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2 years ago

A train traveling at a constant speed covers a distance of 960 meters in 30 seconds. What is the train's speed?

Physics

2 answers:

madreJ [45]2 years ago

6 0

Answer:

32 m/sec

Explanation:

Given,

⇒ Distance that is covered= 960 meters

⇒ Time taken= 30 seconds

We know that,

⇒ $Speed=\frac{distance}{time taken}$

⇒ $Speed =\frac{960}{30}$

Therefore, speed = 32 m/sec.

balu736 [363]2 years ago

4 0

32 meters per second

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How might you describe the mathematical procedure of finding the displacement when an object travels in two opposite directions?

levacccp [35]

Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.

Displacement = 2 km - 2km = 0

Generally, the equation is:
<span>Displacement = Distance of motion to the right - Distance of motion to the left</span>

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3 years ago

In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, a

scoundrel [369]

Answer:

a) Total mass form, density and axis of rotation location are True

b) I = m r²

Explanation:

a) The moment of inertia is the inertia of the rotational movement is defined as

I = ∫ r² dm

Where r is the distance from the pivot point and m the difference in body mass

In general, mass is expressed through density

ρ = m / V

dm = ρ dV

From these two equations we can see that the moment of inertia depends on mass, density and distance

Let's examine the statements, the moment of inertia depends on

- Linear speed False

- Acceleration angular False

- Total mass form True

- density True

- axis of rotation location True

b) we calculate the moment of inertia of a particle

For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is

I = m r²

4 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

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the sys

Archy [21]

Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.

<h3>Conservation of energy</h3>

The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B$

$v_B = \frac{2v_0}{3} \ m/s \ to \ the \ left$

At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.

The complete question is below:

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?

Learn more about conservation of energy here: brainly.com/question/166559

5 0

2 years ago

Read 2 more answers

Why is it important that radioisotopes used in diagnostic tests have short half-lives? 20 A) These radioisotopes have a greater

LiRa [457]

Answer:

Option B) This minimizes the harmful side effects of the radiations

Explanation:

Half-life is the time taken for the decay of an radio-active atom in which it disintegrates such that it becomes half of its value at the beginning.... The nuclei should be in active mode for a longer duration sufficient for the treatment of the condition but these nuclei should have a sufficient shorter half life so that they don't get enough time to cause any damage to the health of the person other than treating the cause.

A shorter half life gives the assurance that the radiation after the treatment will leave the body without getting accumulated and cause harm to the body cells and other organs.

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3 years ago