A ball is thrown directly upward into the air. The graph below shows the vertical position of the ball with respect to time

By the nucleus, protons and neutrons of an atom does it contain more mass

Temka [501]

Answer:

Virtually all the mass of an atom resides in its nucleus, according to Chemistry

Explanation:

7 0

3 years ago

What will happen for Potential energy of a body if the mass is dobled keeping height constant

4vir4ik [10]

<h3>Given :-</h3>

The mass of the body is doubled
The height of the body is constant

<h3>Solution :- </h3>

We know that ,

Potential energy = mgh

Therefore,

We can say that,

PE is directly proportional to Mass of the body

According to the question,

PE of the body = 2m * g * h. ...eq( I)

From (I) , we can conclude that, If mass of the body get doubled then its PE will also be doubled .

4 0

3 years ago

Please Help Me

JulijaS [17]

1. a=Δv/Δt=(v-vo)/t=(0-25)/5=-25/5=-5 m/s²

The "-" sign shows us that the car has a slow motion

2.a=Δv/Δt=(v-vo)/t=(10-0)/4=10/4=2,5 m/s²

3.they do not have acceleration because they go at constant speed

7 0

4 years ago

A car travels at a speed of 55 km/hr and slows down to 10 km/hr in 20 seconds. What is the acceleration?

WITCHER [35]

Answer:

b

Explanation:

3 0

3 years ago

Read 2 more answers

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

4 years ago