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3 years ago

What is the solution?

Physics

1 answer:

serg [7]3 years ago

5 0

Answer:

1) x = 30 - 8 t

2) x = -10

3) x = -10 + 5 t

4) x = -10 - 4 t

Explanation:

Motion 1: constant negative velocity calculated via the points (0, 30) and (5, -10) rendering the equation of motion x = 30 - 8 t

Motion 2: constant position over time, so the object is not moving, and the equation of motion is x = -10

Motion 3: constant positive velocity estimated via the points (0, -10) and (2, 0), and the equation of motion is: x = -10 + 5 t

Motion 4: constant negative velocity estimated via the points (0, -10) and (5, -30), and the equation of motion is: x = -10 - 4 t

The starting position of motion 1 is 30 meters

the starting position for the other 3 motions is - 10 meters. And none of them is accelerated (acceleration = zero for all).

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

Could a car drive on a frictionless surface? Explain using the terms action

Julli [10]

Answer:

No, it cannot. The car needs the friction of the surface to drive because the car pushes the surface backwards, and the surfaces makes a reaction force pushing the car forward, and that works because of the friction. In a frictionless surface the tires would rotate in the same place

7 0

3 years ago

An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8

Aloiza [94]

Answer:

<em>The mass of the apple is 0.172 kg (172 g)</em>

Explanation:

<u>The Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

$m_2v_2-m_2v'=m_1v'-m_1v_1$

Factoring m2 and m1:

$m_2(v_2-v')=m_1(v'-v_1)$

Solving:

$\displaystyle m_2=\frac{m_1(v'-v_1)}{v_2-v'}$

Substituting:

$\displaystyle m_2=\frac{0.043(16.8-84)}{0-16.8}$

$\displaystyle m_2=\frac{-2.8896}{-16.8}$

$\displaystyle m_2=0.172\ kg$

The mass of the apple is 0.172 kg (172 g)

3 0

3 years ago

A 90-kg astronaut is stranded in space at a point 6.0 m from his spaceship, and he needs to get back in 4.0 min to control the s

Stella [2.4K]

Momentum = 0.5 * 4 = 2
to conclude the man’s velocity after he throws the piece of equipment, divide this number by the man’s mass.

v = 2/90

This is about 0.0222 m/s. To know if he can move 6 meters at velocity in 4minutes, use the following equation.

d = v * t, t = 4 * 60 = 240 s
d = 2/90 * 240 = 5⅓ meters.

This is ⅔ of a meter from the spaceship. To know the velocity that he must have to move 6 meter, use the same equation.

6 = v * 240
v = 6/240
This is about 0.00416 m/s.
His final momentum = 90 * 6/240 = 2.25

To know the velocity of the package, divide this number by the mass of the package.
v = 2.25/0.5 = 4.5 m/s

8 0

3 years ago

Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp

maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = $\dfrac{N}{C}m$

c = Speed of light = m/s

Momentum is given by

$\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s$

The unit of MeV/c in SI fundamental units is kg m/s

5 0

3 years ago