The movements of the tectonic plates
Explanation:
since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.
We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.
range can be calculated by the formula :-

u is the velocity during its take off and
is the angle at which its thrown
Given that
- u = 8m/ s
= 40°
calculating range using the above formula


value of sin 80 = 0. 985



Hence,

Answer:
The number density of the gas in container A is twice the number density of the gas in container B.
Explanation:
Here we have
P·V =n·R·T
n = P·V/(RT)
Therefore since V₁ = V₂ and T₁ = T₂
n₁ = P₁V₁/(RT₁)
n₂ = P₂V₂/(RT₂)
P₁ = 4 atm
P₂ = 2 atm
n₁ = 4V₁/(RT₁)
n₂ =2·V₁/(RT₁)
∴ n₁ = 2 × n₂
Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.
This can be shown based on the fact that the pressure of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.
Answer:
1) Time interval Blue Car Red Car
0 - 2 s Constant Velocity Increasing Velocity
2 - 3 s Constant Velocity Constant Velocity
3 - 5 s Constant Velocity Increasing Velocity
5 - 6 s Constant Velocity Decreasing Velocity
2) For Red and Blue car y₂ = 120 v =
=
= 20 m/s
We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.
3) At t = 2s, the cars are the same position, and are moving at the same rate
Position - same
Velocity - same
The position-time graph shares the same spot for two cars.
Answer:
+ 24 N
Explanation:
the computation is shown below:
Given that
Mass of the block = m = 0.7 kg
Sprint constant = k = 160 N / m
x = 0.15m
Now the force on the block is
F = kx
= (160) (0.15)
= 24 N
As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction
Therefore the third option is correct