The sun is important to is because of the heat it provides. The sun is just one of _______

Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?

Licemer1 [7]

Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

8 0

3 years ago

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

Explanation:

Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

$\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$ ......(1)

Here $\lambda_{s}$ is source wavelength, $\lambda_{o}$ is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, $\lambda_{s}$ = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, $\lambda_{0}$ = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

$555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$

$\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84$

${\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056$

$1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}$

$\frac{v}{c}=\frac{0.2944}{8.056}$

$v = 0.036c=0.036\times3\times10^{8}$

v = 1.08 x 10⁷ m/s

8 0

3 years ago

Which of these factors best explains why liquid water exists on earth ?

Sveta_85 [38]

Answer:

earth strong gravity

Explanation:

The only planet that have water bodies is earth. We know that water is one of the important factors for all living organism. Liquid water exists on the earth's surface is due to the atmospheric pressure. The reason behind this is not completely mentioned.

There are so many hypothesis about it. Out of given options the correct option is earth's strong gravity.

Hence, the correct option is (b) " earth strong gravity ".

3 0

2 years ago

Read 2 more answers

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

Santa doesn't want to push a 25.0 kg wooden box across a wooden floor with a uk of 0.20 at

scZoUnD [109]

Answer:

49 N

Explanation:

In order to move the box at constant speed, the acceleration of the box must be zero (a=0): this means, according to Newton's second law,

F = ma