Physics: Post your doubts here!

[xyn1123]

Related to Kirchoff's first law, coz the resistance is the same, so I is equally divided.

Ohh. Please help me with few questions >>>
http://olevel.sourceforge.net/papers/9702/9702_s13_qp_11.pdfQ8, 16, 17, 18,19,24

 

[periyasamy]

I will help u with those

 

[periyasamy]

Firstly for q 8--make eqn relating same distance for both
s=ut+1/2at^2
(10)t +1/2(0)t^2=(0)(t)+1/2(0.5)t^2
ans=40s

 

[snowbrood]

I think I get Q33
Just find the ratio of cross sectional area I guess, it will be 1:16, so resistance is 0.2*16= 3.2

not exactly u got that wrong
area of 0.5mm =(d^2/16)*pi
area of 1mm=(d^2/4)*pi
comparing both we find that
area of 0.5mm is 1/4 the area of 1mm
since volume of the wire remains the same whether u strech or not
V=l*A
for 0.5mm area decreases by four times so length should increaes by four so that we get the same V
R=pl/A equation 1
p(resistivity) will be constant
the length increases by 4 and area decreases by 4 this increases the R by 16
Rx=4pl/(A/4)
comparing Rx=16pl/A to equation 1 Rx should equal 16R end of story peace

 

[periyasamy]

question 17
Is answer a?
If it is this is the way ,1st mgh=1/2mv^2,,,,,m(9,81)(0.8-0.08)=1/2(m)(v^2)
v=3.76
after bounce given that 1/2mv^2=0.75
so subs into eqn n u will get===>mass is 0.1kg
Then u know conservation of energy so mgh after bounce is (0.1)(9.81)(0.45-0.08)

 

[xyn1123]

Ans:

Q8 D
16 C
17B
18 B
19B
24C​

 

[Jessie N.]

not exactly u got that wrong
area of 0.5mm =(d^2/16)*pi
area of 1mm=(d^2/4)*pi
comparing both we find that
area of 0.5mm is 1/4 the area of 1mm
since volume of the wire remains the same whether u strech or not
V=l*A
for 0.5mm area decreases by four times so length should increaes by four so that we get the same V
R=pl/A equation 1
p(resistivity) will be constant
the length increases by 4 and area decreases by 4 this increases the R by 16
Rx=4pl/(A/4)
comparing Rx=16pl/A to equation 1 Rx should equal 16R end of story peace

Thank you TTvTT！！！！！

 

[mynameisjh]

A small charge q is placed in the electric field of a large charge Q. Both charges experience a force F. What is the electric field strenght of the charge Q at the position of the charge q?

A F/Qq B F/Q C FqQ D F/q

Answer is D but whyyy?????

 

[Talhakhan]

A barrel of mass 50 kg is loaded onto the back of a lorry 1.6 m high by pushing it up a smooth plank 3.4 m long.
What is the minimum work done?
A-80J. B-170J. C-780J. D-1700J.

Explain me the answer, its C.

 

[periyasamy]

A barrel of mass 50 kg is loaded onto the back of a lorry 1.6 m high by pushing it up a smooth plank 3.4 m long.
What is the minimum work done?
A-80J. B-170J. C-780J. D-1700J.

Explain me the answer, its C.

mgh===>50(9.81)(1.6)=780j

 

[moinul]

plz help me with this question :
what is the average KE of an athlete running at maximum speed during a 100 m race?
the answer given is 4000.

 

[periyasamy]

A small charge q is placed in the electric field of a large charge Q. Both charges experience a force F. What is the electric field strenght of the charge Q at the position of the charge q?

A F/Qq B F/Q C FqQ D F/q

Answer is D but whyyy?????

Because it is itself the formula of electric field strength===>E=v/d

 

[mynameisjh]

Because it is itself the formula of electric field strength===>E=v/d

I know but why cant it be B) F/Q???

 

[mynameisjh]

Because it is itself the formula of electric field strength===>E=v/d

I know but why cant it be B) F/Q???

 

[periyasamy]

plz help me with this question :
what is the average KE of an athlete running at maximum speed during a 100 m race?
the answer given is 4000.

guess it like this==>1/2mv^2
1/2*(100)*(10^2)
100===weight of an adult
10ms^-2==velocity of usain bolt in 100m race
My calculation give 5000j
It is the nearest to your answer..
Just an estimation question.

 

[mynameisjh]

Because it is itself the formula of electric field strength===>E=v/d

I know but why cant it be B) F/Q???

 

[periyasamy]

Because i guess the formula states electric field strength=force per unit test charge.Test charge is a small charge.
I guess this is the reason

 

[mynameisjh]

Because i guess the formula states electric field strength=force per unit test charge.Test charge is a small charge.
I guess this is the reason

hmmmm, alright :/ thanks alot!!

 

[periyasamy]

Good luck guys for tomorrow.

 

[snowbrood]

Ans:

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

 

[Abhi Mevawala]

sam 2 u

 

[Abhi Mevawala]

sam 2 u

 

[Talhakhan]

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!!

 

[snowbrood]

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