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2 years ago

10

a car's engine is only 12 % efficient at converting chemical energy in gasoline to mechanical energy. If it takes 18000 N of for

ce to keep the car moving at a constant speed of 21 m/s, how much chemical energy would be needed to move the car a distance of 450m at this speed?

1 answer:

Georgia [21]2 years ago

6 0

I wish i could help but idk the answer

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A stationary block resting on the ground is pulled up with a tension force of 100N, but does not leave the ground.

tekilochka [14]

Answer:

The normal force the ground exerts on the block, F = -300 N

Explanation:

Given data,

The block pulled up with a tension force, T = 100 N

The weight of the block, W = 300 N

The weight of the block is due to the force of attraction of gravitation.

The surface exerts a force that is equal and opposite to the force acting on the block due to gravitation.

The weight of the block,

W = mg

300 N

The normal force the ground exerts on the block,

F = - mg

= - 300 N

Hence, the normal force the ground exerts on the block, F = -300 N

6 0

3 years ago

. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

Dmitriy789 [7]

Answer:

The magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

Explanation:

Given :

Magnitude of magnetic field $B = 0.078$ T

Radius of circle $r = 0.10$ m

Angle between field and surface normal $\theta =$ 25°

From the formula of flux,

$\phi = B.A$

$\phi = BA\cos \theta$

Where $\theta =$ angle between magnetic field line and surface normal, $A =$ area of circular surface.

$A = \pi r^{2}$

$A = 3.14 \times (0.10) ^{2}$

$A = 0.0314$ $m^{2}$

Magnetic flux is given by,

$\phi = 0.078 \times 0.0314 \times \cos 25$

$\phi = 2.22 \times 10^{-3}$ Wb

Therefore, the magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

6 0

3 years ago

A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f

uysha [10]

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

$\Sigma F=ma$

Replacing the values:

$mg\sin 40=ma$

We may cancel out the mass:

$g\sin 40=a$

Using the gravity constant as 9.8 meters per square second:

$9.8\frac{m}{s^2}\sin 40=a$

Solving the operations:

$6.3\frac{m}{s^2}=a$

Therefore, the acceleration is 6.3 meters per square second.

8 0

9 months ago

What is the kinetic energy of a hammer that starts from rest and decreases its potential energy by 10 kJ?

erma4kov [3.2K]

Answer:

final kinetic energy of the hammer is 10 kJ

Explanation:

As we know that there is no non conservative force on the system

So here we can use the theory of mechanical energy conservation

So we will have

$\Delta K + \Delta U = 0$

here we know that

$\Delta U = - 10 kJ$

from above expression now

$K_f - K_i - 10 kJ = 0$

$K_f - 0 = 10 kJ$

so final kinetic energy of the hammer is 10 kJ

3 0

3 years ago

Perhaps to confuse a predator, some tropical gyrinid beetles(whirligig beetles) are colored by optical interference that is duet

gulaghasi [49]

Answer:

The grating spacing of the beetle is $1.078*10^{-6}m$

Explanation:

The concept to solve this problem is relate to interference effect given in the Young's Slits. Here was demonstrated that the length of the side labelled \lambda is known as the path difference. The equation is given by,

$n\lambda = dsin\theta$

Where,

$\lambda$ = wavelenght of light

N = a positive integer: 1,2,3...

$\theta$ = Angle from the center of the wall to the dark spot

d= width of the slit

Replacing our values we have that for n=1,

$dsin\theta = n\lambda$

$dsin(30)=(1)(539*10^{-9})$

$d=1.078*10^{-6}m$

Therefore the grating spacing of the beetle is $1.078*10^{-6}m$

6 0

2 years ago