A 200-kg boulder is 1000-m above the ground. what is its potential energy?

A skydiver of mass 80kg jumps from a slow moving aircraft and reach a terminal speed of 50 m/s. What's her acceleration when her

miskamm [114]

Answer:

a = g = 9.81[m/s^2]

Explanation:

This problem can be solve using the second law of Newton.

We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.

m*g = m*a

where:

g = gravity = 9.81[m/s^2]

a = acceleration [m/s^2]

Note: If the skydiver will be under air resistance forces his acceleration will be different.

7 0

3 years ago

When several radio telescopes are wired together, the resulting network is called a radio

WINSTONCH [101]

Answer:

Interferometer

Explanation:

4 0

3 years ago

A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the

Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

$W_{r} = |F_{r}|*|d|*cos(\theta_{1})$

Where:

$F_{r}$ : is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

$W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J$

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

$W_{g} = |F_{g}|*|d|*cos(\theta_{2})$

The force of gravity is given by the weight of the bike.

$F_{g} = -mgsin(24)$

And the angle between the force of gravity and the direction of motion is 180°.

$W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})$

$W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J$

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.

I hope it helps you!

3 0

3 years ago

A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the

Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

$w sin\theta = m\lambda$

Where,

w = width

$\lambda =$ wavelength

m is an integer, m = 1, 2, 3...

We here know that as $sin\theta$ as w are constant, then

$\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}$

We need to find $w_2$ , then

$w_2 = \frac{w_1}{\lambda_1}\lambda_2$

Replacing with our values:

$w_2 = \frac{4.4*10^{-6}}{487}496$

$w_2 = 4.48*10^{-6}m$

Therefore the width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

3 0

3 years ago

Two cylindrical resistors are made of the same material and have the same resistance. The resistors, R1 and R2, have different r

Paladinen [302]

Answer:

Option d is correct.

Explanation:

We know , resistance of a body is directly proportional to its length and inversely proportional to its area.

$R=\dfrac{\rho\ L}{A}=\dfrac{\rho\ L}{\pi r^2}$ ( Here, $\rho$ is constant dependent on object material )

Writing $R_1 \ and\ R_2$ also :

$R_1=\dfrac{\rho\ L_1}{\pi r_1^2}\ , \ R_2=\dfrac{\rho\ L_2}{\pi r_2^2}$ ( since they are of same material therefore, $\rho$ is same.)

Now , if $r_2=2r_1 \ and \ 4L_1=L_2$ .

Then $R_1=R_2$

Therefore, option d. is correct.

Hence, this is the required solution.

7 0

3 years ago