $\Delta y = 2.4 *10^{-6} \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 480 \ nm = 480*10^{-9} \ m$

The first order of the dark fringe is $m_1 = 16$

The second order of dark fringe considered is $m_2 = 6$

Generally the condition for destructive interference is mathematically represented as

$y = \frac{m \lambda}{2}$

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

$y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=> $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

=> $y_1 - y_2 = \Delta y = 5 (480*10^{-9})$

=> $\Delta y = 2.4 *10^{-6} \ m$

8 0

3 years ago

What if the Earth was moved out to Jupiter's orbit which is about 5

Stels [109]

Answer:

25.0 less force

Explanation:

4 0

3 years ago