Ask question

Ask question

All categories

1 year ago

13

According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car

that accelerates from rest on a flat road. What force did the work that increased the car’s kinetic energy?
1. the force of the car engine

2. air resistance

3. the friction between the road and the tires

4. gravity

1 answer:

Leni [432]1 year ago

6 0

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

Also, W = F.d

$Fd=\dfrac{1}{2}m(v^2-u^2)$

Where

F is the force applied by the engine of car

d is the displacement

m is the mass of an object

u is the initial speed

v is the final speed

So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

You might be interested in

A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o

olya-2409 [2.1K]

Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

m=8/32=0.25

To find the spring constant

Kx=mg (given that c=6 in and mg=8 lb)

K(0.5)=8 (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0

2 years ago

What is the instantaneous velocity of a freely falling object 9.0 s after it is released from a position of rest? Express your a

Umnica [9.8K]

Answer:

So instantaneous velocity after 9 sec will be 88.2 m/sec

Explanation:

We have given time t = 9 sec

As the object is released from rest so its initial velocity u = 0 m/sec

We have to find its final velocity v

Acceleration due to gravity $g=9.8m/sec^2$

From first equation of motion we know that $v=u+gt$

$v=0+9.8\times 9=88.2m/sec$

So instantaneous velocity after 9 sec will be 88.2 m/sec

5 0

1 year ago

The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

irina [24]

Answer:

a = 15.1 g

Explanation:

The relation between mass and acceleration is given by :

$m\propto \dfrac{1}{a}$

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂

So,

$\dfrac{m_1}{m_2}=\dfrac{a_2}{a_1}\\\\a_2=\dfrac{a_1m_1}{m_2}\\\\a_2=\dfrac{0.8g\times 1510}{80}\\\\a_2=15.1g$

So, the car's acceleration would be 15.1 g.

6 0

1 year ago

A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the

Zina [86]

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

<u>Concepts and Principles </u>

The intensity at a distance r from a point source that emits waves of power P is given as:

I=P/4π*r^2 (1)

<u>Given Data</u>

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.

<u>Required Data</u>

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.

<u>solution:</u>

(a)

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:

I_1/I_2=(r_2/r_1)^2 (2)

Solve for I_2 :

I_2=I_1(r_2/r_1)^2

I_2=0.0625 I_1

(b)

Solve Equation (2) for r_2:

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

=3.16 m

3 0

1 year ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

1 year ago