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elena55 [62]
3 years ago
10

A car traveled at an average speed of 60 mph for two hours. How far did it travel?

Physics
2 answers:
Virty [35]3 years ago
7 0
Speed=distance/time so answer is 120 miles
Akimi4 [234]3 years ago
6 0
The answer is C- 120 Miles.
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Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li
melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate Q of a hypodermic needle:

Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}  (1)

Where the dimensions of each one is:

Volume flow rate Q=\frac{L^{3}}{T}

Radius of the needle R=L

Length of the needle L=L

Pressures at opposite ends of the needle P_{2} and P_{1}=\frac{M}{LT^{2}}

Viscosity of the liquid \eta=\frac{M}{LT}

We need to find the value of n whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}  (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}  (3)

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}  (4)

As we can see n must be 4 if we want the exponent to be 3:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}  (5)

Finally:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}  (6)

8 0
3 years ago
Explain why you can only change one variable in an experiment
ohaa [14]
The purpose of an experiment is to LEARN the EFFECT of something.

The way you do that is to CHANGE the thing and see what happens.

You can change as many things as you want to.  But If you change
TWO things and observe the result, then you don't know which one
of them caused the effect you see. 

Or maybe BOTH of them working together caused it.  You don't know. 

So your experiment is not really much good.  You need to do it again.
5 0
3 years ago
You need to connect your doorbell to a battery. You should select wire that will
Harlamova29_29 [7]
Run electrity through or is postive to the circuit
6 0
3 years ago
An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1
goblinko [34]

Answer:

v"_{1} = v_{1} tanΘ

v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

Θ = tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )

Explanation:

Applying the law of conservation of momentum, we have:

Δp_{x = 0}

p_{x} = p"_{x}

m_{1}v_{1} = m_{2}v"_{2} cosΘ (Equation 1)

Δp_{y} = 0

p_{y} = p"_{y}

0 = m_{1} v"_{1} - m_{2} v"_{2} sinΘ (Equation 2)

From Equation 1:

v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

From Equation 2:

m_{2} v"_{2}sinΘ = m_{1} v_{1}

v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }

Replacing Equation 3 in Equation 4:

v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}

v"_{1}=v_{1}\frac{sinΘ}{cosΘ}

v"_{1}=v_{1}tanΘ (Equation 5)

And we found Θ from the Equation 5:

tanΘ=\frac{v"_{1}}{v_{1}}

Θ=tan^{-1}(\frac{v"_{1}}{v_{1}})

7 0
3 years ago
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