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levacccp [35]
3 years ago
14

Need help with this! just scienceeeeeee

Physics
1 answer:
LiRa [457]3 years ago
8 0

Answer:

7)  λ = 0.5 m,  8)  f = 4.8 10¹⁴ Hz

Explanation:

The speed of an electromagnetic wave is

          c = λ f

where c is the speed of light in vacuum c = 3 10⁸ m / s

7) indicate the frequency f = 6.0 10⁸ Hz

we do not know the wavelength

         λ = c / f

       

we calculate

        λ = 3 10⁸ / 6.0 10⁸

        λ = 0.5 m

8) indicate the wavelength  λ = 6.25 10-7 m

we do not know the frequency

         f = c / λ

we calculate

        f = 3 10⁸ / 6.25 10⁻⁷

        f = 0.48 10¹⁵ Hz

        f = 4.8 10¹⁴ Hz

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Do you think seismographs predict earthquakes or measure earthquakes, explain your answer?
Yuki888 [10]

Answer:

No. Neither the USGS nor any other scientists have ever predicted a major earthquake. We do not know how, and we do not expect to know how any time in the foreseeable future.

5 0
3 years ago
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What is an example of a sting nuclear force ?
slavikrds [6]

Answer:

A.) stability of an atom

Explanation:

5 0
3 years ago
Can someone plz help me with this
Elena-2011 [213]
1st Law: Objects that are in motion tend to stay in motion. This motion can change with external forces. 

<span>If you were to stop pedaling on bike while in motion, you will notice that you will keep moving. This is because a moving body (you) has inertia. If there wasn't any friction between the tires and the ground, between the axles and wheel, any air resistance, or any other force that acts against you, then you could be coasting indefinitely! </span>

<span>2nd Law: Force is equal to the mass times acceleration. </span>

<span>When you pedal, you are applying a force onto the pedal. This force is then translated through tension to apply torque onto the wheel. Turning the wheel will make you accelerate in the lateral direction. </span>

<span>3rd Law: For every action, there is an equal and opposite reaction. </span>

<span>Without this, you could pedal and pedal, but you will be not go anywhere! It is essentially the friction between the tires and the ground that propels you forward. If the ground did not apply to the tire the same amount of force that the tire was applying to the ground, the tire would not "catch" and no friction would be applied. And if there was no third law, the weight of you and your bike would "sink" into the ground because the ground would not be applying a normal force back onto you.

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3 0
3 years ago
What is the velocity of a narwhal that swims 76 kilometers North in 3 hours
tino4ka555 [31]

Answer:

<h3>25km/hr</h3>

Explanation:

Velocity is the change in displacement of a body with respect to time.

Velocity - Displacement/time

Given

displacement = 76km

Time = 3hours

Substitute the given parameters into the formula;

Velocity = 75km/3hrs

Velocity = 25km/hr

Hence the velocity of the narwhal is 25km/hr

4 0
4 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
4 years ago
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