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Ahat [919]
3 years ago
8

A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi

ng it a speed of 0.17 m/s directly away from the shuttle. Seven-and-a-half seconds later the astronaut comes into contact with the shuttle.
Required:
What was the initial distance from the shuttle to the astronaut?
Physics
1 answer:
WARRIOR [948]3 years ago
8 0

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

m1v1 + m2v2 = m1'v1' + m2'v2'

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

0 + 0 = 91*v1' + 1300*0.17

v1' = -1300*0.17/91 = -2.4286\ m/s

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

distance = 2.4286 * 7.5 = 18.2145\ meters

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An American traveler in China carries a transformer to convert China's standard 220 V to 120 V so that she can use some small ap
Arisa [49]

Answer:

(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833

(b) The ratio of input to output current is 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

Explanation:

Given;

input voltage, V_p = 220 V

output voltage, V_s = 120 V

General transformer equation is given as;

\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}

where;

Np is number of turns in the primary coil

Ns is number of turns in the secondary coil

Is - is the secondary current or output current

Ip - is the primary current or input current

(a) The ratio of turns in the primary and secondary coils of her transformer;

\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833

(b) The ratio of input to output current;

\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

8 0
3 years ago
H H H O H O N C C N C C H O H H O H H C H H H ALANINE GLYCINE
Trava [24]
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6 0
2 years ago
A measurement must include both a number and a(an)
IrinaK [193]
<span>A measurement must include both a number and an unit of measurement. </span>
7 0
3 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
Lisa is conducting an investigation to determine the coefficient of friction between two surfaces. She uses a 25 kg block. What
Paladinen [302]

Answer:

245.25

Explanation:

1 kg is 9.81 newtons

4 0
3 years ago
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