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3 years ago

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A red car with a mass of 3.0 kg traveling at 8 m/s collided with a blue car with a mass of 2.0 kg, which is at rest. The velocit

y of the blue car after the elastic collision is 9.6 m/s. What is the velocity of the red car?

1 answer:

anygoal [31]3 years ago

8 0

Answer:

460.8

Explanation:

just combine like terms

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Please help on this one? PLEASE.

Fantom [35]

Answer: The answer is D

Explanation:

6 0

3 years ago

A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg

dezoksy [38]

Answer:

For cast iron we have

$h = 0.92 m$

For copper

$h = 1.05 m$

For Lead

$h = 1.23 m$

For Zinc

$h = 2.43 m$

Explanation:

As we know that final speed of the block is calculated by work energy theorem

$W_f + W_g = \frac{1}{2}mv^2$

now we have

$-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2$

now we have

$v^2 = 2gh - 2\mu_k g h cot\theta$

$v = \sqrt{2gh(1 - \mu_k cot\theta)}$

For cast iron we have

$4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}$

$h = 0.92 m$

For copper

$4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}$

$h = 1.05 m$

For Lead

$4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}$

$h = 1.23 m$

For Zinc

$4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}$

$h = 2.43 m$

4 0

3 years ago

An object travels at a speed of 7500 cm/sec . how far will it travel in kilometers in one day

DiKsa [7]

Answer:

6480 km

Explanation:

The speed of the object is

v = 7500 cm/sec

We need to convert centimetres into kilometers and seconds into days. We have:

$1 cm = 1\cdot 10^5 km$

$1 s = \frac{1}{60\cdot 60 \cdot 24}d$

Using these conversion factors, we find:

$v=7500 \frac{cm}{s} \cdot 1\cdot 10^{-5} \frac{km}{cm}\cdot (24)(60)(60) \frac{s}{d}=6480 km$

3 0

3 years ago

You drop a pencil from your desk, which is 1 meter above the floor. How long does it take for the pencil to hit the floor? How f

vova2212 [387]

Answer:

1. 0.45 s.

2. 4.41 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) = 1 m

Time (t) =?

Velocity (v) =?

1. Determination of the time taken for the pencil to hit the floor.

Height (h) = 1 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.8

t² = 1/4.9

Take the square root of both side

t = √(1/4.9)

t = 0.45 s.

Thus, it will take 0.45 s for the pencil to hit the floor.

2. Determination of the velocity with which the pencil hit the floor.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 0.45 s.

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 0.45)

v = 0 + 4.41

v = 4.41 m/s

Thus, the pencil hit the floor with a velocity of 4.41 m/s

6 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