<span>braking and returning suddenly to the roadway</span>
3. A
4.B
3.Since the Earth rotates through two tidal “bulges” every lunar day, we experience two high and two low tides every 24 hours and 50 minutes
4.An eclipse of the Sun happens when the New Moon moves between the Sun and Earth, blocking out the Sun's rays and casting a shadow on parts of Earth
Answer:
Spring constant of the spring will be equal to 9.255 N /m
Explanation:
We have given mass m = 0.683 kg
Time taken to complete one oscillation is given T = 1.41 sec
We have to find the spring constant of the spring
From spring mass system time period is equal to
, here m is mass and K is spring constant
So 

Squaring both side


So spring constant of the spring will be equal to 9.255 N /m
Answer:
part (a)
towards north east direction.
part (b) s = 46.60 m
Explanation:
Given,
- velocity of the river due to east =

- velocity of the boat due to the north =

part (a)
River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,
Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 
Let 'v' be the velocity of the boat relative to the shore.

Let
be the angle of the velocity of the boat relative to the shore with the horizontal axis.
Direction of the velocity of the boat relative to the shore.
part (b)
- Width of the shore = w = 300m
total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken
Let 't' be the total time taken by the boat to cross the width of the river.
Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.