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tangare [24]
3 years ago
6

Explain why a reference points that are stationary are usually chosen to determine whether an object is in motion.

Physics
1 answer:
weeeeeb [17]3 years ago
4 0

Answer: Stationary reference points are used to determine if a object is in motion because if the reference point is still,you will be able to see if the object is in motion. If you had a reference point that wasn't stationary, you wouldn't be able to tell if the object was in motion.

Explanation:

Stationary reference points are used to determine if a object is in motion because if the reference point is still,you will be able to see if the object is in motion. If you had a reference point that wasn't stationary, you wouldn't be able to tell if the object was in motion.

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Balance the equation <br> CuCl2 + H2S → CuS + HCl
Leviafan [203]

Answer:

CuCl2 + H2S -> CuS + 2HCl

Explanation:

in the reactants the 2 in 2HCl multiplies to give you 2 Hydrogens and 2 Chlorides.

5 0
2 years ago
Read 2 more answers
Quasars are thought to be the nuclei of active galaxies in the early stages of their formation. A typical quasar radiates energy
uranmaximum [27]

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

6 0
3 years ago
If the acceleration of motorboat is 4m/s^2 and the motorboat stsrtsfrol.Rest what is velocity after 6.0 s
UkoKoshka [18]

Answer:

24 m/s

Explanation:

Using v = u + at where u = initial velocity of the motorboat = 0 m/s (since the boat starts from rest), a = acceleration = 4 m/s², t = time = 6 s and v = velocity of the motorboat after 6.0 s.

Substituting the values of the variables into the equation, we have

v = u + at

= 0 m/s + 4 m/s² × 6.0 s

= 0 m/s + 24 m/s

= 24 m/s

3 0
3 years ago
A spring-loaded toy dart gun is shot to a height h. The same dart is shot straight up a second time from the same gun, but this
yarga [219]

Answer:

The height reached by the dart in the second shot is (4 H).

Explanation:

It is given that, a spring-loaded toy dart gun is shot to a height h. In this case, all the potential energy stored in the spring is converted to potential gravitational energy at the maximum height.

\dfrac{1}{2}kx^2=mgH........(1)

At the second shot, the spring is compressed twice as far before firing. x' = 2x

\dfrac{1}{2}kx'^2=mgh

\dfrac{1}{2}k(2x)^2=mgh.........(2)

h is the height reached by the dart in the second shot.

Dividing equation (1) and (2) as:

4=\dfrac{h}{H}

h = 4H

So, the height reached by the dart in the second shot is (4 H). Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
Question 2 - If Juan starts out at 10m/s, and in 15 s speeds up to 60 m/s, what is his acceleration?
34kurt

Answer:

B) 3.33 m/s²

Explanation:

Given that,

The initial velocity of Juan, u = 10 m/s

Final velocity of Juan,  v = 60 m/s

Time taken, t = 15 s

The acceleration is given by the relation

                          a = v-u/t  m/s²

Substituting in the above equation

                           a = 60 - 10 / 15

                           a = 3.33 m/s²

Hence, the acceleration of the Juan is, a = 3.33 m/s²

5 0
3 years ago
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