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FromTheMoon [43]
3 years ago
9

The si unit of measurement is a way for scientists to?

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
7 0
Have a universal record base. Everyone is able to understand the data compiled since the same measurement systems are being used around the world. This is just to simplify all of the information.
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Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , mea
lapo4ka [179]

by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

m*2*R + Iw = 0

w = -\frac{2mR}{I}

so merry go round will turn in opposite direction with above speed

6 0
3 years ago
Which planet is smaller than Earth’s core?
makvit [3.9K]

Answer:

The answer <em><u>is C. Mars</u></em>. Mars and Mercury are both smaller than Earth's core. Hope this helps you :)

7 0
3 years ago
Read 2 more answers
a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o
Stels [109]

Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

F = ma

F = (1210 kg) (-4.10 m/s²)

F = -4960 N

The magnitude of the force is 4960 N.

4 0
3 years ago
Read 2 more answers
What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

7 0
3 years ago
30 points plz help ill do anything... literally anything.
ruslelena [56]

Answer:

1. 2.5s

Explanation:

1. For time, divide Distance / speed

25m / 10

=2.5s

3 0
3 years ago
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