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Physics: Post your doubts here!

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Q8 D
16 C
17B
18 B
19B
24C
for q8
for one train to overtake the other it must travel the same distance as the other
s=ut+1/2at^2
u=0 for express train and a=0 for the train traveling at constant speed
0.5*0.5*t^2=10t
t=40s
for q 16
both weight and contact forces are acting downwards while the tension is acting upwards.
in other words the vertical component of tension should equal to sum of vertical components of the two and horizontal one should equals to that of the contact force at hinge
T is greatest of the force only C seems to fit this
 
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A hose ejects the water with speed of 20m/s on the vertical wall.The cross-sectional area of the jet is 5x10^-4. If the density of the water is 1000kg/m3, find the force acting on the wall.(water comes to rest after hitting the wall)

Need help!!
 
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887
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Ans:

Q8 D
16 C
17B
18 B
19B
24C
for q17
since diameter is 0.08m
the bottom of ball is at 0.8-0.08=0.72
kinectic energy of ball before hitting ground=gravitional potiential energy at 0.72m from ground
kincectic energy after hitting ground=gravitional potiential energy at 0.45-0.08=0.37m
m*g*0.72=0.75
m=0.75/0.72*g
(0.75/0.72*g)*g*0.37
(0.75/0.72)*0.37=0.39j
 
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An object of mass m passes a point P with a velocity v and slides up a frictionless incline to stop at point Q, which is at a height h above P.

When a second object of mass 1/2m passes P with velocity of 1/2v, it will rise to
A-1/4h. B-1/2h.
 
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A 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

Ans is D
 
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From marking scheme . But I need explanation of those question

Found a formula ...
P = 0.5*ρ*A*v^3

P = power (W), ρ = density of air (kg/m^3), A = area wind passing through perpendicular to the wind (m^2), v = wind velocity (m/s)

It can be used to find the answer :)
 
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Hey, i think its like this: s=v/t . Since the wave is being reflected, s =150 x 2. Then v is the speed of light, c, which is 3.0x10^8. This will give you the answer of D. I think this is how to do it :)

I took the speed as Speed of light, 3.0 x 10^11 and s= 300. The answer that comes, however, is wrong. Can you specify the steps?
 
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HELP PLS question 8 Summer 2013 paper 1/1
u see that train will overtake the other train when there distances will equal.
the 1st train is travelling with constant speed hence no acceleraion
the 2nd train is acclerating from rest hence no inital speed
using the formula
s=ut+0.5at^2
for the first train
s=10t
for the train starting from rest
s=0.5*0.5t^2
placing both equations equal to each other.
10t=0.5*0.5*t^2
solve nad get 40s
 
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A hose ejects the water with speed of 20m/s on the vertical wall.The cross-sectional area of the jet is 5x10^-4. If the density of the water is 1000kg/m3, find the force acting on the wall.(water comes to rest after hitting the wall)

Need help!!
 
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