1. Gwen exerts a 16 N horizontal force as she pulls a 32N sled across a cement

An 8.2 kg object accelerates at 8.0 m/s^2. What is the acceleration?

ZanzabumX [31]

Answer:

8.0m/s²

Explanation:

meters per second squared is a unit of acceleration so 8.0m/s² is the answer

6 0

2 years ago

Starting from rest, a dragster travels a straight 1/4 mi racetrack in 7.10 s with constant acceleration. What is its velocity wh

Gennadij [26K]

268.6567 mph is its velocity when it crosses the finish line

d=(v1+v2 /2) x t

.25=(0+v2 /2) x 6.7/3600 hours

900=v2/2 x 6.7

v2=268.6567 mph as the speed with which the dragster crosses the finish

<h3>When acceleration is not zero, can speed remain constant?</h3>

The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.

Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.

Learn more about velocity refer

brainly.com/question/24681896

#SPJ9

5 0

1 year 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,

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