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:
3.258 m/s
Explanation:
k = Spring constant = 263 N/m (Assumed, as it is not given)
x = Displacement of spring = 0.7 m (Assumed, as it is not given)
= Coefficient of friction = 0.4
Energy stored in spring is given by
As the energy in the system is conserved we have
The speed of the 8 kg block just before collision is 3.258 m/s
You can't. Velocity and acceleration measure two different things, so their units are incompatible. It's like asking, "How many meters does this book weigh?"
Maybe you mean "find" acceleration using given velocities, or a velocity function?
Answer:
138.3 days
Explanation:
Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.
The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.
T^2 is proportional to r^3
That is,
T^2/r^3 = constant
98^2 / 62^3 = T^2 / 78^3
Make T^2 the subject of formula.
T^2 = 98^2 / 62^3 × 78^3
T^2 = 19123.2
T = sqrt ( 19123.2 )
T = 138.2867 days
Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.
20/9.8 = 2.0 seconds. The ball stops after 2 seconds.
