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BabaBlast [244]
3 years ago
14

two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f

orward on the rope by moving hand over hand and gradually reeling in more of the rope. In doing so, he exerts a force of 35N [backwards] on the rope. This causes him to accelerate toward the other person. Assuming that the friction acting on the skaters is negligible, how long will it take for them to meet? Explain your reasoning.
Physics
1 answer:
Zina [86]3 years ago
5 0

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

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Please help
natali 33 [55]

Answer:

Thermosphere

temperature continue to rise up to 1500*C

Ionosphere

Mesosphere

temperature falls to -93*C

Stratosphere

protective ozone layer

UV radiation causes temperature to rise

Troposphere

78% nitrogen, 21% oxygen

temperature drops from 17 to -52*C

Explanation:

3 0
3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
3 years ago
Read 2 more answers
4. Which are characteristics of an atom? (Choose all that apply.)
Zarrin [17]

Answer:

4.

Explanation:

7 0
3 years ago
Read 2 more answers
explain a ball is pitched east at a speed of 40 km/h. the batter hits it west at a speed of 40 km/h.did the ball accelerate
Slav-nsk [51]
Yep, uh-huh, indeed, indubitably, and you better believe it.
The ball did accelerate.

Acceleration means any one of these:
-- speeding up
-- slowing down
-- changing direction .

The direction of the ball changed from east to west. 
That's acceleration right there.
_____________________________________

Acceleration means any change of velocity.
Velocity is speed and direction.

The velocity of the ball changed from (40 km/hr east) to (40 km/hr west).

The change in velocity was 80 km/hr to the west.
5 0
3 years ago
A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

7 0
3 years ago
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