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Alex_Xolod [135]
4 years ago
12

How much time would it take for an

Physics
1 answer:
Elina [12.6K]4 years ago
4 0

36 and a half hours

Explanation:

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How has two dogs go awnser my other qestion first one will get brainly anser
aksik [14]
Yes, yea, yep, you don't make much sense.

7 0
4 years ago
A 210 Ohm resistor uses 9.28 W of
IgorLugansk [536]
  • Resistance=R=210Ohm
  • Power=9.28W=P

Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}

\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}

\\ \sf\longmapsto I\approx\dfrac{2}{10}

\\ \sf\longmapsto I\approx0.2A

4 0
3 years ago
A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee
Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2}  = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\

<u>v = 7.69 x 10³ m/s</u>

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v =  \frac{2\pi r}{T}\\\\

where,

T = Time Period of Satellite = ?

Therefore,

T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\

<u>T = 5500 s = 91.67 min = 1.53 h</u>

7 0
3 years ago
After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc
lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

f = \frac{1}{P}

f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + 0 = \frac{1}{200}

d_i = 200 cm

so his far point according to this pair of glass is 200 cm

7 0
3 years ago
1.20 Newton force is working on a 250 gram object. What is the acceleration?
Leokris [45]

Answer:

The answer is B

Explanation:

250g = 0.25kg

F = m × a

a = F/m

= 1.2/0.25

= 4.8m/s²

7 0
3 years ago
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