Physics: Post your doubts here!

[ftfayek]

You should be more specific otherwise we cannot help..

SINGAPORE-CAMBRIDGE AS LEVEL PHYSICS 2005-2015 QUESTION PAPER ?

 

[Hamnah Zahoor]

can anyone help me with Question 7 .....have exam tomorrow!

 

[MShaheerUddin]

can anyone help me with Question 7 .....have exam tomorrow!

Initial speed u=50 ms–1
Final speed v=100 ms–1
Time t=20 s
Find deceleration a using v = u + at.
10=50 + a(20)
a=-2 ms–2.

Find marker distance using s=ut + ½ at2.
s=(50)(20) + ½ (-2)(202)
s=600 m


Answer: D

 

[saraawan]

can someone explain how will we do this question
A car of mass 1400kg is travelling on a straight, horizontal road at a constant speed of 25ms–1. The output power from the car’s engine is 30kW. The car then travels up a slope at 2° to the horizontal, maintaining the same constant speed.
What is the output power of the car’s engine when travelling up the slope?
A 12kW B 31kW C 42kW D 65kW
ans = C

 

[MShaheerUddin]

can someone explain how will we do this question
A car of mass 1400kg is travelling on a straight, horizontal road at a constant speed of 25ms–1. The output power from the car’s engine is 30kW. The car then travels up a slope at 2° to the horizontal, maintaining the same constant speed.
What is the output power of the car’s engine when travelling up the slope?
A 12kW B 31kW C 42kW D 65kW
ans = C

When Car is travelling at horizontal road it's speed is 25 m/s .. The car maintains this speed at slope WHICH MEANS THE CAR ALSO MAINTAINS 30kW OUTPUT POWER AT THE SLOPE (with new power at slope which is calculated below) AND SPEED OF 25 m/s.

At Slope the horizontal component of weight is forward driving force which is [mgSin(2)]=[1400x9.8Sin2]=478.8 N
Power is Force x Velocity ----> 478.8 x 25 = 11970.52 W
Since Car maintains the Output power of 30K at slope also so Output power of the car's engine is 11970.52 + 30kW Which is equals to 41970.5 W
Round it off to get 42kW of Power. (C ANS)

 

[MShaheerUddin]

Need help..

 

[amina1300]

View attachment 62947
Need help..

V in km/h * 1000m/km * 1h/3600s = V/3.6 in m/s
(1 / 3.6)² ≈ (1 / 13.0) ≈ 0.0772

so, A is the answer.

 

[MShaheerUddin]

explained at
http://physics-ref.blogspot.com/2018/01/when-constant-braking-force-is-applied.html

 

[MShaheerUddin]

http://physics-ref.blogspot.com/2018/01/a-digital-caliper-is-used-to-measure.html

 

[Cutilixious Alizeh]

View attachment 62962

is the ans B?

 

[Cutilixious Alizeh]

View attachment 62953

C?

 

[Cutilixious Alizeh]

View attachment 62947
Need help..

B

 

[MShaheerUddin]

B

Everybody knows that answer is B as it is clearly stated in Marking Scheme
But explanation is what i require

 

[Hamnah Zahoor]

Everybody knows that answer is B as it is clearly stated in Marking Scheme
But explanation is what i require

For finding total displacement we subtract the area under graph that is area of the graph above the x axis minus area of graph below the x axis.

(0.5*1*10)-(0.5*0.5*5)=3.75

 

[MShaheerUddin]

 

[MShaheerUddin]

 

[Laiba inayat]

June 2004
P1 Mcq no. 6 and 7

 

[MShaheerUddin]

June 2004
P1 Mcq no. 6 and 7

AS or A2?

 

[corona]

View attachment 62963

The answer is C.

The only forces acting on an object after release are drag and gravity, but in this context drag can be ignored as the stone is thrown on the Moon, where there is no air resistance.

Hence, there is no horizontal force but a single constant vertical force (gravity) acting downwards.

 

[corona]

View attachment 62965

B.

Areas under the graph:

Cheetah: 0.5*(15+17)(29) = 464
Antelope:0.5*(14+16)(25) = 375

Distance = 100+375-464 = 11m.