All categories

3 years ago

A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =

1 answer:

8 0

   Kinetic energy = (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.

KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

   = (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²

   = (725.5) · ( 40/3 )² · ( kg·m² / sec²)

   = 128,978 joules (rounded)

You might be interested in

Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the

kifflom [539]

Answer:

Height will be 3.8971 m

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity $\omega =4.46rad/sec$

Moment of inertia is given by $I=\frac{1}{2}mr^2$

Transnational Kinetic energy is given by $KE=\frac{1}{2}mv^2$ as we know that v = $v=\omega r$

So $KE=\frac{1}{2}m(\omega r)^2$

Rotational kinetic energy is given by $KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2$

Potential energy is given by mgh

According to energy conservation

$mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2$

$h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m$

3 0

3 years ago

Which of the following is a pair of vector quantities?

hammer [34]

Answer:

velocity and displacement answer

Explanation:

thanks me

6 0

3 years ago

Read 2 more answers

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

4 years ago

Read 2 more answers

Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.91 m, a 2.95-kg object is at +2.43 m, a 2.41-kg

Anna [14]

Answer:

0.958 m

Explanation:

So the total mass of the system is

M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28 kg

let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation

$My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4$

$11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806$

$y = \frac{10.806}{11.28} = 0.958 m$

So the center of mass is 0.958 m from the origin

3 0

3 years ago

Unpolarized light with an average intensity of 845 W/m2 enters a polarizer with a vertical transmission axis. The transmitted li

RideAnS [48]

The concept to develop this problem is the Law of Malus. Which describes what happens with the light intensity once it passes through a polarized material.

Mathematically this can be expressed as

$I = I_0 cos^2\theta$

Where

I = New intensity after pass through the Polarizer

$I_0$ = Original intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

When the light passes perpendicularly through the first polarizer, the light intensity is reduced by half which will cause the intensity to be $225W / m ^ 2$ at the output of the new polarizer, mathematically:

$I= \frac{I_0}{2} cos^2\theta$

$225 = \frac{845}{2}cos^2\theta$

Solving to find the angle we have

$\theta = 43.11\°$

The orientation angle of the second polarizer relative to the first one is 43.11°

5 0

3 years ago