1. A 2-kg bowling ball sits on top of a building that is 40 meters tall. Circle one: KE / GPE / both Show your work for finding the values of each type of energy the object has:
KE = 0 (because velocity is 0)
PE = mgh = 2kg*9.8m/s^2*40m= 784 joule
2. A 2-kg bowling ball rolls at a speed of 5 m/s on the roof of the building that is 40 tall. Circle one: KE / GPE / both Show your work for finding the values of each type of energy the object has:
BothKE = [1/2]m*v^2 = [1/2](2kg)(5m/s)^2 = 25 joulPE = mgh = 40 joule
3. A 2-kg bowling ball rolls at a speed of 10 m/s on the ground. Circle one: KE / GPE / both Show your work for finding the values of each type of energy the object has:
KE = [1/2]m(v^2) = [1/2]2kg*(10m/s)^2 = 100 joule
GPE = mgh = 0
1. A 1,000-kg car has 50,000 joules of kinetic energy. What is its speed?
KE = [1/2]m(v^2) => v = √[2KE/m] = √[2*50,000joules/1000kg] = 10m/s
2. A 200-kg boulder has 39,200 joules of gravitational potential energy. What height is it at?
GPE = mgh => h = GPE / (mg) = 39,200 joules / (200kg * 9.8m/s^2) = 20m
3. A 1-kg model airplane has 12.5 joules of kinetic energy and 98 joules of gravitational potential energy. What is its speed? What is its height?
KE = [1/2]m(v^2) => v = √ [ 2KE/m] = √[2*12.5 j / 1kg] = 7.9 m/s
GPE = mgh => h = GPE/(mg) = 98/(1kg*9.8m/s^2) = 10 m
It will be choice e.100 because 500/5=100 km per hour
Answer:
see attachment
Explanation:
all four questions have been explained in attachment
Answer:
1. 2.67 s
2. 0.1 m/s²
Explanation:
1. Determination of the time taken for the penguin to fall.
Height (h) of cliff = 35 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
35 = ½ × 9.8 × t²
35 = 4.9 × t²
Divide both side by 4.9
t² = 35 / 4.9
Take the square root of both side
t = √(35 / 4.9)
t = 2.67 s
Thus, it will take 2.67 s for the penguin to fall onto the head of a napping polar bear.
2. Determination of the acceleration of the penguin.
Initial velocity (u) = 0 m/s.
Final velocity (v) = 2 m/s.
Time (t) = 20 s
Acceleration (a) =?
a = (v – u)/t
a = (2 – 0)/ 20
a = 2 / 20
a = 0.1 m/s²
Thus, the acceleration of the penguin is 0.1 m/s²
Explanation :
It is given that, q and 4q are placed at a distance of l.
Let x is the distance where third charge is placed so that the entire three charge system is in static equilibrium.
Equilibrium means the net force acting on the system of charges is equal to zero. Let Q is the third charge.
So,
On solving,
For magnitude :
using
So, a charge of -4q/9 is at a distance of l/3 is placed. It is placed to the right of +q.