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Katarina [22]
4 years ago
9

A car (m = 1302 kg) traveling along a road begins accelerating with a constant acceleration of 1.50 m/s2 in the direction of mot

ion. After traveling 392 m at this acceleration, its speed is 35.0 m/s. Determine the kinetic energy of the car when it began accelerating.
Physics
1 answer:
antiseptic1488 [7]4 years ago
5 0

First we will use the concepts of motion kinetics for which the final speed is defined as shown below,

v_f^2=v_i^2+2as

Here,

v_f= Final velocity

v_i= Initial velocity

a = Acceleration

s = Distance

Replacing,

(35)^2 = v_i^2+2(1.5)(392)

v_i = 7m/s

Using the conservation of energy for kinetic energy we have,

KE = \frac{1}{2}mv_i^2

KE = \frac{1}{2}(1302)(7)^2

KE = 31900J

Therefore the kinetic energy of the car is 31900J

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Convert the arc length of 107869221.61 revolutions into rotation
Genrish500 [490]

Answer:

1 r = 6.283185 rad

677,762,308.31685 rad

Explanation:

8 0
3 years ago
A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
4 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
3 years ago
If yall friend me ill give lots of points like this in the future
Ne4ueva [31]

Answer:

ok......................

3 0
3 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
4 years ago
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