Answer:
A. 12 m/s
Explanation:
Let’s remember that the definition of velocity is the variation of position of an object respect with to time. We know that the boy dropped the stone when the boat was 27 meters from the bridge and the stone hit the water 3 meters in front of the boat. So, the Boat must have traveled x=27 m-3m=24 m. The next step is calculating the amount of time that took the boat to make that travel; coincidentally, it is the same time that takes the stone to reach the water.
The equation that describes the motion of the stone is:
y = y_0 + v_0 * t+1/2 * a * t^2
The boy drops the stone from rest, so we can say that v_0=0. We can fixate the reference line on top of the bridge, so y_0=0 as well. The equation will be then:
-19,6 m = -1/2 * 9,8 m/s^2 * t^2
t^2= -(19,6 m)/(-4,9 m/s^2) = 4,012 s^2
t=√(4,012 s^2) = 2,003 s
Knowing the time that takes the stone to reach the water, that is the same that time that the boat uses to travel the 24 meters. The velocity of the boat is:
v = ∆x/∆t = (27 m-3 m)/(2,003 s-0s) = 11,9816 m/s ≈ 12 m/s
Have a nice day! :D
Answer:
Explanation:
a) Change in momentum, Δp = mΔv = m(v - u) = (15 * 10-3) * (90 - 300) = -3.15 kg-m2
b) Acceleration of the bullet, a = (v2 - u2) / 2s = (902 - 3002) / (2 * 0.02) = -2047500 m/s2
So, the bullet is in contact with the plastic for the time, 
c) Average force, 
Answer:
v_y = 12.54 m/s
Explanation:
Given:
- Initial vertical distance y_o = 10 m
- Initial velocity v_y,o = 0 m/s
- The acceleration of object in air = a_y
- The actual time taken to reach ground t = 3.2 s
Find:
- Determine the actual speed of the object when it reaches the ground?
Solution:
- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:
y = y_o + v_y,o*t + 0.5*a_y*t^2
0 = 10 + 0 + 0.5*a_y*(3.2)^2
a_y = - 20 / (3.2)^2 = 1.953125 m/s^2
- Use the principle of conservation of total energy of system:
E_p - W_f = E_k
Where, E_p = m*g*y_o
W_f = m*a_y*(y_i - y_f) ..... Effects of air resistance
E_k = 0.5*m*v_y^2
Hence, m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2
g*(10) - (1.953125)*(10) = 0.5*v_y^2
v_y = sqrt (157.1375)
v_y = 12.54 m/s
Answer: (d) It is !3 times the original speed.
Explanation: The rms speed of a gas is related to its temperature by the formulae below;
U(r.m.s) =√(3RT)/M
Where;
T represents the temperature.
R represents the gas constant.
M represents the molar mass of the gas.
Therefore, if the temperature increases from 200k to 600k
The temperature has then increased by a factor of 3,
However, we must note that temperature in the formulae is included in the square-root
Recall,
U(r.m.s) =√(3RT)/M
Consequently, temperature (T) can now be represented by (3T).
The inference drawn from this is that the root-mean-square speed would increase by a factor of √3
Therefore, option (d) is correct.
In the system described above we will have four forces that is acting on the puck. These are the weight, the normal force, the frictional force, and the force applied by the player. To determine the force applied by the player, we need to calculate first for the frictional force which is equal to the product of the coefficient of friction and the normal force. We do as follows:
Summation of forces in the y-direction:
W = Fn
Fn = 1.70 N
Summation of force in the x-direction
F = Fr = 0.06Fn
F = 0.06 (1.70) = 0.102 N
