A 3.5 Newton force (directed at a 30 degree angle with the vertical) to

If a photon has a frequency of 1.15 × 1015 hertz, what is the energy of the photon? Given: Planck's constant is 6.63 × 10-34 jou

Vika [28.1K]

5.82 x 10-49 joules7.62 x 10-19 joules8.77 x 10-12 joules1.09 x 10-12 joules answer is b

8 0

3 years ago

Read 2 more answers

Infer why the output force exerted by a rake must be less than input force?

adell [148]

<h3>Answer and Explanation;</h3>

input force refers to the force exerted on a machine, also known as the effort, while the output force is the force machines produce or the Load. The ratio of output force to input force gives the mechanical advantage of a simple machine
The output force exerted by the rake must be less than the input force because one has to use force while raking. The force used to move the rake is the input force. 
The rake is not going to be able to convert all of the input force into output force, the force the rake applies to move the leaves, because of friction.

5 0

4 years ago

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

Bezzdna [24]

Answer:

Train Bike Truck

Explanation:

3 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

3 years ago

If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co

sergij07 [2.7K]

Answer:

Angular velocity, $\omega=35.36\ rad/s$

Explanation:

It is given that,

Maximum emf generated in the coil, $\epsilon=20\ V$

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, $B=9\times 10^{-3}\ T$

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

$\epsilon=NBA\omega$

$\omega=\dfrac{\epsilon}{NBA}$

$\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}$

$\omega=35.36\ rad/s$

So, the angular velocity of the circular coil is 35.36 rad/s. Hence, this is the required solution.

6 0

3 years ago