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Minchanka [31]
2 years ago
8

How is the independent variable affected by the dependent variable

Physics
1 answer:
stiv31 [10]2 years ago
5 0

Answer:

A dependent variable is a variable that is tested in an experiment. An independent variable is that can be modified. Depending on what you are testing, the dependent variable will change accordingly to the dependent variable.

- I'm reading this back and it doesn't make much sense, if you want me to reword this I can

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Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0
2 years ago
a body that has a mass of 0.04 k g is thrown up with a speed of 60m/s. Kinetic energy? a) in the begining,b)after 6 seconds​
Katena32 [7]

Answer:

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8 0
3 years ago
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
2 years ago
Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? S
Zepler [3.9K]

Answer:

About two kilometers away

\rm distance=2.401\ km

Explanation:

Given:

The time gap between the light and sound to travel to the house, t=7\ s

<em>Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.</em>

<u>∴Distance between the lightning-strike and the house:</u>

\rm distance=v\times t

we have the speed of sound as: v=343\ m.s^{-1}

So,

\rm distance=343\times 7

\rm distance=2401\ m

\rm distance=2.401\ km

6 0
2 years ago
A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?
alexandr1967 [171]

Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

v=u+at

Here, train starts from rest so, u = 0

v=0+1.5\ m/s^2\times 5\ s  

v = 7.5 m/s

So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.

7 0
2 years ago
Read 2 more answers
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