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Alinara [238K]
3 years ago
10

A bicycle traveled 150 meters west from point A to point B. Then it took the same route and came back to point A. It took a tota

l of 2 minutes for the bicycle to return to point A. What is the average speed and average velocity of the bicycle? A. The average speed is 2.5 meters/second, and the average velocity is 2.5 meters/second east. B. The average speed is 0 meters/second, and the average velocity is 2.5 meters/second east. C. The average speed and average velocity are both 0 meters/second. D. The average speed is 2.5 meters/second, and the average velocity is 0 meters/second.
Physics
1 answer:
Dennis_Churaev [7]3 years ago
6 0

Answer

D. The average speed is 2.5 meters/second, and the average velocity is 0 meters/second.

The difference between speed and velocity is that, speed is a scalar quantity while velocity is a vector quantity.

Average speed = total distance/Total time

                          = (150 + 150) / (2×60)

                          = 300/120

                           = 2.5 m/s

Average velocity = Total displacement/ Total time

                             = (150 + -150) / (2 × 60)

                            = 0/120

                            = 0 m/s  

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The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a vel
Brilliant_brown [7]

Answer:

Explanation:

We shall write the velocities given in vector form to make the solution easy.

The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form

waV(e) = 2.2 cos 60 i + 2.2 sin 60 j

waV(e) = 1.1 i + 1.9 j

Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows

wiV(e) =  - 4.5 cos 50 i - 4.5 sin 50 j

wiV(e) = - 2.89 i - 3.45 j

Now we have to calculate velocity of wind with respect to water that is

wiVwa

wiV( wa) = wiV ( e)+ eV(wa)

=  wiV( e)- waV(e)

- 2.89 i - 3.45 j - 1.1 i -  1.9 j

= - 3.99 i - 5.35 j

Magnitude of this relative velocity

D² = 3.99² + 5.35²

d = 6.67 m /s

6 0
3 years ago
Una esfera de aluminio de 0,5cm de radio se introduce en alcohol. Calcular la indicación que marca el diámetro cuando esta dentr
klemol [59]

Answer:

sry I just wanted the points I'm in middle school so I don't know this stuff either but can you give free brainlyest I'm soo close to my next rank I'd really appreciate it if you would

6 0
3 years ago
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Vocabulary
katrin2010 [14]
We don’t have the terms
8 0
2 years ago
Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24
Oksi-84 [34.3K]
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}
Now At xp we have:
\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0
\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0
Which is a second order equation, using the quadratic formula to solve for xp would give us:
xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
or
xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

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Answer:

60ml

Explanation:

I'm going to assume you mean 210ml not centimeters. To find the volume all we do is subtract both values or with the formula [ f - i = v ]  where f = final amount and i = initial amount.

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3 years ago
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