3.

How much power does a 2000 kg car need to accelerate from 20 m/s to 35 m/s in 7 seconds?

Alexus [3.1K]

firstly you get your acceleration with the formula, a=v-u/t. Then you use the formula for kinetic energy 1/2mv^2

then you can finally get the answer for power by dividing your previous answer by the time

3 0

3 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

If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s

Vlad1618 [11]

The magnitude of the air drag is 784 N

Explanation:

An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:

$F_D = mg$

where

$F_D$ is the magnitude of the air drag

m is the mass of the object

g is the acceleration of gravity

In this problem, we have

m = 80 kg is the mass of the airman

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting into the formula, we find:

$F_D = (80)(9.8)=784 N$

Learn more about forces here:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

4 0

2 years ago

PLease YAlL i NEed IT- DOnt DO me Like that-

Natasha2012 [34]

Answer:

what does your parents do at home

Example

A person who is adamant about something has formed an opinion or taken a position that is not going to change because the person is determined to keep that opinion or position. ... The noun adamant comes from a Latin word meaning "material of extreme hardness, diamond

7 0

2 years ago

Read 2 more answers

81. A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a

Irina18 [472]

Answer:

Explanation:

Given

mass of squirrel $m=560 gm$