Answer
given,
initial velocity of skateboard = 5.1 m/s
angle above the horizontal = 55°
height of the ramp = 1 m
a) maximum height of projectile
H = 0.889 m
the maximum height of the skateboard above the ground
= 1 + 0.889
= 1.889 m
b) time to reach the height
t = 0.426 s
horizontal distance = u cos θ × t
= 5.1 × cos 55° × 0.426
horizontal distance = 1.25 m
Answer:
the tangential velocity of the student is 4.89 m/s.
Explanation:
Given;
the radius of the circular path, r = 3.5 m
duration of the motion, t = 4.5 s
let the student's tangential velocity = v
The tangential velocity of the student is calculated as follows;
Therefore, the tangential velocity of the student is 4.89 m/s.
Answer:
Heat capacity, Q = 2090 Joules.
Explanation:
Given the following data;
Mass = 100 grams
Specific heat capacity = 4.18 J/g°C.
Temperature = 5°C
To find the quantity of heat required;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
t represents the temperature of an object.
Substituting into the formula, we have;
Heat capacity, Q = 2090 Joules.