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Readme [11.4K]
3 years ago
8

How do you draw two vectors to represent two forces that balance each other?

Physics
1 answer:
puteri [66]3 years ago
5 0

If two forces "balance" each other, that means that they add up to zero.

THAT means that they must have exactly equal magnitudes and exactly opposite directions.

If you want to draw a picture of a pair of two balanced force vectors . . . .

1).  Draw two arrows with exactly the same length.

2).  Flip one of them over, so that it points in exactly the opposite direction compared to the first one.

3).  Place them tail-to-tail.

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OlgaM077 [116]
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6 0
3 years ago
What is the mass of a liquid having a density of 1.50 g/ml and a volume of 3.5 liters?
PIT_PIT [208]

We know the formula for density = Mass/ volume

So     Mass, M = Volume * Density

         Volume = 3.5 L= 0.0035m^3

         Density = 1.50 g/ml  = 1500 kg/m^3

         Mass, M =  0.0035*1500 = 5.25 kg

So mass of liquid having density 1.50 g/ml and volume 3.5 liters is 5.25 kg.

6 0
4 years ago
A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass
stellarik [79]
Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s²  (nearest hundredth)

7 0
4 years ago
Помогите плз!
Softa [21]

Все написано в скобках правильно

3 0
3 years ago
If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1
Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

I=\frac{Q}{t}

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
3 0
1 year ago
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