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givi [52]
3 years ago
7

How would the number 13,900 be written using scientific notation?

Physics
1 answer:
Roman55 [17]3 years ago
3 0
1.39*10^4
...........
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A ball with a weight of 0.5 N is submerged under water and then released. There is a net force of 5 N upwards. what is the buoya
Natasha2012 [34]

Answer:

4.5 N upward

Explanation:

You take the net force and subtract it from the weight

~DjMia~

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1) If you release a rubber band that had 10 units of elastic energy, 12 units of movement energy cannot be produced. Why not?
Mila [183]

Answer:

press dat crown for meh

Explanation:

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Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock
pashok25 [27]

Answer:

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

c) H=294300\ m

d) t_T=544.95\ s

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, a=2g=2\times 9.81=19.62\ m.s^{-2}

time for which the rocket accelerates, t_a=100\ s

<u>For the course of upward acceleration:</u>

using eq. of motion,

s_a=ut+\frac{1}{2}at_a^2

where:

u= initial velocity of the rocket at the launch =0

s_a= height the rocket travels just before its fuel finishes off

so,

s_a=0+\frac{1}{2}\times 19.62\times 100^2

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

v_a=u+at_a

v_a=0+19.62\times 100

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

v^2=v_a^2-2gh

where:

g= acceleration due to gravity

v= final velocity of the rocket at the top height

0^2=1962^2-2\times 9.81\times h

h=196200\ m

c) So the total height at which the rocket gets:

H=h+s

H=196200+98100

H=294300\ m

d)

Time taken by the rocket to reach the top height after the fuel is over:

v=v_a+g.t

0=1962-9.81t

t=200\ s

Now the time taken to fall from the total height:

H=v.t'+\frac{1}{2}\times gt'^2

294300=0+0.5\times 9.81\times t'^2

t'=244.95\ s

Hence the total time taken by the rocket to strike back on the earth:

t_T=t_a+t+t'

t_T=100+200+244.95

t_T=544.95\ s

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0
3 years ago
Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

4 0
3 years ago
ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo
vaieri [72.5K]
Distance is the total length covered = 2m + 3m = 5m

Displacement is his distance from original position.

Displacement =  2m + (-3)m.               Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement  is 1m behind his original starting point.
4 0
3 years ago
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