We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.
Where,
- F is force
- m is mass
- a is acceleration
In our case,
- F = ?
- m = 2500 kg
- a = 20m/s
<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>
Answer:
Dimension of cardboard is 22 m by 16 m
Explanation:
Given that,
Area = 352 cm²
Side of each square cutting from corner = 2 cm
Volume of box = 432 cm³
Let the two sides are x and y.
The area of the rectangular piece is
-------- (1)
The volume of the rectangular piece
x=16,22
Put the value of x in the equation (I)
For x = 16
For x = 22
Dimension of cardboard is 22 m by 16 m
A scientific law is the simple mathematical expression of the relationship involved. A principle is the same relationship expressed in words. A theory is the explanation of the facts that make up the relationship.
Answer:
a) Pb= 200 PA
b).work done= -3600 joules
c).3600joules
D).the system works under isothermal condition so no heat was transferred
Explanation:
2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a). PbVb= PaVa
Pb= (PaVa)/VB
Pb= (600*3)/9
Pb= 1800/9
Pb= 200 PA
b). work done= n(Pb-Pa)(Vb-Va)
Work done= 2*(200-600)(9-3)
Work done= -600(6)
Work done=- 3600 Pam³
work done= -3600 joules
C). Change in internal energy I the work done on the system
= 3600joules
D).the system works under isothermal condition so no heat was transferred
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.
The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds) = 4,000 foot-pounds of work.
If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds) = 645.2 foot-pounds per second.
The rate of doing work is called "power".
(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)
So back to our problem.
John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.
Well, 550 foot-pounds per second is called 1 "horsepower".
So as John runs up the steps to the balcony, he's doing the work
at the rate of
(645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)
= 1.173 Horsepower. GO JOHN !
(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________
Oh my gosh ! Look at #26 ! There are the metric units I was talking about.
Do you need #26 ?
I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.
a). 5
b). 750 Joules
c). 800 Joules
d). 93.75%
You're welcome.
And #27 is 0.667 m/s .
