For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:
a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time
The solution is as follows;
a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²
2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
The choices are:
a. Normal Force
b. Gravity Force
c. Applied Force
d. Friction Force
e. Tension Force
f. Air Resistance Force
Answer:
The answer is letter e, Tension Force.
Explanation:
Force refers to the "push" and "pull" of an object, provided that the object has mass. This results to acceleration or a change in velocity. There are many types of forces such as <em>Normal Force, Gravity Force, Applied Force, Friction Force, Tension Force and Air Resistance Force.</em>
The situation above is an example of a "tension force." This is considered the force that is being applied to an object by strings or ropes. This is a type force that allows the body to be pulled and not pushed, since ropes are not capable of it. In the situation above, the tension force of the rope is acting on the bag and this allows the bag to be pulled.
Thus, this explains the answer.
Answer:
It is called so because it is applicable on all bodies having mass, and the bodies will be governed by the same law, that is newton's law of gravitation. Thus, as it is applicable universally, it is called as universal law.
Answer:
11 m/s south
Explanation:
The velocity of the passenger relative to the river bank is equal to the velocity of the passenger relative to the ferry, plus the velocity of the ferry relative to the river, plus the velocity of the river relative to the river bank.
v_passenger,bank = v_passenger,ferry + v_ferry,river + v_river,bank
If we take north to be positive and south to be negative:
v = 1.0 m/s + (-10 m/s) + (-2 m/s)
v = -11 m/s
v = 11 m/s south
