Answer:
Expression of work done is

Work done to move the sled is given as 187.2 J
Explanation:
As we know that the formula of work done is given as

here we know that
F = 12.6 N
d = 15.4 m

so we will have


Answer: The energy absorbed by the reaction from the water is 996 Joules.
Explanation:
Energy absorbed by the reaction or energy lost by the water to the reaction,Q.
Mass of the the reaction ,m = 60 g
Specific heat of water = c = 4.15 J\g ^oC
Change is temperature=

Negative sigh indicates that energy was given by the water to the reaction.
The energy absorbed by the reaction from the water is 996 Joules.
Answer:
option (B)
Explanation:
Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.
Its unit is N/m².
The formula for the Young's modulus is given by

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.
It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.
Answer:

Explanation:
Given:
temperature at the hotter end, 
temperature at the cooler end, 
length of rod through which the heat travels, 
cross-sectional area of rod, 
mass of ice melted at zero degree Celsius, 
time taken for the melting of ice, 
thermal conductivity k=?
By Fourier's Law of conduction we have:
......................................(1)
where:
=rate of heat transfer
dT= temperature difference across the length dx
Now, we need the total heat transfer according to the condition:
we know the latent heat of fusion of ice, 



Now the heat rate:



Now using eq,(1)


Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>