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dimulka [17.4K]
2 years ago
6

A space probe lands on a newly discovered planet. A small canister is released from the probe and falls a distance of 3 m in 0.5

sec. What is the value for the acceleration of gravity on this new planet?
Physics
1 answer:
ivanzaharov [21]2 years ago
3 0

Answer:

24m/s²

Explanation:

Given

Distance S = 3m

Time of fall = 0.5sec

Required

Acceleration due to gravity

Using the equation of motion

S = ut+1/2gt²

Substitute the given values

3 = 0+1/2g(0.5)²

3 = 1/2(0.25)g

3 = 0.125g

g = 3/0.125

g = 24

Hence the value for the acceleration of gravity on this new planet is 24m/s²

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podryga [215]

Answer:

The lightbulb will NOT light.

Explanation:

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This is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

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A 7.0kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal
sukhopar [10]

Answer:

16.1 N

Explanation:

From the question,

F = ma.............................. Equation 1

Where F = horizontal force, m = mass of the object, a = acceleration .

Given: m = 7.0 kg, a = 2.3 m/s²

Substitute this values into equation 1

F = (7.0×2.3)

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When a star is moving away the light waves appear to be....
BabaBlast [244]
When a star is moving away from earth it appears blue
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3 years ago
7. What does the slope of a position-time graph represent?
allochka39001 [22]

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7 0
3 years ago
How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
solong [7]

Answer:

71 rpm

Explanation:

Given that:

Angular momentum (L) = 0.26

Diameter = 25cm = 0.25 cm

Radius, r = (d/2) = 0.125m

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Moment of inertia (I) = 2mr² / 5

I = (2 * 5.6 * 0.125^2) / 5

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Angular speed (w) ;

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Angular speed = 70.94 rpm

= 71 rpm

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