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Y_Kistochka [10]
2 years ago
6

Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle? they are high

enough above the earth's surface that they have escaped the pull of earth's gravity. the space shuttle has anti-gravity devices on it that allow the astronauts to float inside it. the astronauts are falling at the same rate as the space shuttle as it orbits around earth. orbiting objects do not experience any gravity?
Physics
1 answer:
Maslowich2 years ago
7 0

If we see the forces on the astronauts then there is only one force on each astronaut while they are in air.

This force is due to gravity of earth.

Now while the astronauts are in air and doing some activities then the net force on them is counterbalanced on them by centrifugal force.

so we can say the rate of fall of astronauts due to gravity is at same rate as the orbiting rate of the space shuttle.

this is given by

w^2R = g

so all astronauts will experience the situation of free fall.

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A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magne
Taya2010 [7]

Answer:

Explanation:

a )  No of turns per metre

n = 450 / .35

= 1285.71

Magnetic field inside the solenoid

B = μ₀ n I

Where I is current

B = 4π x 10⁻⁷ x 1285.71 x 1.75

= 28.26 x 10⁻⁴ T

This is the uniform magnetic field inside the solenoid.

b )

Magnetic field around a very long wire at a distance d is given by the expression

B = ( μ₀ /4π ) X 2I / d

= 10⁻⁷ x 2 x ( 1.75 / .01 )

= .35 x 10⁻⁴ T

In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .

6 0
2 years ago
Read 2 more answers
An engineer is designing a process for a new transistor. She uses a vacuum chamber to bombard a thin layer of silicon with ions
adelina 88 [10]

Answer:

E=5.7\times 10^{-3}\ V/m

Explanation:

Given that

m_p=5.18\times 10^{-26}\ kg

re= 46 cm

Vp= 180 m/s

We know that

E=\dfrac{\Delta V}{r}

\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}

So

E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}

Now by putting the all given values in the questions

E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}

E=5.7\times 10^{-3}\ V/m

So the average electric field is E=5.7\times 10^{-3}\ V/m.

6 0
1 year ago
A fisherman is watching water waves ripple past his boat. How could he determine the wavelength of the water waves?
natta225 [31]
Wavelength can be calculated using the following formula: wavelength = wave velocity/frequency. Wavelength usually is expressed in units of meters.
7 0
1 year ago
B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find
lozanna [386]

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

4 0
1 year ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
1 year ago
Read 2 more answers
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