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Stella [2.4K]
1 year ago
5

PLEASE HELP

Physics
1 answer:
Paul [167]1 year ago
7 0
B would be your right answer. Because Jupiter’s gravity pull is much stronger than earths
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Inessa [10]

Its A according to Ed Genuity

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3 years ago
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What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?
Gekata [30.6K]

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

p = mv

Let's replace what we know and solve for whatever's left

100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s

7 0
3 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
3 years ago
Intelligence tests that are given to participants in groups
White raven [17]

<span><span>a. are typically paper-and-pencil measures.</span>


Similar with psychological tests, mostly structured personality tests.

Psychological tests comes two ways: </span>
<span>The structure psychological tests or, objectives tests and </span>unstructured psychological tests or, also called projective tests<span>. By what you are referring the responder strongly asserts a projective tests which in definition comes with an unambiguous stimuli or no paper test just drawings and images. If what the responder’s suggesting is correct you are referring to the </span>Rorschach projective tests, these tests are a figure symmetrically placed in an inkblot that lets you visualize or create a mental picture out of it, and makes you describe what you in see much detail as you can.<span>
</span>


4 0
3 years ago
A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much
Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0
3 years ago
Read 2 more answers
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