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madam [21]
3 years ago
5

Identify the constants and variables in 2cm +4y​

Physics
1 answer:
Harman [31]3 years ago
4 0

constant = 2 and 4

variable = y

Explanation:

A constant is usually a symbol with a known value. In the expression, the constants are 2 and 4.

Their values are accurately known.

A variable is place holder and the value is not known. Variables are not static. They can take different numerical values as defined by an equation:

                 2cm +4y​

  the cm is the a unit

      2 and 4 are constants

          y is a variable that can assume any value

Learn more:

Experiment brainly.com/question/1621519

#learnwithBrainly

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For every action, there is an equal and opposite reaction.

Explanation:

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3 years ago
A current of 12 amps is measured in a circuit with a total resistance of 9.0 ohms. What is the size of the voltage source that s
Serjik [45]
Data:
i (current) = 12 A
R (resistance) = 9.0 Ω
V (voltage) = ? (volts)

Formula:
V = R*i

Solving:
V = R*i
V = 9.0*12
\boxed{\boxed{V = 108\:volts}}\end{array}}\qquad\quad\checkmark
6 0
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If 0.250 L of a gas weighs 0.308 g under normal conditions of pressure and temperature, what is its molecular weight?
Phantasy [73]

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=28

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8 0
3 years ago
Two cars are heading towards one another. Car A is moving with an acceleration of aA = 4 m/s2. Car B is moving with an accelerat
Paladinen [302]

Answer:

Car B reaches car A in 19.7 s.

Explanation:

Hi there!

The equation of the position of an object moving in a straight line at constant acceleration is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration

When both cars meet, their positions are the same. At the meeting point:

position of car A = position of car B

xA = xB

x0A + v0A · t + 1/2 · aA · t² = x0B + v0B · t + 1/2 · aB · t²

Let´s place the origin of the frame of reference at the point where A is located. In that case x0A = 0 and x0B = 2900 m. Since both cars are initially at rest, v0A and v0B = 0. So, the equation gets reduced to this:

1/2 · aA · t² = x0B + 1/2 · aB · t²  

If we replace with the data we have and solve for t:

1/2 · 4 m/s² · t² = 2900 m - 1/2 · 11 m/s² · t²

2 m/s² · t² =  2900 m - 5.5 m/s² · t²

5.5 m/s² · t² + 2 m/s² · t² = 2900 m

7.5 m/s² · t² = 2900 m

t² = 2900 m / 7.5 m/s²

t = 19.7 s

Car B reaches car A in 19.7 s.

4 0
3 years ago
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