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Yuri [45]
3 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]3 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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A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
Evaporation of sweat cools the skin because
Law Incorporation [45]

Answer: c. the molecules with the highest energy evaporate first, lowering the temperature of the sample

Explanation:

The process by which liquid starts to change into vapor phase at any temperature is known as evaporation.

During evaporation , the molecules which possess higher energies escape from the upper layer into vapor phase. the molecules which escape draw energy from surroundings and thus decrease the energy of the surroundings and hence lead to decrease in temperature.

As temperature of the system is directly proportional to the energy of the system , thus decrease in energy leads to decrease in temperature.

K.E=\frac{3RT}{2}

K.E. = Kinetic energy

T = temperature

R= gas constant

6 0
3 years ago
Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER
lukranit [14]

Answer:

Explanation:

I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.

First row, KE = 0

Second row, KE = 220500 J

Third row, KE = 183750 J

Fourth row, KE = 205800 J

That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.

3 0
2 years ago
A 45-mH inductor is connected in series with a 60-Ω resistor through a 15-V dc power supply and a switch. If the switch is close
egoroff_w [7]

Answer:

The current is 0.248 A

Explanation:

Given that,

Inductor L= 45\times10^{3}\ H

Resistance R= 60\Omega

Voltage = 15 volt

Time t =7.0\times10^{-3}\ sec

We need to calculate the current

Using formula of current

I=\dfrac{V}{R}(1-e^{\dfrac{-R}{L}}t)

Where, V = voltage

R = resistance

L = inductance

T = time

Put the value into the formula

I=\dfrac{15}{60}(1-e^{\dfrac{-60}{45\times10^{3}}}\times7\times10^{-3})

I=0.248\ A

Hence, The current is 0.248 A.

4 0
3 years ago
What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an
Sindrei [870]

Answer:

a = 12.78 g's

Explanation:

Height reached by the object after push off is given as

H = 2.3 m

v_f^2 - v_i^2 = 2 a s

now we have

0 - v^2 = 2(-9.81)(2.3)

v = 6.72 m/s

now we know that this push last for total distance of 0.18 m

so during the push we will have

v_f^2 - v_i^2 = 2 a d

6.72^2 - 0 = 2a(0.18)

a = 125.35 m/s^2

now in terms of g = 9.81 m/s/s we have

a = \frac{125.35}{9.81} gs

a = 12.78 g's

6 0
3 years ago
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