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3 years ago

Which is an outer planet?

Physics

2 answers:

saw5 [17]3 years ago

8 0

Answer:

pluto

Explanation:

Dwarf because it is very minut

EleoNora [17]3 years ago

4 0

Answer:

I think Saturn.

Explanation:

I think the inner planets are Mercury, Venus, Earth and Mars whereas the outer ones are Jupiter, Saturn, Uranus and Neptune.

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A 10.-kilogram object, starting from rest, slides down a frictionless incline with a constant acceleration of 2.0 m/sec2 for fou

aniked [119]

Answer:

C) 100N

Explanation:

Formula for calculating the Weight of an object is expressed as;

Weight = mass × acceleration due to gravity

Given

Mass of the object = 10kg

Acceleration due to gravity = 9.81m/s²

Substitute into the formula above

Weight = 10×9.81

Weight = 98.1N

Hence the approximate weight of the object is 100N

5 0

3 years ago

If a dog has the mass of 14.5kg and accelerates at a rate of 8m/sec2 →, If he runs into you what is the force he hits you with?

pashok25 [27]

Answer:

F=ma

m=14.5kg

a=8m/s^2

F=14.5×8

F=116N

Explanation:

PLS MARK BRAINLIEST

7 0

3 years ago

A train travels due south at 20 m/s. It reverses its direction and travels due north at 20 m/s. What is the change in velocity o

luda_lava [24]

Answer:

40 m/s due north

Explanation:

Consider that the south direction a negative Y axis and north direction as + Y axis

v1 = 20 m/s South = 20 (-j) m/s

v2 = 20 m/s North = 20 j m/s

Change in velocity = v2 - v1 = 20 j - 20 (-j) = 40 j m/s

So, change in velocity is 40 m/s due north.

5 0

3 years ago

Which feature of a balanced chemical equation demonstrates the law of conservation of mass?

svetlana [45]

A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow

6 0

3 years ago

A bat, flying at 5.00 m/s, emits a chirp at 40.0 kHz. If this sound pulse is reflected by a wall, what is the frequency of the e

fredd [130]

Answer:

The answer is below

Explanation:

Firstly, the frequency is received by the wall and then it is reflected and received by the bat.

The frequency received by the wall (f') is given by:

f' = $f(v\pm v_o)/v\\\\$

The - sign is used when the observer is moving away from the source and + sign when the observer is moving towards the source.

Since the bat is moving towards the wall, we use a positive sign. Hence:

f' = $f(v+ v_o)/v\\\\$

The frequency reflected and received by the bat f" is:

f'' = $f'\frac{v}{ (v\pm v_s)}\\\\$

- sign is used when the source moves toward the observer and + is used when the source moves away

since the bat moves towards the wall, then::

f'' = $f'\frac{v}{(v-v_s)} =\frac{f(v+v_o)}{v}*\frac{v}{(v-v_s)} =f\frac{(v+v_o)}{(v-v_s)} \\\\$

v = speed of sound in air = 331 m/s, vo = velocity of observer = 5 m/s, vs = velocity of source = 5 m/s. Therefore:

$f"=f\frac{(v+v_o)}{(v-v_s)} =40\ kHz\frac{(331\ m/s+5\ m/s)}{(331\ m/s+5\ m/s)} \\\\f"=41\ kHz$

3 0

3 years ago