Ask question

Ask question

All categories

4 years ago

14

An 8000-N car is traveling at 12 m/s along a horizontal road. When the brakes are applied, the car skids (slides) to a stop in 4

.0 s. Find the work done on the car.

1 answer:

kicyunya [14]4 years ago

5 0

Answer:

192000 J or 192 kJ

Explanation:

Work done = Force × distance moved along the direction of force.

From the question,

W = F×d............... Equation 1

Where W = work done by the car, F = Force of the car, d = distance move by the car along the direction of the force.

We can calculate for d using the equation of motion

d = (v+u)t/2..................... Equation 2

Where v = Final velocity of the car, u = Initial velocity of the car, t = time.

Given: u = 12 m/s, v = 0 m/s(skids to a stop), t = 4.0 s

Substitute into equation 2

d = (12+0)4/2

d = 24 m.

Also given: F = 8000 N

Substitute into equation 1

W = 8000×24

W = 192000 J

W = 192 kJ.

You might be interested in

Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de

ahrayia [7]

Answer:

a) El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

$\Delta z = \frac{v_{out}^{2}}{2\cdot g}$

Donde:

$\Delta z$ - Diferencia de altura, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

$v_{out} = \sqrt{2\cdot g \cdot \Delta z}$

Si $g = 9.807\,\frac{m}{s^{2}}$ y $\Delta z = 0.3\,m$ , entonces la rapidez de salida del chorro es:

$v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}$

$v_{out} \approx 2.426\,\frac{m}{s}$

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

$\dot V_{out} = v_{out}\cdot A_{t}$

Donde:

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

$A_{t}$ - Área transversal del orificio, medido en metros cuadrados.

$\dot V_{out}$ - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que $v_{out} = 2.426\,\frac{m}{s}$ y $A_{t} = 5\,cm^{2}$ , el caudal de salida del chorro es:

$\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)$

$\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}$

El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

5 0

3 years ago

Simon puts new batteries in his radio-controlled car and its controller. He activates the controller, which sends a radio signal

Kaylis [27]

Chemical potential energy to kinetic energy

3 0

4 years ago

A train travels 55km, south along a straight track in 34minutes. What is the train Ls average velocity in kilometers per hour

Westkost [7]

Answer:

96.5 km/h

Explanation:

The average velocity of the train is given by:

$v=\frac{d}{t}$

where

d is the displacement

t is the time taken

For this train, we have:

d = 55 km south (displacement is a vector, so we must also consider the direction)

$t= 34 min \cdot \frac{1}{60 min/h}=0.57 h$

Substituting into the equation, we find the average velocity:

$v=\frac{55 km}{0.57 h}=96.5 km/h$

5 0

3 years ago

A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr

MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

1/v=-1/24

v= - 24cm

This implies that the image is on the same side as the object and it is real

7 0

3 years ago

Read 2 more answers

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

4 years ago