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blondinia [14]
3 years ago
15

Hi, could someone please answer the first 3 questions of each planet

Physics
1 answer:
larisa86 [58]3 years ago
7 0

Answer:

Hey bro im here to help it looks like for 3 years you get slow answer if you want your question to be answer in seconds have this in your title

=------------------------------------------------------=

I WILL MARK YOU BRAINLYLISET

=-------------------------------------------------------=

you will get answer fast

Explanation:

You might be interested in
I need it in the next hour or so!
PSYCHO15rus [73]

The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is

∑ F = F[a] - F[f] - F[air] = ma

3100 N - 200 N - F[air] = (650 kg) (3 m/s²)

Solve for F[air] :

F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)

F[air] = 3100 N - 200 N - 1950 N

F[air] = 950 N

3 0
2 years ago
There are two main groups of recyclable material,
Illusion [34]

Answer:Paper and plastic!

Explanation:

3 0
3 years ago
Jen pushed a box for a distance of 80m with 20 N of force. How much work did she do?
Drupady [299]
The answer is 1,600 J.

A work (W) can be expressed as a product of a force (F) and a distance (d):
W = F · d<span>

We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
8 0
3 years ago
BRAINLEST FOR CORRECT ANSWER
xxMikexx [17]

Answer:

3kg sledgehammer swung at 1.5 m/s

Explanation:

Small Sledgehammer:

Mass:3.0

Velocity:1.5

MASS×VELOCITY=MOMENTUM

3.0×1.5= 4.5 (momentum)

Large Sledgehammer:

Mass:4.0

Velocity:0.9

4.0×0.9=3.6 (momentum)

higher momentum is the smaller Sledgehammer.

3 0
3 years ago
Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +18 pC, fairly typical values.
muminat

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

V = \frac{kq}{r}

k = Coulomb's constant

q = Charge

r = Distance between them

q = 18 pC \rightarrow q = 1.8*10^-11 C

d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m

Replacing,

V = \frac{kq}{r}

V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}

V = 135 V

Therefore the potential at the surface of the raindrop is 135 V

3 0
3 years ago
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