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2 years ago

8

Horizontal circular motion questions:

1 answer:

NemiM [27]2 years ago

7 0

1. D, constant
Force must be in the direction of the velocity to change speed.

2. B. to the center of the curvature
This allows the vehicle to complete the turn.

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An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate

riadik2000 [5.3K]

Answer: $6.408(10)^{-19} C$

Explanation:

This problem can be solved by the following equation:

$\Delta K=q V$

Where:

$\Delta K=7.37(10)^{-17} J$ is the change in kinetic energy

$V=115 V$ is the electric potential difference

$q$ is the electric charge

Finding $q$ :

$q=\frac{\Delta K}{V}$

$q=\frac{7.37(10)^{-17} J}{115 V}$

Finally:

$q=6.408(10)^{-19} C$

4 0

2 years ago

(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

2 years ago

The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.

Eva8 [605]

Answer: 29.17m/s^2

Explanation:

Given the following :

Velocity = 525 m/s

Time = 18 seconds

Acceleration = change in Velocity with time

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity and t = time

Plugging our values

525 = 0 + a × 18

525 = 18(a)

a = 525 / 18

a = 29.166666

a = 29.17 m/s^2

8 0

3 years ago

How do you think heat gets from a filament to the kernel?

PilotLPTM [1.2K]

Answer:

Explanation:

It goes thru the fluma

8 0

2 years ago

Rectangular plate has a voltage of +180V and plate to the first plate) has a 'voltage of -5V. Determie another rectangular plate

3241004551 [841]

For a Rectangular plate has a voltage of +180V and a 'voltage of -5V. , the second plate has the Electric field mathematically given as

E=21.5*10^3v/m

<h3>What is the field strength?</h3>

Generally, the equation for the Electric field is mathematically given as

E=v/d

Where

v={180-(-5)}v

v=185v

Therefore

E=185/8.6*10^{-3}

E=21.5*10^3v/m

In conclusion, Electric field

E=21.5*10^3v/m

Read more about electric field

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4 0

2 years ago