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1 year ago

A periodic wave has a wavelength of 0,50 m and a speed of 20 m/s, What is the wave frequency?

Physics

1 answer:

faltersainse [42]1 year ago

6 0

Answer:

u=speed, w=wavelenght, f=frequency

It's known that u=w*f => f=u/w

u=20m/s ==> f=20/0,5 => f=40 Hz

w=0,50m

Explanation:

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What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?

rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

$E_{k}=\frac{1}{2} *m*v^{2}$

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

For the person running

$E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]$

For the bullet

$E_{k} =\frac{1}{2} *m*v^{2}$

$E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]$

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0

2 years ago

Pls help me fast and correct

bekas [8.4K]

Answer:

Explanation:

due to position at height is the answer

4 0

2 years ago

One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from

Luba_88 [7]

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

$s_1 = v_1t = 8t$

Similarly the distance traveled by the 2nd skater after t seconds is

$s_2 = v_2t = 10t$

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

$s_2 = s_1 + 80$

We can substitute $s_1 = 8t, s_2 = 10t$

$10t = 8t + 80$

$2t = 80$

$t = 80 / 2 = 40 s$

7 0

2 years ago

A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and

MariettaO [177]

Answer:

$f=140\ N$

Explanation:

Given:

mass of the object on a horizontal surface, $m=50\ kg$

coefficient of static friction, $\mu_s=0.3$

coefficient of kinetic friction, $\mu_k=0.2$

horizontal force on the object, $F=140\ N$

Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:

$F_s=\mu_s.N$

where:

$N=$ normal force of reaction acting on the body= weight of the body

$F_s=0.3\times (50\times 9.8)$

$F_s=147\ N$

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

$f=140\ N$

8 0

3 years ago

The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of

Korvikt [17]

Answer:

The mass of the earth, $M=6.023\times 10^{24}\ kg$

Explanation:

It is given that,

Time taken by the moon to orbit the earth, $T=27.3\ days=2358720\ m$

Distance between moon and the earth, $r=384000\ km=384\times 10^6\ m$

We need to find the mass of the Earth using Kepler's third law of motion as :

$T^2=\dfrac{4\pi^2}{GM}r^3$

$M=\dfrac{4\pi^2r^3}{T^2G}$

$M=\dfrac{4\pi^2\times (384\times 10^6)^3}{(2358720)^2\times 6.67\times 10^{-11}}$

$M=6.023\times 10^{24}\ kg$

So, the mass of the earth is $6.023\times 10^{24}\ kg$ . Hence, this is the required solution.

7 0

3 years ago