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pshichka [43]
2 years ago
14

A 1.00-kilogram ball is dropped from the top of a building. just before striking the ground, the ball's speed is 12.0 meters per

second. what was the ball's gravitational potential energy, relative to the ground, at the instant it was dropped? [neglect friction.]
Physics
1 answer:
Anarel [89]2 years ago
6 0
During the fall, the potential energy stored in the ball is converted into kinetic energy.
Thus,
PE = KE before hitting the ground
= 1/2 • mv^2
= 1/2 • 1 • 12^2
= 72J
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What is NOT one of the three primary resources that families have to reach financial goals? It is c) education
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2 years ago
Kai takes a hamburger off of the grill and puts a piece of cheese on it. Explain how and why the cheese melts when Kai puts it o
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If Kai takes the burger off of the grill and puts cheese on it, it melts because the burger still is hot from being on the grill. The heat in and on the burger doesn’t go away immediately, so that is how and why the cheese melts.
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1 year ago
Read 2 more answers
A non-uniform rod 60cm long and weighs 32N is balanced at the 45cm mark. A load of 2N is hung on the zinc rod at the 25cm mark.
Pavlova-9 [17]

Answer:

The second knife-edge must be placed 46.2 cm from the zero mark of the rod.

Explanation:

From the law of equilibrium, ΣF = 0 and ΣM = 0.

Let R be the reaction at the knife edge. Since the weight of the rod and zinc load act downward, and we take downward position as negative

-32 N - 2 N + R = 0

-34 N = -R

R = 34 N

Also, let us assume the knife-edge is x cm from the zero mark. Taking moments about the weight and assuming the knife-edge is right of the weight of the rod. Taking clockwise moments as positive and anti-clockwise moments as negative,

-(45 - 25)2 + (x - 45)R = 0

-(20)2 + (x - 45)34 = 0

-40 = -(x - 45)34  

x - 45 = 40/34

x - 45 = 1.18

x = 45 + 1.18

x = 46.18 cm

x ≅ 46.2 cm

The second knife-edge must be placed 46.2 cm from the zero mark of the rod.

7 0
2 years ago
An object of mass 100 kg is initially at rest on a horizontal frictionless surface. At time t = 0, a horizontal force of 10 N is
satela [25.4K]

Answer:

(D) It is moving at a constant speed

Explanation:

Before t = 1s. Due to the force, albeit small, acting on the object, since there's no static friction stopping the object from moving, this mass object would have a constant acceleration and it's velocity would be increasing.

According to Newton's 1st law, an object will stay at a constant speed if the net force acting on it is 0. After t = 1s, horizontally speaking there's no other force exerting on the mass object. There is no friction force at play here as the surface is frictionless.

Therefore the correct statement is (D) It is moving at a constant speed

8 0
2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
2 years ago
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