All categories

3 years ago

How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?

Physics

1 answer:

krok68 [10]3 years ago

7 0

Explanation:

We have,

Mass of an automobile is 1150 kg

The automobile traveling at 86 km/h and then it comes to stop.

86 km/h = 23.88 m/s

It is required to find work done by the automobile.

Concept used : Work energy theorem

Th change in kinetic energy of an object is equal to the work done by it. The work done is then given by :

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

Here, v = 0

$W=-\dfrac{1}{2}mu^2\\\\W=-\dfrac{1}{2}\times 1150\times (23.88)^2\\\\W=-327896.28\ J$

$W=3.27\times 10^5\ J$

Therefore, the work done by the automobile is $-3.27\times 10^5\ J$ .

You might be interested in

Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2

lesya692 [45]

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

$F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C$

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

7 0

3 years ago

Wнιcн deмonѕтraтeѕ condυcтιon

jekas [21]

C.) cool feet walking across a hot pavement.

The reason because the other ones deals with radiation. Only C.) is the right answer because the feet is touching the hot pavement which is conduction.

3 0

3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

In general, how did the water pressure in the tank change when mass was added to the fluid?

MissTica

Answer:

As the height increases the pressure must increase.

Explanation:

When we add masses to the fluid, the amount of fluid in the tank increases, therefore its height increases and the pressure is described by the expression

P = ρ g h

where rho is constant for a given fluid and h is the height measured from the surface of the fluid.

As the height increases the pressure must increase.

3 0

3 years ago

22) The sensory cells associated with the deep layers of the epidermis are

finlep [7]

Answer:B. Merkel Cells

Explanation:

3 0

3 years ago

How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?​

How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?