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Nat2105 [25]
2 years ago
14

A galaxy containing substantial amounts of dark matter will.

Physics
2 answers:
pogonyaev2 years ago
5 0

Answer:

A galaxy containing substantial amounts of dark matter will spin faster

AveGali [126]2 years ago
4 0

Answer:

spin faster

Explanation:

-

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A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei
Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

F = (150\ N)Sin\ 20^o

<u>F = 51.3°</u>

8 0
2 years ago
What’s the answer for this problem?
pickupchik [31]
The answer is always true a



4 0
3 years ago
What is the relationship between distance and sound wave return time?
nekit [7.7K]
The time for the echo to return is directly proportional to the distance. vw = fλ. In a given medium under fixed conditions, vw is constant, so that there is a relationship between f and λ; the higher the frequency, the smaller the wavelength.
6 0
2 years ago
What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

6 0
3 years ago
How much tension is in a rope if it pulls a 5-kg bucket filled with water with an upward acceleration of 1m/s^2
liraira [26]
F=ma
Tension - weight = mass x acceleration
T - 5(9.81) = 5 x 1
T = 5 + 5(9.81)
T = 54.05 N
T ≈ 54 N
4 0
3 years ago
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