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777dan777 [17]
3 years ago
15

A car of mass m accelerates from speed v1 to speed v2 while going up a slope that makes an angle θ with the horizontal. The coef

ficient of static friction is μs, and the acceleration due to gravity is g. Find the total work W done on the car by the external forces. Express your answer in terms of the given quantities. You may or may not use all of them.
Physics
1 answer:
nlexa [21]3 years ago
7 0

Answer:

The answer is W=\frac{1}{2} m (v_2^{2} -v_1^{2} )

Explanation:

Here you have to take into account the theorem of work and energy.  This theorem says that the total work done by external forces on a body is used to modify the kinetic energy. So the only thing you have to do is determine the kinetic energy in the initial (K_1) and final moment (K_2), then the diference between them is the amount of net work that act over the body.

K_1=\frac{1}{2} m v_1^{2}

K_2=\frac{1}{2} m v_2^{2}

W=K_2-K_1=\frac{1}{2} m v_2^{2} - \frac{1}{2} m v_1^{2}=\frac{1}{2} m (v_2^{2} -v_1^{2} )

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A travelling disturbance that carries energy through matter or space.
dimulka [17.4K]
The answer is a Wave
5 0
3 years ago
The idealized body that emits the maximum amount of radiation at a given temperature and absorbs all the radiation incident on i
Sidana [21]

Answer:

Black body.

Explanation:

It's the property of the blackbody that it emits the maximum amount of radiation at a given temperature and also absorbs the radiation which incident upon it.

3 0
3 years ago
Calculate the number of atoms contained in a cylinder (1 m radiusand1 m deep)of (a) magnesium (b) lead.
Jobisdone [24]

Answer:

The question is incomplete,below is the complete question

"Calculate the number of atoms contained in a cylinder (1μm radius and 1μm deep)of (a) magnesium (b) lead."

Answer:

a. 1.35*10^{11} atoms

b. 1.03*10^{11} atoms

Explanation:

First, we determine the volume of the magnesium in the cylinder container

using the volume of a cylinder

V=\pi r^{2}h\\ r=10^{-6}m\\ h=10^{-6}m\\V=\pi *10^{-6*2}*10^{-6}\\V=\pi *10^{-18}\\V=3.14*10^{-18}m^{3}\\

a. Next we determine the mass of the magnesium ,

using the density=mass/volume

since density of a magnesium

the density of magnesium =1.738*10^{3}kg/m^{3}  \\mass=density * volume \\mass=1.738*10^{3}*3.14*10^{-18}\\mass=5.46*10^{-15}kg\\ \\mass=5.46*10^{-12}g\\

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

mole=5.46*10^{-12}/ 24.305\\mole=0.225*10^{-12}mol\\

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

number of atoms = 0.225*10^{-12}*6.02*10^{23}\\number of atoms =1.35*10^{11} atoms

b. for he lead, we determine the mass of the lead  ,

using the density=mass/volume

since density of a magnesium

the density of lead =11.34*10^{3}kg/m^{3}  \\mass=density * volume \\mass=11.34*10^{3}*3.14*10^{-18}\\mass=35.60*10^{-15}kg\\ \\mass=35.60*10^{-12}g\\

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

mole=35.60*10^{-12}/ 207.2\\mole=0.1718*10^{-12}mol\\

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

number of atoms = 0.1718*10^{-12}*6.02*10^{23}\\number of atoms =1.03*10^{11} atoms

3 0
3 years ago
Examples of rectilinear motion<br>​
EleoNora [17]

Answer:

Examples for Rectilinear Motion

1) The use of elevators in public places is an example of rectilinear motion.

2) Gravitational forces acting on objects resulting in free fall is an example of rectilinear motion.

3) Kids sliding down from a slide is a rectilinear motion.

4) The motion of planes in the sky is a rectilinear motion.

7 0
3 years ago
Read 2 more answers
A 1400-kg car moving with a speed of 32 m/s is approaching a stoplight. The light turns yellow and the car makes an abrupt stop
inna [77]

Answer:

c. 716, 800 J

Explanation:

t = Time taken

u = Initial velocity = 32 m/s

v = Final velocity = 0

s = Displacement = 60 m

a = Acceleration

m = Mass of car = 1400 kg

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-32^2}{2\times 60}

Work done is given by

W=Fscos\theta\\\Rightarrow W=mascos\theta\\\Rightarrow W=1400\times \dfrac{0^2-32^2}{2\times 60}\times 60\times cos0\\\Rightarrow W=-716800\ J

The amount of work done to stop the car is 716800 J

8 0
3 years ago
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