Answer:
1) 
& 
east of sign post
2) 
east of sign post
3) 
east of the signpost.
4) 
Explanation:
Given:
- position of motorcyclist on entering the city at the signpost,
- time of observation after being at x=5m east of the signpost,
- constant acceleration of the on entering the city,
- distance of the motorcyclist moments later after entering,
- velocity of the motorcyclist moments later after entering,
<u>Now the initial velocity on at the sign board:</u>
where:
initial velocity of entering the city at the signpost
Putting respective values:
1)
Position at time 
sec.:
Using equation of motion,
because it has already covered 5m before that point
east of sign post
Velocity at time sec.:
2)
Position when the velocity is 
:
using equation of motion,
east of sign post
3)
Given that:
acceleration be, 
time, 
Position after the new acceleration and the new given time:
using equation of motion,
east of the signpost.
4)
now time of observation, 
Answer:
D. increases by a factor of 4.
Explanation:
General equation of SHM
Lets taken the general equation of the displacement given as
x = A sinω t
A=Amplitude ,t=time ,ω=natural frequency
We know that speed V
V= A ω cosωt
The mechanical energy of spring mass system
K=Spring constant
Now when Amplitude A become 2 times then the mechanical energy will become 4 times.
Therefore the answer is D.
As per the question Bob drops the bag full with feathers from the top of the building.
The mass of the bag(m)= 1.0 lb
Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.
Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2
Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s
Hence time t= 1.5 s
From equation of kinematics we know that -
S=ut + 0.5at^2 [ here S is the distance travelled]
For motion under free fall initial velocity (u)=0.
Hence S= 0×1.5+{0.5×(-9.8)×(1.5)^2}
⇒ -S =0-11.025 m
⇒ S= 11.025 m
=11 m
Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .
Hence the correct option is B.
Answer:
The expression of gravitational field due to mass 
at a distance 
Explanation:
We have given mass is 
Distance of the point where we have to find the gravitational field is
Gravitational constant G
We have to find the gravitational filed
Gravitational field is given by 
This will be the expression of gravitational field due to mass at a distance
Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0
This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u />
<u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u />![-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]](https://tex.z-dn.net/?f=-k%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20x%7D%5D_%7Bx%3DL%7D%3Dh%5BT%28L%2Ct%29-T_%7B%5Cinfty%7D%5D)
<u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
