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Oxana [17]
2 years ago
11

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

ation so she accelerates at 1.50 m/s2. What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall? Give your answer relative to an observer who is space walking and therefore does not accelerate with the space station due to the push.
Physics
1 answer:
Aleonysh [2.5K]2 years ago
3 0

Answer:

a = 5.83 \times 10^{-4} m/s^2

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = 1.50 m/s^2

now we know that

F = ma

F = 70(1.50) = 105 N

now for the space station will be same as above force

F = ma

105 = 1.8 \times 10^5 (a)

a = \frac{105}{1.8 \times 10^5}

a = 5.83 \times 10^{-4} m/s^2

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For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical
nydimaria [60]

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

7 0
2 years ago
Which of the following four circuit diagrams best represents the experiment described in this problem?
Valentin [98]

We don't see any circuit diagrams.  

This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.

6 0
2 years ago
The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan
miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴  × 1000 m

r = 1.08  × 10⁷ m, G = 6.67  × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08  × 10⁷ )²/75( 6.67  × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0
2 years ago
Physics Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
djverab [1.8K]
Good work on solving part a).
b) may look complicated, but it's not too bad.

It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do.  (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
breathe hot air out,and perspire.)

You already know how much mechanical energy the climber needed to lift
himself to the top of the mountain... 2.4x10⁶ joules.
That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.

-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.

-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules

-- Portion of a kilogram he needs to use =   (9.6 x 10⁶) / (3.8 x 10⁷)

Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
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5 0
2 years ago
Read 2 more answers
A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put
olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

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PE = \frac{1}{2} kx^2

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

6 0
2 years ago
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