The electric resistance in a length of wire is doubled when the wire is ________

A piece of metal has attained a velocity of 107.8 m/sec after fallinf for 10 seconds what is its initial velocity

soldi70 [24.7K]

Answer:

7.8 m/s

Explanation:

Here object is falling with a gravitational acceleration there for we can take acceleration = 10 m/ s² and its constant through out the motion there for we can use motion equation

V = U + at

V - Final velocity

U - Initial velocity

a - acceleration

t - time

V=U+at

107.8=U + 10×10

= 7.8 m/s

4 0

3 years ago

Why is energy transferred from the substance to the surroundings when a substance freezes

jeyben [28]

Zalice sucks chakra woman trAnsfer to the fezze y try e

7 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

. Friction is a rubbing force that ___________ a spinning yo-yo.

Advocard [28]

The yo-yo speeds up when you rub it

3 0

3 years ago

Read 2 more answers

A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.

irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ( $\tau$ ), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

$\alpha$ - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$ , $\omega_{o} = 0\,\frac{rad}{s }$ , $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$ , then the moment of inertia of the motor is: