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2 years ago

Describe the energy transformations that occur from the time a skydiver jumps out of a plane until landing on the ground.

2 answers:

7 0

I think, the skydiver has maximum potential energy before jumping. However, as the skydiver jumps and decreases in height, he increases the velocity. Thus, as soon as the skydiver has jumped he is experiencing kinetic energy. The velocity of the skydiver increases as he moves down wards due to gravitational acceleration until he attains terminal velocity.

icang [17]2 years ago

4 0

Sample Response: Before jumping from the plane, the skydiver has potential energy. When the skydiver jumps, the potential energy is transformed into kinetic energy, which increases until the skydiver reaches terminal velocity. Potential energy is then transformed into thermal energy.

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Monochromatic light is incident on a grating that is 75 mm wide and ruled with 50,000 lines. the second-order maximum is seen at

DIA [1.3K]

Answer:

The wavelength of the incident light is $\lambda =$ 400 nm

Explanation:

Given data

Distance between the sits

$d = \frac{0.075}{50000}$

d = 1.5 × $10^{-6}$ m

$\theta = 32.5$ °

m = 2

We know that the wavelength of the incident light is given by

$\lambda = \frac{d\sin \theta}{m}$

Put all the value in above formula we get

$\lambda = \frac{1.5 (\sin 32.5)}{2}$ × $10^{-6}$

$\lambda =$ 4 × $10^{-7}$ m

$\lambda =$ 400 nm

Therefore the wavelength of the incident light is $\lambda =$ 400 nm

4 0

2 years ago

An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector

Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

$v_{x}$ = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

$t = \frac{X}{v_{x}}$

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel = 3.67 x 10⁻⁸ sec

$v_{y}$ = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = $v_{y}$ t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0

2 years ago

The velocity of a 10.0 kg object that has 720 J of kinetic energy is m/s. (Report the answer to two significant figures.)

nikdorinn [45]

Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. It is expressed as:

KE = mv^2 /2

720 = 10.0v^2 /2
v = 12 m/s

Hope this answers the question. Have a nice day.

4 0

2 years ago

Read 2 more answers

Consider a loop of wire placed in a uniform magnetic field. Which factors affect the magnetic flux Φm through the loop?

dimaraw [331]

Answer:

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

Explanation:

Magnetic flux is the scalar product of the magnetic field over the area

Ф = ∫ B. dA

where B is the magnetic field and A is the area

Let's look at stationary, for which factors affect flow

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

4 0

2 years ago

What is the average momentum of a 70-kg runner who covers 400 m in 50 s?

kykrilka [37]

Answer:

560 kg m/s

Explanation:

First of all, we have to find the velocity of the runner, which is given by the ratio between the distance covered (400 m) and the time taken (50 s):

$v=\frac{d}{t}=\frac{400 m}{50 s}=8 m/s$

And now we can calculate the average momentum of the runner, which is equal to the product between the mass of the runner (70 kg) and its velocity, that we have previously calculated:

$p=mv=(70 kg)(8 m/s)=560 kg m/s$

8 0

2 years ago