I need help with my science.

A big lump of meat of mass 5Kg is hung from a spring balance in an elevator. Find the reading of the balance of (I) the elevator

Mrrafil [7]

The reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

T = m g + m a

T = Force

m = Mass

g = Acceleration due to gravity

a = Acceleration

m = 5 kg

g = 10 m / s²

I ) If the elevator is moving with a steady speed,

At steady speed, a = 0

T = ( 5 * 10 ) + ( 5 * 0 )

T = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s²,

a = 0.2 m / s²

T = ( 5 * 10 ) + ( 5 * 0.2 )

T = 50 + 1

T = 51 N

Therefore, the reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

To know more about reading on a spring balance

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7 0

1 year ago

The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo

weeeeeb [17]

Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

<u>Explanation:</u>

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since $L=I \times \omega$ , the turning rate of the two-puck framework must stay unaltered.

4 0

3 years ago

True or false The momentum of a 100kg object traveling at 3 m/s is 300 N

melomori [17]

Answer:

True

Explanation:

Momentum of an object can be defined as the product of its mass and velocity at which it is travelling. With that in mind, momentum = 3*100=300(kg⋅m/s).

One thing to note is the units mentioned. The SI unit of momentum is kg * m/s as it is the product of mass(kilograms) and velocity(meter per second) and not Newton.

5 0

3 years ago

What percent of sample of AS-198 to decay to 1/8 its original

denis-greek [22]

Answer:

bannana

Explanation:

5 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

4 years ago