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Yuri [45]
4 years ago
13

An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o

bject suspended from it. Assume a wire of mass 3.60 g is 1.60 m long and has a 3.00-kg object suspended from it. A pulse requires 60.1 ms to traverse the length of the wire. Calculate gplanet from these data. (You may neglect the mass of the wire when calculating the tension in it.)
gplanet = m/s2
Physics
1 answer:
Alekssandra [29.7K]4 years ago
8 0

Answer:

0.53m/s^2

Explanation:

We are given that

Mass of wire=m=3.6 g=3.6\times 10^{-3} kg

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms=60.1\times 10^{-3} s

1 ms=10^{-3} s

Speed,v=\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s

g=\frac{v^2m}{m'l}

Using the formula

g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2

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Ball A was carried to the top of a hill in a straight line, while ball B was carried in a longer, zigzag path. At the top of the
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Answer: equal

Explanation:

Don’t know why honestly

5 0
3 years ago
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To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-tr
REY [17]

Answer:

Explanation:

Electric field talks about a region around a charged particle or object within which a force would be exerted on other charged particles or objects. to find the electric field inside the bulb we will apply the electric filed formula.

Please kindly check attachment for step by step explaination.

6 0
3 years ago
What is my displacement if i walked 200 meters west then turned around and walked 150 meters east
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You're walking in one direction, and then the exact opposite of that direction, so you simply have to subtract the two distances. 
200-150=50
You're 50 meters west of where you originally started.
You're west because 200 meters west is greater than 150 meters east. If the distance walked east was greater than the distance walked west, you would've been east of your starting position.
6 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
2 years ago
The figure above shows 4forces 3N, 10N, 3√3N, and 6N acting on a particle P. The resultant of the four forces is.
Brilliant_brown [7]

As you gave no pic I took them on one lined

  • F_1=3N
  • F_2=10N
  • F_3=3root 3 N
  • F_4=6N

\\ \sf\longmapsto F_{net}=F_1+F_2\dots

\\ \sf\longmapsto F_{net}=3+10+6+3\sqrt{3}

\\ \sf\longmapsto F_{net}=19+3(1.732)

\\ \sf\longmapsto F_{net}=19+5.196

\\ \sf\longmapsto F_{net}=24.196N

6 0
3 years ago
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