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Len [333]
4 years ago
14

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains

497 cm^3 of air at atmospheric pressure (1.01×10^5 Pa) and a temperature of 27.0°C. At the end of the stroke, the air has been compressed to a volume of 47.0 cm^3 and the gauge pressure has increased to 2.80×10^6 Pa.
Required:
Compute the final temperature.
Physics
1 answer:
Colt1911 [192]4 years ago
4 0

If only I was smart then I could help you :/ no but like for real im madddd dumb sorry :(

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Alona [7]
Gravity as all bodies whether large or small have mass.
7 0
4 years ago
Arm abcd is pinned at b and undergoes reciprocating motion such that θ=(0.3 sin 4t) rad, where t is measured in seconds and the
storchak [24]
<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s
 
vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

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8 0
3 years ago
In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second pist
marin [14]

Answer:

847.45 N

Explanation:

F₁=force of exerted by smaller piston

F₂=force of exerted by larger piston=1.44×10⁴ N

A₁=Area of smaller piston= 2.62 cm =0.0265 m

A₂=Area of larger piston= 10.8 cm =0.108 m

Pressure exerted by both the pistons will be equal

P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times  0.108^2}\pi\times  0.0262^2\\\Rightarrow F_1=847.45\ N

Hence, force exerted to lift a 14400 N car is 847.45 N

8 0
4 years ago
What is the acceleration of a 45 kg mass pushed with 15 N of force? What is the acceleration if the force changes to 25 N? What
Rama09 [41]
F = ma
Rearrange this so acceleration is the subject:
A = f/m

Now input your values into this equation for the first question :
A = f/m
A = 15/45
A = 0.33333333 m/s

Then using this same equation change the force of 15 to 25 :
A = f/m
A = 25/45
A = 0.55555556 m/s

For the last question keep the force of 15N but change the mass to 70kg into the same equation :
A = f/m
A = 15/70
A = 0.21 m/s (rounded)
8 0
3 years ago
Read 2 more answers
Two charged balls are 10 cm apart. One is positively charged, and the other is negatively charged.
asambeis [7]

Answer:Explanation: The field lines of the electric charges extend from the positive to negative side charge. ... The force between the two charged balls is attractive. It is directly proportional to the product of their charges and inversely proportional to the magnitude of the distance between them.

Explanation:

hope this helps

8 0
3 years ago
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