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0.99 is this the answer ?help!
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0.99 is this the answer ?help!
no its 30.99 is this the answer ?
thnxWe know that the breaks were applied when the skid marks appeard. We also know the final speed would be zero. Use
2as = vsquare - u square
for second part he is asking the time interval before the brakes were applied
Just use s = ut + 1/2atsquare ( no acceleration)
Which year paper ? Post the linkno its 3
Which year paper ? Post the link
Yeahhh!!, thanks
The questions says "thnx
but how do u know that the skid marks appeared when the breaks were applied
max time =help!
For the ball to reach max height
Here
agha saad 22A strong wind of speed 33 m s–1 blows against a wall. The density of the air is 1.2 kg m–3. The wall
has an area of 12 m2
at right angles to the wind velocity. The air has its speed reduced to zero
when it hits the wall.
What is the approximate force exerted by the air on the wall?
A 330 N B 400 N C 480 N D 16 000 N
can someone explain me this
As CB also has resistance, just think of it as another resister in series with the two parallel resistors. now if the emf of the batter is 6v (dont remember the question), some of the voltage would be dissipated across the CB resistor first, and the rest of 6v is then equally dissipated across the two parallel resistorsthanks but i'm still confused a bit. can you please explain to me the full concept of this question (about potential divider?) ?
SORRY FOR LATE ANSWER ASMA!! WO JANA PARA LET MEE SEE WATS IT!! MMM ANSWER "A" hai?
I dunno the answer I think Its A thts y I asked u :/SORRY FOR LATE ANSWER ASMA!! WO JANA PARA LET MEE SEE WATS IT!! MMM ANSWER "A" hai?
Mass of wind coming in contact with the blades per second = density * speed * area = 475.2 kg/sA strong wind of speed 33 m s–1 blows against a wall. The density of the air is 1.2 kg m–3. The wall
has an area of 12 m2
at right angles to the wind velocity. The air has its speed reduced to zero
when it hits the wall.
What is the approximate force exerted by the air on the wall?
A 330 N B 400 N C 480 N D 16 000 N
can someone explain me this
I too got DMass of wind coming in contact with the blades per second = density * speed * area = 475.2 kg/s
F = mv-mu/t As it is 475.2 kg/s, we will use
F = 475.2*0 - 475.2*33/1 = D
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