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Harlamova29_29 [7]
3 years ago
11

Lowest frequency?_______​

Physics
1 answer:
Sergio039 [100]3 years ago
8 0

Answer:

Radio waves are the lowest frequency

Explanation:

Here is the order of frequencies highest to lowest to help you in the future:

gamma rays

X-rays

ultraviolet radiation

visible light

infrared radiation

radio waves

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Mass is a measurement of
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D is the amount of space object takes up
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3 years ago
You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do
melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

W = F_g .d

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

8 0
3 years ago
Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

8 0
3 years ago
What would be a good reason to increase friction between surfaces? to allow objects to slide past each other easily to reduce we
Lelu [443]
So basically the objects would be sandpaper and smooth metal, the sandpaper can indirectly touch the metal since it’s so smooth and it won’t cause any temp change either
5 0
3 years ago
Read 2 more answers
If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c
hichkok12 [17]

Answer:

4

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m_1 = Mass of Earth

m_2 = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

F_o=\dfrac{Gm_1m_2}{r^2}

New gravitational force

F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}

Dividing the equations

\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4

The ratio is \dfrac{F_n}{F_o}=4

The new force would be 4 times the old force

7 0
3 years ago
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