The correct option is this: IT WOULD MOVE IN A CURVED CIRCULAR PATH.
Objects that are travelling in circular paths change directions all the time as they move round the circle, but they are prevented from moving off in a straight line by centripetal force. The centripetal force keeps pulling the objects towards the center of the circle. <span />
Answer:
270 m
Explanation:
We can find the distance travelled by the car by using the following suvat equation:
where
s is the distance travelled
u is the initial velocity
v is the final velocity
t is the time
For the car in this problem,
u = 0
v = 45 m/s
t = 12 s
Substituting into the equation, we find:
Answer:
True
Explanation:
When non-conservative forces are present, the amount of work done increases with the length of the path, this is true because, when a force is applied, the force does when and the non-conservative forces also do work. Since the non-conservative force work against the force applied, this tend to increase the net work done by the applied force to compensate for the loss in energy due to the work done by the non-conservative forces.
In a transverse wave the particle displacement is perpendicular to the direction of wave propagation. The animation below shows a one-dimensional transverse plane wave propagating from left to right. The particles do not move along with the wave; they simply oscillate up and down about their individual equilibrium positions as the wave passes by. Pick a single particle and watch its motion.
The S waves (Secondary waves) in an earthquake are examples of Transverse waves. S waves propagate with a velocity slower than P waves, arriving several seconds later.
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Answer:
v = 3.24 m/s
Explanation:
Since we don't have time, we can use the formula;
(Final distance - initial distance)/time = (initial velocity + final velocity)/2
Thus;
(x_f - x_i)/t = ½(v_xi + v_xf)
We are given;
x_i = 0 m
x_f = 5 m
v_xi = 0 m/s
v_xf = 5 m/s
Thus, plugging in the relevant values;
(5 - 0)/t = (0 + 5)/2
5/t = 5/2
t = 2 s
Using Newton's first law of motion, we can find the acceleration.
v = u + at
Applying to this question;
5 = 0 + a(2)
5 = 2a
a = 5/2
a = 2.5 m/s²
To get the speed, in m/s, of the block when it had traveled only 2.1 m from the top, we will use the formula;
v² = u² + 2as
v² = 0² + 2(2.5 × 2.1)
v² = 10.5
v = √10.5
v = 3.24 m/s
