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Pachacha [2.7K]
3 years ago
10

Two examples of inventions that increase friction?

Physics
1 answer:
Ivan3 years ago
3 0

Answer:

My hard pp...............

Explanation:

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List the types of electromagnetic radiation in order from lowest energy photons to highest energy photons.
frutty [35]

radio waves,X-rays,

Explanation:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

8 0
2 years ago
This diagram of the first right-hand rule relates which two quantities?
AURORKA [14]

Option A, current (thumb) to magnetic field (fingers)

As per the First right-hand rule,

Using right hand, if we suppose that thumb points towards the electric current

fingers curl towards the magnetic field

7 0
3 years ago
Read 2 more answers
A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding
sveta [45]

Answer:

F1= 588 N

F2= 784 N

Explanation:

Please see the attached file.

5 0
3 years ago
Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t
givi [52]
For t1:

t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec

For t2:

t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec

Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)

d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m

Where:

d = hor. distance

ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25

The answer is 3.19 : 2.25
8 0
3 years ago
Read 2 more answers
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
3 years ago
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