If the forces on an object are not balanced than the object must be...

Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e

Anon25 [30]

The block has the greatest average power provided is bock m.

<h3>What is instantaneous power?</h3>

This is the product of force and velocity exerted on an object.

Mathematically instantaneous power is calculated as;

P = Fv

where;

F is the applied force
v is the velocity

Both blocks (m and 2m) will experience the same force but different velocity.

The smaller block (m) will experience greater velocity.

Thus, the block has the greatest average power provided is bock m.

Learn more about instantaneous power here: brainly.com/question/8893970

7 0

2 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

Select the three correct parts that make up the cell theory.

Lana71 [14]

Answer:

Options A, D and E....make up cell theory

8 0

3 years ago

5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)

jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

3 0

3 years ago

The term test variable refers to:

timofeeve [1]

Answer:

answer should be C.) Independent variable since its the one being changed

Explanation:

4 0

3 years ago

Read 2 more answers