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horsena [70]
2 years ago
7

A bungee jumper attains a speed of 30 m/s just as the bungee cord begins to stretch. If the period of stretch is 2 s while comin

g to a halt, the jumper's average deceleration in g’s is about ______?
Physics
1 answer:
Bogdan [553]2 years ago
4 0

Answer:1.53g

Explanation:

average deceleration= ?

inial velocity: u=0

final velocity: v=30m/s

time: t=2seconds

The first law of kinematics :

v=u+at

find a the subject of the formula

a=v-u/ t

a=\frac{0-30} 2

a=-30/2

a=-15m/s^{2}

The deceleration about g(acceleration due to gravity) will be:

15/9.8

1.53g

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The chemical formula for water, H2O, means that each water molecule contains
Ainat [17]
The chemical formula for water, H2o means that each water molecule contains one oxygen atom and two hydrogen atoms. This is the formula for water which has a liquid form, a solid form as ice, and also a gaseous form as water vapor. 
5 0
2 years ago
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Scotesia swims from the north end to the south end of a 50.0 m pool in 20.0 s. As she begins to make the return trip , Sean, who
slega [8]

Answer:

a) 2.5m/s

b) 0.91m/s

c) 0m/s

Explanation:

Average velocity can be said to be the ratio of the displacement with respect to time.

Average speed on the other hand is the ratio of distance in relation to time

Thus, to get the average velocity for the first half of the swim

V(average) = displacement of first trip/time taken on the trip

V(average) = 50/20

V(average) = 2.5m/s

Average velocity for the second half of the swim will be calculated in like manner, thus,

V(average) = 50/55

V(average) = 0.91m/s

Average velocity for the round trip will then be

V(average) = 0/75, [50+25]

V(average) = 0m/s

3 0
2 years ago
Suppose you want to make a scale model of a hydrogen atom. You choose, for the nucleus, a small ball bearing with a radius of 1.
RoseWind [281]

Answer:

A)  x _electron = 0.66 10² m , B)   x _Eart = 1.13 10² m , C)  d_sphere = 1.37 10⁻² mm

Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

                x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

               x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

               R_{E} = 6.37 10⁶ m

                tex]R_{Sum}[/tex] = 6.96 10⁸ m

                Distance = 1.5 10¹¹ m

                x_Earth = 1.5 10¹¹ / 6.96 10⁸  1.5

                x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

            d_sphere = 1.5 / 6.96 10⁸  6.37 10⁶

            d_sphere = 1.37 10⁻² mm

5 0
2 years ago
Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13
Julli [10]
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
                 m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
                    (85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
3 0
2 years ago
A horizontal spring with spring constant 85 n/m extends outward from a wall just above floor level. a 3.5 kg box sliding across
Rina8888 [55]

k = spring constant of the spring = 85 N/m

m = mass of the box sliding towards the spring = 3.5 kg

v = speed of box just before colliding with the spring = ?

x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m

the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

Kinetic energy of spring before collision = spring energy of spring after compression

(0.5) m v² = (0.5) k x²

m v² = k x²

inserting the values

(3.5 kg) v² = (85 N/m) (0.065 m)²

v = 0.32 m/s

8 0
2 years ago
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