Anytime an object is moving, it has what kind of energy

Pls help fast !!!!!!!!!

Elodia [21]

Answer:

P and R me think

Explanation:

7 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

Name three types of energy that exist in a large piece of charcoal on a grill in the sunlight. Explain why the charcoal has each

zloy xaker [14]

Answer:

* thermal energy

potential energy,

potential energy

Explanation:

The ship has various types of energy,

* thermal energy. This energy is associated with the temperature of the coal, the hotter the greater its internal energy,

* potential energy this energy is stored in the constituent atoms within carbon

* potential energy. It is due to the configuration of the system, in this case the sun heats the coal

5 0

3 years ago

Bart runs up a 2.91 meters high flight of stairs at a constant speed in 2.15 seconds. If barts mass is 65.9kg determine the work

Tanzania [10]

Answer:

1879.33J

874.1W

Explanation:

Given parameters:

Distance covered = 2.91m

Time taken = 2.15s

Mass of Bart = 65.9kg

Unknown:

Work done = ?

Power rating = ?

Solution:

Here, the work done is related to the the potential energy in climbing this flight of stairs.

Work done = Potential energy = mgH

where m is the mass

g is the acceleration due to gravity

H is the height

Work done = 65.9 x 9.8 x 2.91 = 1879.33J

Power is defined as the rate at which work is being done.

Power = $\frac{Work done }{time taken}$

= $\frac{1879.33}{2.15}$

= 874.1W

6 0

3 years ago

Attached here is the question:

ICE Princess25 [194]

Answer:

t = 6 [s]

Explanation:

In order to solve this problem we must first use this equation of kinematics.

$v_{f}^{2}=v_{o}^{2} +2*a*x$

where:

Vf = final velocity = 0 (the car comes to rest)

Vo = initial velocity = 72 [km/h]

a = acceleration [m/s²]

x = distance = 60 [m]

First we must convert the velocity from kilometers per hour to meters per second.

$72 [\frac{km}{h}]*\frac{1000m}{1km} *\frac{1h}{3600s} =20 [m/s]$

$0=(20)^{2} -2*a*60\\400 = 120*a\\a=3.33[m/s^{2} ]$

Now using this other equation of kinematics.

$v_{f}=v_{o}-a*t$

0 = 20-3.33*t

t = 6[s]

5 0

3 years ago