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3 years ago

How far does light travel in the time it takes sound to travel 1 cm in air at 20°c?

1 answer:

8 0

The speed of sound at $20^{\circ}C$ is approximately v=343 m/s. The distance covered by the sound wave is
$s=1 cm=0.01 m$
And the time it takes is
$t= \frac{S}{v}= \frac{0.01 m}{343 m/s}=2.9 \cdot 10^{-5} s$

Now we want to find how far the light travels during this time. Light travels at speed $c=3 \cdot 10^8 m/s$ , therefore the distance it covers during this time is
$S=ct = (3 \cdot 10^8 m/s)(2.9 \cdot 10^{-5} s)=8700 m$

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The body weighing 2 kg moves through the horizontal surface and crosses the path x = 75 cm The coefficient of friction of the bo

Taya2010 [7]

The kinetic energy of the body in definitive position is 4.24 J.

Explanation:

As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.

Frictional force = Coeffficent of friction × Normal force.

As the weight of the body is 2 kg, the normal force acting on it will be mass multiplied with acceleration due to gravity.

Frictional force = - 0.8×9.8 × 2 =-15.68 N

So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.

Work done = Frictional force × Displacement

Work done = -15.68×0.75 = -11.76 J.

So the work is done by the object.

If the kinetic energy of the body at starting is 16 J, then the kinetic energy of the body at definitive position will be obtained as below.

Work done = change in kinetic energy

-11.76 J = Final kinetic energy-16 J

Final Kinetic energy = - 11.76+16

Final kinetic energy = 4.24 J

Thus, the kinetic energy of the body in definitive position is 4.24 J.

3 0

3 years ago

A radio station broadcasts at a frequency of 600 kHz. Knowing that radio waves have a speed of 300 000 000 m/s, what is the wave

soldier1979 [14.2K]

Answer:

ccvtesgdujtdchgdrgggggggfrrrtyfaasdddfffghgdshh

6 0

3 years ago

4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?

aleksandrvk [35]

The resultant force on the object is

∑ F = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N

which has a magnitude of

F = √((6 N)² + (8 N)²) = √(100 N²) = 10 N

By Newton's second law, the acceleration has magnitude a such that

F = m a

10 N = (2 kg) a

a = (10 N) / (2 kg)

a = 5 m/s²

so the answer is B.

4 0

3 years ago

a rectangular tank measures 12.5 metres long 10.0 wide and 2.0 metre high calculate the mass of the water in the tank when it is

Tom [10]

Answer:

250000kg

Explanation:

Mass =density * volume

Density=1000kg/m³

Volume=l*w*h

12.5*10.0*2.0=250m³

Mass=1000 *250

=250000kg

7 0

3 years ago

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

3 years ago