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zmey [24]
3 years ago
7

A 7kg object is dropped on Earth,. Assuming it has not yet hit the ground, what is the velocity of the object after 2 seconds of

falling?

Physics
2 answers:
kirza4 [7]3 years ago
7 0
The answer would be 9.8 meters/ sec^2.

kap26 [50]3 years ago
3 0
Based on my knowledge the answer would be 9.8 sec
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Monochromatic electromagnetic radiation illuminates an area of a surface. The electric and magnetic fields of the waves are then
emmasim [6.3K]

Answer:

c) True. factor four

Explanation:

The energy density of an electromagnetic wave is given by

             u = ½ ε₀ E² = B² / 2μ₀

Where ε₀ is the dielectric constant and μ₀ the magnetic permittivity.

Let's apply this equation to the present case

 

If we double the electro field

               E ’= 2 E₀

               u’= ½  ε₀ (2E₀)²

               u’= ½ ε₀ Eo² 4

               u’= 4 u₀

Therefore the energy is multiplied by four

Let's check the answers

a) False

b) False

c) True

d) False

e) False

6 0
3 years ago
Ley de Coulomb para describir y predecir fuerzas gravitacionales y electromacneticas​
chubhunter [2.5K]

Answer:

Explanation:

i dont speak spanish :P

4 0
3 years ago
PLEASEE I NEED HELP FAST!!! .Study the scenario.A small container of water with a low temperature is poured into a large contain
RSB [31]

Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

             Q_c = m ce (T_{f}-T₀)

Heat released by the largest container is

              Q_a = M ce (T_{i}-T_{f})

how

        Q_c = Q_a

       m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

Of the final statements, the one that best describes the process is C

since it talks about the thermal energy and the heat that is transferred in the process

8 0
4 years ago
A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass
arlik [135]

Answer:

a) v_{60^{o}} =4.98 m/s

b) \theta_{max}=79.34^{o}

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

K_{lowest}=K_{60^{o}}+U_{60^{0}}

So now, we can substitute the respective equations for kinetic and potential energy so we get:

\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}

we can divide both sides of the equation into the mass of the pendulum so we get:

\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}

and we can multiply both sides of the equation by 2 to get:

v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}

so we can solve this for v_{60^{o}}. So we get:

v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

cos \theta = \frac{y}{4m}

so we solve for y to get:

y = 4cos \theta

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

h_{60^{o}}=4m-2m

h_{60^{o}}=2m

So now we can substitute the data in the velocity equation we got before:

v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}

v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}

so

v_{60^{o}}=4.98 m/s

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

K_{lowest}+U_{lowest}=K_{max}+U_{max}

In this case, we know that U_{lowest} will be zero and K_{max} will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

K_{lowest} =U_{max}

And now we substitute for the respective kinetic energy and potential energy equations.

\frac{1}{2}mv_{lowest}^{2}=mgh_{max}

again, we can divide both sides of the equation into the mass, so we get:

\frac{1}{2}v_{lowest}^{2}=gh_{max}

and solve for the height:

h_{max}=\frac{v_{lowest}^{2}}{2g}

and substitute:

h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}

to get:

h_{max}=3.26m

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

cos \theta_{max}= \frac{0.74}{4}

\theta_{max}=cos^{-1}(\frac{0.74}{4})

so we get:

\theta_{max}=79.34^{o}

5 0
3 years ago
What is the wavelength of an electromagnetic wave?
gizmo_the_mogwai [7]

Answer: (A) The number of waves that pass a point in a certain amount of time.  

Explanation:

Electromagnetic waves are categorized according to their frequency f or, equivalently, according to their wavelength λ = c/f. Visible light has a wavelength range from ~400 nm to ~700 nm.

Frequency: Is the number of waves that pass a certain point in a specified amount of time.

Trough: The low point of the wave cycle.

Wavelength: The distance between two successive peaks.

Electromagnetic wave: One of the waves that are propagated by simultaneous periodic variations of electric and magnetic field intensity.

6 0
2 years ago
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