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Vlad1618 [11]
3 years ago
6

What two forces are there when you skydive

Physics
2 answers:
dusya [7]3 years ago
7 0
Gravity is pulling you down and friction is slowing you down so you don't plummet to the ground at super high speeds.
Usimov [2.4K]3 years ago
7 0
1). The force of bravado, machismo, and testosterone, propelling you forward, to DO THIS and impress your woman.

2). The force of reason and self preservation, telling you that this is crazy, and you must NOT jump out of a perfectly good airplane that's not on the ground.
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Prank text my sister, I wanna see her reaction.<br><br> ‪(346) 298-3870‬
dlinn [17]

Answer:

she is going to be mad dude

4 0
2 years ago
1) According to the law of conservation of energy, A) an object loses most of its energy as friction. B) the total amount of ene
Blababa [14]
Your answer is C) <span> the potential energy of an object is always greater than its kinetic energy </span>
5 0
3 years ago
In his psychology class, Professor Adams said that fear of spiders and snakes are common in humans, but fear of automobiles is n
Svetradugi [14.3K]

Answer:

Evolutionary

Explanation:

Because our ancestors were afraid of these poisonous spiders and snakes which lead to them to telling their kids of these dangers. Their kids will fear them without even encountering any and this will also be passed on to their future generations.

5 0
3 years ago
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total di
S_A_V [24]

Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.

Now the spring is in series combination

F\alphae

F=ke

k=f/e.........*

where k is the force constant or the constant of proportionality

k=f/e

f_{eff} =f_{1} +f_{2}............................1

also for effective force constant

divide all through by extension

1) Total force is

Ft=F1+F2

Ft=k1e1+k2e2

F = k(e1+e2) 2)

Since force on the 2 springs is the same, so

k1e1=k2e2

e1=F/k1 and e2=F/k2,

and e1+e2=F/keq

Substituting e1 and e2, you get

1/keq=1/k1+1/k2

Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.

4 0
3 years ago
An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0
3 years ago
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