Answer:
The tension in the cord is
Explanation:
Given:
M = mass
b = radius
R = spool of radius
The equation is:
(eq. 1)
The sum of forces in y:
∑Fy = Mg - T = Ma
Replacing in eq. 1
Power is the A. Rate of energy delivery (work done per second) B. Amount of energy available C. Capacity to do work (force over
1 answer:
Answer:
A. Rate of energy delivery (work done per second)
Explanation:
Work is the amount of force applied over a displacement. Joule is the SI unit
Energy is the quantity that must be transferred to the object in order to produce the desired effect. The effect can be heating the object or in order work to be performed on the object.
The amount of work done with respect to time is known as power i.e.,
The unit is
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Explanation:
It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².
The second equation of kinematics gives the relationship between the height reached and time taken by it.
Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.
We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :
t is time taken by the ball to hit the ground
is initial speed of the ball
So, the correct option is (A).
Answer:
- 59.97 º at 18.23 mph
Explanation:
To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.
To add the two vectors analytically you decompose each vector into their vertical and horizontal components.
<u>1. 18 mph 327º</u>
- Horizontal component: 18 mph × cos (327º) = 15.10 mph
- Vertical component: 18 mph × sin (327º) = - 9.80 mph
- Vector notation:
<u>2. 4 mph 60º</u>
- Horizontal component: 4 mph × cos (60º) = 2.00 mph
- Vertical component: 4 mph × sin (60º) = 3.46 mph
- Vector notation:
<u>3. Addition:</u>
You add the corresponding components:
To find the magnitude use Pythagorean theorem:
<u>4. Direction:</u>
Use the tangent ratio:
Find the inverse:
- arctan (1.73) ≈ 59.97º