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3 years ago

A ______ is a tiny packet of energy

Physics

1 answer:

Nataliya [291]3 years ago

5 0

<span>Individual packets or quanta of light energy are called Photons.
The answer is photons.
</span>

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Chanice drives her scooter 7 kilometers north. she stops for lunch and then drives 5 kilometers and then 1 km east again. what d

lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

$d = d_1 + d_2 + d_3$

$d = 7 km + 5 km + 1 km$

$d = 13 km$

Now in order to find the displacement we will show all with their directions

$\vec d_1 = 7 + 5 = 12 km$ towards North

$\vec d_2 = 1 km$ towards East

So total displacement is

$\vec d = \vec d_1 + \vec d_2$

$\vec d = 12 \hat j + 1 \hat i$

so net displacement will be

$d = \sqrt{12^2 + 1^2} = 12.04 km$

so displacement is 12.04 km

6 0

3 years ago

When an electron in a certain excited energy level in a one-dimensional box of length 2.00 Å makes a transition to the ground st

Fiesta28 [93]

Answer:

Calculate the wavelength associated with an electron with energy 2000 eV.

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J

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2 years ago

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

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2 years ago

I need helpppppp asap

asambeis [7]

It’s c because it’s not Control so that means that it would be broken and non fix able

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2 years ago

Calculate the speed of an 8.0x10^4 kg airliner with a kinetic energy of 1.1x10^9 j ...?

Over [174]

Kinetic energy is the energy possessed by an object on motion. it is expressed as follows:

KE = 0.5mv^2

where m is the mass and v is the velocity of the object. We calculate as follows:

KE = 0.5mv^2
1.1x10^9 J = 0.5(8.0x10^4 kg) v^2
v = 165.83 m/s

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3 years ago