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marishachu [46]
4 years ago
5

A diver dives off of a raft - what happens to the diver? the raft? how does this relate to newton's third law? action force: ___

____________ reaction force: _____________
Physics
2 answers:
svetlana [45]4 years ago
7 0

1.      The diver transfers forward and dives into the water.

2.      The raft transfers backwards in the water for the reason that of the reaction force.

3.      The action force is the diver pushing off of the raft; and the

4.      reaction force is the raft pushing back on the diver (causing the diver to go forward and into the water)

kondaur [170]4 years ago
5 0
<span>Actually newtons third law says for every action there is an equal and opposite reaction, Hence here in this case, the diver diving of a raft is the action, after which surely reaction should come in the form where the raft and the driver will rebound with same speed back, and hence here the action force is diving and reaction force is rebounding from the diving place, with same intensity.</span>
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A train travels 90 kilometers in 1 hours, and then 60 kilometers in 3 hours. What is its average speed
Serhud [2]
Average speed = (total distance) / (total time to cover the distance)

                       = (90km + 60km) / (1hr + 3hr)

                       =       (150 km)   /   (4 hr)

                       =          37.5 km/hr      . 
7 0
3 years ago
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

6 0
2 years ago
Write 8.54 x 103 in standard notation<br> a. 854,000<br> b. 85,400<br> c. 8,540<br> d. 854
weeeeeb [17]
B. 8540, (assuming you meant 8.54 x 10 to the power 3) you move the decimal point 3 places to the right, if the decimal point can't pass any more numbers add zeros. If it confuses you to think of it this way do 10 cubed first, i.e. 10 x 10 x10 = 1000 and then do 8.54 x 1000 if thats easier for you.
5 0
3 years ago
Melting Point of Substances
frez [133]
<span>(kg) Melting Point of Tin

</span>
5 0
3 years ago
Read 2 more answers
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.5 m&gt;s. Two seconds later
ivanzaharov [21]

Answer:

4.28 s

Explanation:

after two seconds (2 s) His friends is

d = 3.5 m/s x 2 s = 7 meter ahead.

in this state, a bicylist start from initial velocity vo = 0 m/s and accelerat 2.4 m/s²

then, when bicylist reach his friend

t friend = t bicyclist = t

d bicylist = d friend + d

-------

d friend = 3.5 . t

d bicylist = vo . t + ½ a t²

d friend + d = vo . t + ½ a t²

3.5 t + 7 = 0 . t + ½ . 2.4 . t²

3.5 t + 7 = 1.2 t²

0 = 1.2 t² - 3.5 t - 7

t = -1.363 and t = 4.28

take the positive one

6 0
3 years ago
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