Describe what happens to the magnitude of the net electrostatic force on the electron as the electron

A 49 N coconut is in a 15 m tree. What is te potential energy of the coconut?

eduard

Answer:

735 J

Explanation:

From the question given above, the following data were obtained:

Weight (W) = 49 N

Height (h) = 15 m

Potential energy =?

Potential energy is simply defined as the product of weight of the object and height to which the object is raised. Mathematically, it is expressed as:

Potential energy = weight × height

With the above formula, we can obtain the potential energy of the coconut as follow:

Weight (W) = 49 N

Height (h) = 15 m

Potential energy =?

Potential energy = weight × height

Potential energy = 49 × 15

Potential energy = 735 J

Thus, the potential energy of the coconut is 735 J

4 0

3 years ago

Which equation is most likely used to determine the acceleration from a velocity vs:time graph?

tresset_1 [31]

Acceleration, a = (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.

7 0

3 years ago

Read 2 more answers

How independent are we on technology?

Leona [35]

Technology makes us feel more alone because we are more dependent on social media connections than real life connections. Simply put, technology-independent architecture means not being biased towards any particular platform or software language.

6 0

3 years ago

Read 2 more answers

The highest speed attained by a cyclist on level ground is 105 km/h. To attain this speed, the cyclist used a streamlined recumb

omeli [17]

Answer:

The acceleration is 0.133 m/s²

The time taken to build the final speed is 219.32 s

Explanation:

Given;

final speed of the cyclist, v = 105 km/h = 29.17 m/s²

distance traveled by the cyclist, d = 3.2 km = 3,200 m

initial speed, u = 0

Apply the following kinematic equation to determine the acceleration of the cyclist;

v² = u² + 2as

(29.17 )² = 0 + (2 x 3200)a

850.89 = 6400a

a = (850.89 ) / (6400)

a = 0.133 m/s²

The time taken to build up the final speed is given by;

v = u + at

29.17 = 0 + (0.133)t

t = 29.17 / 0.133

t = 219.32 s

4 0

3 years ago

Consider two identical objects released from rest high above the surface of Earth. (Neglect air resistance for this question.)

Nadusha1986 [10]

The question is missing its alternatives. Here is the complete question.

Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.

1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.

a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;

Answer: C) K2 = (4/3)K1

Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - $\frac{G.Me.m}{r}$ + U₀

In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.

The potencial energy of an object of mass m on the surface of the Earth is:

Usurface = - $\frac{G.Me.m}{Re}$