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Sergio [31]
3 years ago
8

Since the Sun has more mass, why do objects on earth not move closer to the Sun instead of staying put on Earth?

Physics
1 answer:
maw [93]3 years ago
6 0

Answer:

Because the Earth has it's own gravity that keeps us put, and we also have the moon.

Explanation:

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The manufacturer of a 6V dry-cell flashlight battery says that the battery will deliver 15 mA for 60 continuous hours. During th
Vaselesa [24]

Answer:

16200 J

Explanation:

t = Time the battery is on = 60 hours

I = Current = 15\times 10^{-3}\ A

Average voltage

V=\dfrac{6+4}{2}=5\ V

Energy is given by

E=V\times I\times t

\\\Rightarrow E=5\times 15\times 10^{-3}\times 60\times 3600

\\\Rightarrow E=16200\ J

The energy delivered in the given time is 16200 J

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3 years ago
Explain how wind erosion changes land forms
valkas [14]

Answer:

the wind carries abrasive materials

Explanation:

such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first

8 0
2 years ago
A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?
Alinara [238K]

Answer:

400cm^2

Explanation:

sides are 20cm long Area for a square is a squared

since all the lides are of equal length you can just choose one side.

20squared is 400

20 x 20 = 400cm squared

Hope this helps :)

6 0
3 years ago
Determine the kinetic energy of 1000-kg roller coaster car that is moving with speed of 20.0m/s
nevsk [136]
B, i got the same question
7 0
3 years ago
2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th
Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = 5.33 × 10^{-5} T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ B = \frac{\mu_{0}i }{2\pi R}

Where B = magnetic field due to long wires, \mu_{0} = 4\pi \times10^{-7}, R = perpendicular distance from wire to given point

From any one wire R_{1}  = 5 cm, R_{2}  = 3 cm

so we write,

∴ B = B_{1} + B_{2}

 B = \frac{\mu_{0} i}{2\pi R_{1} } +  \frac{\mu_{0} i}{2\pi R_{2} }

 B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]

 B = 5.33\times10^{-5}  T

Therefore, the magnitude of magnetic field at given point = 5.33\times10^{-5} T

3 0
3 years ago
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