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BabaBlast [244]
3 years ago
12

The winning time for a 500.0 mile circular race track was 3 hours and

Physics
1 answer:
lyudmila [28]3 years ago
3 0

Explanation:

We have,

Distance traveled in a circular track is 500 miles

The winning time was 3 hours and  13 minutes. It means time is 3.217 hours.

The driver's average speed is given by total distance divided by total time taken. Its formula can be written as :

v=\dfrac{d}{t}\\\\v=\dfrac{500\ miles}{3.217\ h}\\\\v=155.42\ mph

At the end of the race, the driver reaches the point form where he has started. It means the displacement of the driver is equal to 0. Hence, driver's average velocity is equal to 0.

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Explanation:

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When cleaning up a local beach, students found many different particles that were in the water that affected the shoreline, like
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The correct answer is D,

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Movement of a molecule against its concentration gradient can occur through Select one or more: a. facilitated diffusion b. pass
uranmaximum [27]

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The answer to your question is:

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5 0
3 years ago
~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

8 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
IgorLugansk [536]

Answer:

photo one- refraction

photo two- diffraction

photo three- reflection

8 0
3 years ago
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