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Irina-Kira [14]
2 years ago
7

The ISS can, at times, be seen from Earth. True or false

Physics
2 answers:
WARRIOR [948]2 years ago
5 0
True, it can be seen from Earth at times
irakobra [83]2 years ago
4 0

Answer:

True but mostly false.

Explanation:

If you squint enough at night, you'd be able to see it. Otherwise it's not really possible.

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A mouse is sitting on a record player. The mouse is 10 cm away from the center of the record. The record player plays at 45 rpm
DIA [1.3K]

Answer:

       v = 0.4712 m / s       In the opposite direction of the turntable

Explanation:

For this exercise the mouse to remain motionless must run at the same speed as the record player point moves,

Let's use rotational kinematics to find the speed of the record player at the point where the mouse is

          v = w r

Let's reduce the magnitude to the SI system

         w = 45 rpm (2π rad / 1rev) (1 min / 60s) = 4.712 rad / s

         r = 10 cm = 0.10 m

Let's calculate

         v = 4.712 0.10

         v = 0.4712 m / s

In the opposite direction of the turntable

5 0
3 years ago
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Virty [35]

Answer:

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3 0
4 years ago
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A child wants to pump up a bicycle tire so that its pressure is 1.2 × 105 pa above that of atmospheric pressure. if the child us
WINSTONCH [101]
General expression is;
Pressure = Force/Area
In which,
Pressure = Required pressure + Atmospheric pressure = (1.2*10^5) + (101325) = 221325 Pa = 221325 N/m^2

Area = πD^2/4 = π*0.035^2/4 = 9.621*10^-4 m^2

Therefore,
Force, F = Pressure*Area = 221325*9.621*10^-4 = 212.94 N
7 0
3 years ago
Read 2 more answers
Information from which evidence led scientists to believe that the continents were parts of larger plates? A. data from hot spot
andrew-mc [135]
D mid ocean ridges and earthquakes
6 0
4 years ago
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A total charge of –6.50 µC is uniformly distributed within a sphere that has a radius of 0.150 m. What is the magnitude and dire
Marrrta [24]

Answer:

A) 2.89 × 105 N/C, radially inward

Explanation:

The magnitude of the electric field generated by a charged sphere is given by the following formula:

E=k\frac{Q}{r^2}  ( 1 )

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

Q: charge of the sphere = -6.50*10^-6 C

r: distance from the center of the sphere to the point in which E is calculated

r = 0.150m + 0.300m = 0.450 m

You replace the values of Q and r in the equation (1):

E=(8.98*10^9Nm^2/C^2)\frac{6.50*10^{-6}C}{(0.450m)^2}=288246.91\frac{N}{C}\\\\E\approx2.89*10^5\frac{N}{C}

The electric field points radially inward because the charge is negative.

hence, the answer is:

A) 2.89 × 105 N/C, radially inward

7 0
4 years ago
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