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faust18 [17]
3 years ago
8

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for

ce against the wheels?
Physics
1 answer:
Artemon [7]3 years ago
6 0
Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks. 

W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ

Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)
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The graph below is called a heating curve. It shows how water changes from one state of matter to another based on temperature a
Neporo4naja [7]

Explanation :

The heating curve shows how water changes from one state of matter to another based on temperature and the addition or removal of heat over time.

Initially, ice is heated until its temperature reaches 0^0\ C and changes to liquid state.

From the attached graph it is clear that until 100^0\ C the temperature will rise steadily. Here, the liquid begins to vaporize. Vaporization is the state of matter at which liquid state changes to the gaseous state.

So, E is the point which shows the gaseous state.

4 0
3 years ago
Read 2 more answers
What is the resultant of a and b if a = 3i 3j and b = 3i − 3j?
ss7ja [257]
 <span>a+b= ? 
3i +3j + (3i -3j) = ? 
3i + 3j + 3i -3j =? 
= 6i + 0j</span>
3 0
3 years ago
Read 2 more answers
A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

3 0
3 years ago
I don’t know if these are correct please help <br> Will mark brainliest :)
Licemer1 [7]

Im not to good at geography but I think you are correct :)

7 0
3 years ago
Please help!
andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0
2 years ago
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