What is the kinetic energy of a 8kg cat running 5m/s?

A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

where k is the coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

7 0

3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

Why would the planets move in a straight path if there was no gravitational energy from the sun

omeli [17]

Explanation:

The sun's gravitational force is very strong. If it were not, a planet would move in a straight line out into space. The sun's gravity pulls the planet toward the sun, which changes the straight line of direction into a curve. This keeps the planet moving in an orbit around the sun

6 0

3 years ago

Read 2 more answers

A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy

GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0

2 years ago

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

c = speed of light = $3 \times 10 ^{8}m/s$

Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0

3 years ago