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2 years ago

Boyle's Law states that as the pressure on a gas increases at constant temperature the volume of the gas

1 answer:

4 0

Boyle’s law is P1V1 = P2V2. This shows that pressure and volume have an inverse relationship, so increase in pressure causes a decrease in volume. B is your answer

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If the ball reaches the ground in 8 seconds and the velocity of the ball before it hits the ground is 78.4. What is its accelera

Natasha2012 [34]

Answer:

9.8 m / s^2

Explanation:

Assuming free fall====> there is no initial downward/upward velocity

Assuming metric units 78.4<u> m/s </u>

vf = a t

78.4 = a (8) shows a = 9.8 m/s^2

6 0

2 years ago

Plz i need help for the 5 problems. plz show the work!!!

Artemon [7]

Answer:

1. 3 $m/s^{2}$

2. 1.5 $m/s^{2}$

3. 3 seconds

4. 0 $m/s^{2}$

5. 2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = $\frac {30-0}{10}=3 m/s^{2}$

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=10m/s, v=22m/s and t=8 seconds

a = $\frac {22-10}{8}=1.5 m/s^{2}$

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=0m/s since at rest, v=15m/s and a=5 $\frac {m}{s^{2}}$

= $\frac {15-0}{5}=3s$

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = $\frac {20-20}{10}=0 m/s^{2}$

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 $\frac {m}{s^{2}}$

= $\frac {0-9}{-4.1}=2.2s$

8 0

2 years ago

If pizza is round will it roll off a cliff?

Rina8888 [55]

If a pizza is round like a ball it will roll off a cliff.If it’s just flat and round it won’t.

6 0

2 years ago

The birth defect caused by the Zika Virus to unborn infants is called__________?

lions [1.4K]

It is B, Microcephaly. Hope I helped! :))

3 0

3 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

3 years ago