All categories

2 years ago

I needed ASAP!!!

Physics

1 answer:

11111nata11111 [884]2 years ago

7 0

They are moving in opposite direction, cant find out anything else because not enough informations is provided, like time .

You might be interested in

Instantaneous angular speed is: Group of answer choices the rate at which the angular acceleration is changing total angular dis

Ludmilka [50]

Answer: magnitude of the instantaneous angular velocity

Explanation:

Instantaneous angular speed is refered to as the magnitude of the instantaneous angular velocity. We should note that the instantaneous angular velocity is the rate that has to do with the rotation of an object in circular path.

5 0

3 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 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

Why does the current is reduced as electrons move through a conductor

hammer [34]

Because the electrons collide with the particles inside the conductor so are therefore slowed down seen as current is the rate of flow of electrons

3 0

3 years ago

In complete sentences describe the energy flow between the Sun, the Earth, and space

agasfer [191]

Answer: The energy from the sun passes through space in the form of invisible waves to the earth surface. It heats up the earth’s surface causing variation in climate.
Explanation:
The amount of incoming energy from the Sun decides the weather and climate of earth. If the energy that is incoming and outgoing on the earth, then climate is in equilibrium. The balance is depending on the scattering, absorption, reflection and transformation of energy.
The energy from sun passes through space and reaches the earth’s surface. On reaching surface, the solar energy warms the atmosphere releasing heat energy which gets transferred throughout the planets system by radiation, conduction and convection. Conduction happens in the atmosphere within first several millimeters close to the surface. This heated air expands as it is dense and rises causing transfer of heat to atmosphere through convection process. It results in formation of clouds.
The radiant energy from sun is transmitted via space in form of invisible waves. But much of the suns radiant energy, is transmitted back to atmosphere. The objects on earth like land, plants, animals absorb radiant energy as heat of which one third gets re-radiated back to atmosphere that is absorbed by carbon dioxide and water vapor. The atmosphere radiates heat energy back to earth increasing the earth temperature. This trapping of radiation is greenhouse effect.
The thermal energy obtained by convection currents are responsible for wind, cloud formation, and weather formation. The hydrosphere that comprises of 70% of earth’s surface absorbs solar energy.
On the basis of the above explanation is:
The energy from the sun passes through space in the form of invisible waves to the earth surface. It heats up the earth’s surface causing variation in climate.

6 0

3 years ago