Answer:
Differences can be :
1. Lever is onr of the six simple machines but pulley is a simple machine which has a wheel and axle .
2. Lever is of 3 types (first class , second class , third class ) but pulley is of only 2 types
I hope this helps you
Answer: 3.92 N.
Explanation:
Your box weighs 400g, or 0.4kg. In order to lift it, you need to overcome the force of gravity. F = ma, and acceleration due to gravity is -9.8 m/s^2. So gravity acts on the box with a force of 0.4 kg * -9.8 m/s^2 = -3.92 N. A force of +3.92 N is required to overcome this.
Answer:
Explanation:
a )
hear energy required to melt 1 g of ice = 340 J ,
hear energy required to melt 80 g of ice = 340 x 80 J = 27220 J .
b ) energy gained by the melted ice ( water at O°C ) = m ct
where m is mass of water , s is specific heat and t is rise in temperature
= 80 x 4.2 x ( 8°C - 0°C)
= 2688 J .
c )
energy lost by lime juice = energy gained by ice and water
= 27220 J + 2688 J .
= 29908 J .
d )
Let specific heat required be S
Heat lost by lime juice = M S T
M is mass of lime juice , S is specific heat , T is decrease in temperature
= 320 g x S x ( 29 - 8 )°C
= 6720 S
For equilibrium
Heat lost = heat gained
6720 S = 29908 J
S = 4.45 J /g °C .
Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative
