Physics: Post your doubts here!

[hellodjfos;s'ff]

This question makes no sense. Can someone solve it?

 

[IN NEED OF HELP]

In still air, a bird can fly at a speed of 10 m s–1. The wind is blowing from the east at 8.0 m s–1. In which direction must the bird fly in order to travel to a destination that is due north of the bird’s current location?
A 37° east of north
B 37° west of north
C 53° east of north
D 53° west of north

 

[Kanekii]

Is ans DD?

 

[khadijaimran]

i cant seem to understand these types of questions please help

 

[Kanekii]

This question makes no sense. Can someone solve it?

A?

 

[hellodjfos;s'ff]

In still air, a bird can fly at a speed of 10 m s–1. The wind is blowing from the east at 8.0 m s–1. In which direction must the bird fly in order to travel to a destination that is due north of the bird’s current location?
A 37° east of north
B 37° west of north
C 53° east of north
D 53° west of north

So they want the destination to be north of the bird's current location meaning that the resultant velocity must be vertically upwards. You draw a vector triangle based on the info that the resultant is directly upwards and the wind is blowing from the east. Then you use sin, cos or tan to find the angle. I would use sin since no further calculations would be needed to find the answer using sin. The angle will turn out to be 53, and you can determine the direction using your vector triangle(it will turn out to be west of north making the answer D).

 

[hellodjfos;s'ff]

A?

It's C. The question is really weird

 

[Kanekii]

i cant seem to understand these types of questions please help

BB?

 

[hellodjfos;s'ff]

Is ans DD? View attachment 64164

Yup, answer is D because x=lamda(D)/a. So fringe separation is proportional to the lamda, so if you use a greater wavelength, you get greater fringe separation.

 

[Kanekii]

Yup, answer is D because x=lamda(D)/a. So fringe separation is proportional to the lamda, so if you use a greater wavelength, you get greater fringe separation.

What if the width of the slits increased would that effect fringe separation?

 

[Hisham Khan]

 

[hellodjfos;s'ff]

i cant seem to understand these types of questions please help

The increase in current suggests that the total resistance of the circuit increased(R=V/I). Now the only way the resistance can increase is through the variable resistor. So the resistance of the variable resistor was increased, and you can determine how much pd it takes up using the potential divider formula,(r1/r1+r2)*Vtotal. You can use some values to figure out that the pd of the variable resistor would increase while pd for fixed resistor would decrease. So the voltmeter readings on Q increase and for P it decreases(B).

 

[hellodjfos;s'ff]

What if the width of the slits increased would that effect fringe separation?

Width of slits don't effect anything in the double slit formula. Width of the slits only impacts the intensity of the wave/fringes.

 

[IN NEED OF HELP]

The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the upthrust force. The hot-air balloon descends vertically at constant speed. The force of air resistance on the balloon is F. Which weight of material must be released from the balloon so that it ascends vertically at the same constant speed?
A F
B 2F
C 3F
D 4F

 

[Kanekii]

View attachment 64166

I=P/S
I=E/ST
4I*0.5S=E/T
E/t=2S

 

[khadijaimran]

The increase in current suggests that the total resistance of the circuit increased(R=V/I). Now the only way the resistance can increase is through the variable resistor. So the resistance of the variable resistor was increased, and you can determine how much pd it takes up using the potential divider formula,(r1/r1+r2)*Vtotal. You can use some values to figure out that the pd of the variable resistor would increase while pd for fixed resistor would decrease. So the voltmeter readings on Q increase and for P it decreases(B).

thankyou so much

 

[hellodjfos;s'ff]

The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the upthrust force. The hot-air balloon descends vertically at constant speed. The force of air resistance on the balloon is F. Which weight of material must be released from the balloon so that it ascends vertically at the same constant speed?
A F
B 2F
C 3F
D 4F

I answered this a couple of days ago, so just gonna copy paste my previous explanation:
8 is a slightly tougher question. Throughout this question, the balloon is moving at a constant speed meaning the balloon is in equilibrium so upwards forces must be equal to downward forces. In the first situation, the balloon is going down so the air resistance(AR) and upthrust(UT) are acting upwards while the weight(W) is acting downwards. This making an equation, W=AR+UT. The question states that the AR has a value of F. The final equation becomes W=F+UT. For the 2nd situation, the balloon goes up, hence the W and AR act downwards while UT acts up. The equation for the 2nd equation is UT=W+AR. Now replaced W with the equation you found in the 1st situation and AR with F(as the question says). UT=F+UT+F. If you solve this equation, you will find that it would be 0=2F. This means that there is a resultant force of 2F acting downwards, so to balance it you must either produce a force of 2F acting upwards or remove a force of 2F acting downwards. The question states how much weight must be released from the balloon, so that means weight of 2F must be removed. This makes B the answer. There could be alternative methods to solve this question but I find this one to be the best.

 

[hellodjfos;s'ff]

thankyou so much

Welcome

 

[IN NEED OF HELP]

A car is moving at constant speed in a straight line with the engine providing a driving force equal to the resistive force F. When the engine is switched off, the car is brought to rest in a distance of 100 m by the resistive force. It may be assumed that F is constant during the deceleration. The process is then repeated for the same car with the same initial speed but with a constant resistive force of 0.800F. How far will the car travel while decelerating?
A 120 m
B 125 m
C 156 m
D 250 m

 

[hellodjfos;s'ff]

These are two very tough questions if you wanna practice.