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3 years ago

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In the photo below, astronaut Alan Bean works at the Apollo 12 lander. Describe the horizon and the surface you see. What kind o

f terrain did they land on for this, the second human Moon landing, and why ?

1 answer:

Marianna [84]3 years ago

3 0

Answer:

moon

Explanation:

I dont KNow

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A wrecking ball is suspended by two cables as shown below. If the tension in cable 2 is 12 000 N, what is the weight of the wrec

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Wrecking ball wow get Miley Cyrus also I’m sorry I’m joking bout it and you should report the person who be putting “links”

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3 years ago

A plane mirror and a concave mirror (f = 8.40cm) are facing each other and are separated by a distance of 22.0 cm. An object is

lesya [120]

Answer:

The distance of the image produced by the concave mirror is 11.27 cm.

Explanation:

Given that,

Focal length = 8.40 cm

Distance between object and plane mirror = 11.0 cm

Distance between plane mirror and concave mirror = 22.0 cm

We need to calculate the object distance

$u =11.0+22.0 = 33.0\ cm$

We need to calculate the distance of the image

Using formula of distance of the image

$v=\dfrac{uf}{u-f}$

Where, u = Object distance

f = focal length

Put the value into the formula

$v=\dfrac{33.0\times8.40}{33.0-8.40}$

$v=11.27\ cm$

Hence, The distance of the image produced by the concave mirror is 11.27 cm.

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3 years ago

Consider two balls on a horizontal, frictionless surface. Ball A (with mass ma) is moving toward ball B (with mass mb = 1 kg). B

Sonbull [250]

Answer:

Explanation:

When two ball of identical mass collides perfectly elastically , there is exchange of velocity between the two balls .

Here ball be was at rest initially . After collision ball A comes to rest , so there is complete exchange of velocity . Hence ball A must have same mass as that of B . mass of ball A = 1 kg .

b ) Due to complete exchange of velocity , velocity of ball A will be picked up by ball B . Hence velocity of ball B = 3.1416 m /s . Yes it will be moving in the direction of ball A .

c )

In case of perfectly inelastic collision they will become single mass

total mass = 2 kg

applying conservation of momentum law

their common velocity after collision = 1 x 3.1416 / 2 = 1.57 m /s

d )

Applying conservation of momentum law

initial momentum = Ma x va

they move in opposite direction after collision

their total momentum after collision

1 x va - Ma va

applying law of conservation of momentum

1 x va - Ma va = Ma va

va = 2Ma va

Ma = .5 kg .

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3 years ago

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3 years ago

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A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

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3 years ago