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3 years ago

HELP ASAP

Physics

2 answers:

ludmilkaskok [199]3 years ago

8 0

It would have both KINETIC ENERGY and POTENTIAL ENERGY.

andrezito [222]3 years ago

3 0

It has both kinetic and potential

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For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency

Angelina_Jolie [31]

Answer:

2.7 J

Explanation:

The energy of one photon is given by

$E=hf$

where

h is the Planck constant

f is the frequency

For the photons in this problem,

$f=6.8\cdot 10^9 Hz$

So the energy of one photon is

$E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J$

The number of photons contained in 1.0 mol is

$N_A = 6.022\cdot 10^{23} mol^{-1}$ (Avogadro number)

So the total energy of $N_A$ photons contained in 1.0 mol is

$E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J$

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2 years ago

In terms of newton first law of motion why is it important to wear a seat belt

lilavasa [31]

Answer:

Seatbelts stop you

Explanation:

Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts.

6 0

3 years ago

In this graph, what is the displacement of the particle in the last two seconds?

nikklg [1K]

In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?

0.2 meters
2 meters
4 meters
6 meters
In this graph, the displacement of the particle in the last two seconds is 2 meters.

7 0

3 years ago

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

2 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$