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4 years ago

A roller coaster car picks up speed as it rolls down the slope. As it starts down

Physics

1 answer:

vivado [14]4 years ago

4 0

Answer:

22 m/s

Explanation:

v(t) = Vo + at

v(3) = 4 + 6*3

v = 4 + 18

v = 22 m/s (almost 50 miles/hour)

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Verizon [17]

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wind=atmosphere
ice=cryosphere
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I think!!

8 0

3 years ago

A 1.5-cm object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal l

zlopas [31]

Answer:

The object for the converging lens is upright and 0.429 cm tall, the image of this converging lens is inverted and 1.375 cm high

Explanation:

Let

$d_{o}=distance of object[tex]\\f=focal length\\d_{i}=distance of image\\I_{h}=Image height$

For diverging lens:

$d_{o} = 0.50\\f = -0.20\\\frac{1}{d_{o}}+\frac{1}{-0.20}\\\frac{1}{d_{i}}=\frac{1}{-0.20}-\frac{1}{0.50}=-7\\d_{o}=-\frac{1}{7}$

Magnification = $\frac{d_{i}}{d_{o}}= -\frac{1}{7}÷ 0.5 = -0.286$

Image height $= -0.286 * 1.5 = -0.429 cm$ (negative sign means the image is virtual, inverted.

This image is $\frac{1}{7}$ meter to left of the center of the diverging lens.

The converging lens is located 0.08 m to the right of the diverging lens

The distance between the image of the diverging lens and center of the converging lens = $\frac{1}{7} + 0.08 = 0.229 m$

The image of the diverging lens becomes the object of the converging lens.

$d_{o} = 0.223\\f = 0.17\\\frac{1}{d_{i}}=\frac{1}{0.17}-\frac{1}{0.223}=0.715\\d_{i}=0.715m to the right of the converging lens$

$Magnification =\frac{d_{i}}{d_{o}} = \frac{0.715}{0.223}=3.206\\image height=3.206 * 0.429 = 1.375 cm.$

7 0

4 years ago

100 POINTS QUESTION. PLEASE PROVIDE EXPLANATION

melomori [17]

Answer:

t = 0, p = 12.0, v = 6.00

t = 20, p = 112, v = 56.0

t = 25, p = 62.0, v = 31.0

Explanation:

Impulse = change in momentum

J = Δp

FΔt = mΔv

The impulse equals the area under the F vs t graph.

At t = 0 s, the initial velocity is 6.00 m/s, so the momentum is:

p = (2.00 kg) (6.00 m/s)

p = 12.0 kg m/s

From t=0 s to t=20 s, the impulse is:

J = (20 N) (5 s) + ½ (20 N) (5 s) + ½ (-10 N) (10 s)

J = 100 Ns

So the new momentum is:

p = 12.0 kg m/s + 100 kg m/s

p = 112 kg m/s

And the new velocity is:

v = (112 kg m/s) / (2.00 kg)

v = 56.0 m/s

From t=20 s to t=25 s, the impulse is:

J = (-10 N) (5 s)

J = -50 Ns

So the new momentum is:

p = 112 kg m/s − 50 kg m/s

p = 62.0 kg m/s

And the new velocity is:

v = (62.0 kg m/s) / (2.00 kg)

v = 31.0 m/s

4 0

4 years ago

Read 2 more answers

How does the light-collecting area of an 8-meter telescope compare to that of a 2-meter telescope? How does the light-collecting

IceJOKER [234]

Answer:

The 8-meter telescope has 16 times the light-collecting area of the 2-meter telescope.

Explanation:

Area of 8 meter telescope

$A_1=\pi \frac{D^2}{4}\\\Rightarrow A_1=16\pi$

Area of 2 meter telescope

$A_2=\pi \frac{D^2}{4}\\\Rightarrow A_1=1\pi$

Dividing the equations we get

$\frac{A_1}{A_2}=\frac{16\pi}{1\pi}\\\Rightarrow A_1=16A_2$

Hence, the 8-meter telescope has 16 times the light-collecting area of the 2-meter telescope.

7 0

3 years ago

A particle moves along the x-axis so that at any time t≥0 its position is given by x(t)=12(a−t)2, where a is a positive constant

DanielleElmas [232]

Answer:

The particle moves to the right when 0 ≤ t < a

Explanation:

Since at any time t≥0 its position is given by x(t) = 12(a−t)², the particle moves to the right if x(t) > 0

So, x(t) > 0 ⇒ 12(a−t)² > 0 ⇒ (a−t)² > 0

So, (a−t)² > 0

taking square-root of both sides, we have

√(a−t)² > √0

⇒ (a−t) > 0

⇒ a > t

Since t ≥ 0 ⇒ a > t ≥ 0 ⇒ 0 ≤ t < a

<u>So, the particle moves to the right when 0 ≤ t < a</u>

8 0

3 years ago