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3 years ago

Which of the following would describe a length that is 2.0×10^-3 of a meter? a: 2.0 kilometers

Physics

1 answer:

natita [175]3 years ago

5 0

Answer:

Option (c) $2\times 10^{-3}\ m=2\ millimeters$

Explanation:

Here, it is required to describe the given length in a particular unit. Firstly, we need to see the following conversions as :

1 meter = 0.001 kilometers

1 meter = 10⁻⁶ megameters

1 meter = 1000 millimeters

1 meter = 1000000 micrometers

From the given option, the correct one is (c) because, 1 meter = 1000 millimeters

So, $2\times 10^{-3}\ m=2\ millimeters$

Hence, the correct option is (c). Hence, this is the required solution.

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An airplane flies horizontally at constant speed in a straightline direction. Its state of motion is unchanging. In other words

Alex_Xolod [135]

Answer:

sum of all forces on the air plane must be ZERO

So both forces must be of same magnitude

Explanation:

As we know that airplane is moving with uniform speed is horizontal plane is a straight line

so the motion of the air plane is uniform without any acceleration

So we will have

$a = 0$

acceleration must be zero

now by Newton's law

$F_{net} = 0$

$F_1 + F_2 = 0$

so sum of all forces on the air plane must be ZERO

7 0

3 years ago

Where is parallel circuits used?

suter [353]

Answer:

if one bulb burns out, the other bulbs in the fixture continue to operate. Other uses include an electronic OR gate, where two switches are in a parallel circuit: one of the switches must be closed for the circuit to function.

8 0

4 years ago

A force of 6.0 N pulls a box 0.40 m along a frictionless plane that is inclined at 36°. What work is being done by the pulling f

lys-0071 [83]

Answer:

Expression of work done is

$W = Fd cos\theta$

Work done to move the sled is given as 1.94 J

Explanation:

As we know that the formula of work done is given as

$W = Fd cos\theta$

here we know that

F = 6 N

d = 0.4 m

$\theta = 36 degree$

so we will have

$W = 6 \times 0.4 cos36$

$W = 1.94 J$

7 0

3 years ago

Can any kind soul help me

Wittaler [7]

Answer:

2917.4 m/s

Explanation:

From the question given above, the following data were:

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

Radius (r) of the Moon = 1737 Km

Escape velocity (v) =?

Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:

Gravitational acceleration of the earth = 9.8 m/s²

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

= 0.25 × 9.8 = 2.45 m/s²

Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:

1 Km = 1000 m

Therefore,

1737 Km = 1737 Km × 1000 m / 1 Km

1737 Km = 1737000 m

Thus, 1737 Km is equivalent to 1737000 m

Finally, we shall determine the escape velocity of the rocket as shown below:

Gravitational acceleration of the Moon (g) = 2.45 m/s²

Radius (r) of the moon = 1737000 m

Escape velocity (v) =?

v = √2gr

v = √(2 × 2.45 × 1737000)

v = √8511300

v = 2917.4 m/s

Thus, the escape velocity is 2917.4 m/s

4 0

3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

Hoochie [10]

Answer:

False

Explanation:

The ground or surface cannot give off energy

(correct me if i am worng) :)

8 0

3 years ago