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velikii [3]
2 years ago
10

The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

ject will be dropped with no initial speed. The objects are initially 179 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects?
Physics
1 answer:
Crazy boy [7]2 years ago
4 0

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

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Explanation:

Since, it is given that one hand completes 3.19 vibrations in 8.46 sec. Therefore, in one second the number of vibrations will be as follows.

              \frac{3.19}{8.46}

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               v = f \times \lambda

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                = 1.57 cm

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3 years ago
A 1000 kg rocket ship is traveling at 40 m/s. If the velocity changes to 70
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Answer: A

Explanation:

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3 years ago
The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance
alukav5142 [94]

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.  

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

1\ light\ year=3.72\times 10^{17}\ inches

So, 4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches

4.218\ light\ year=1.56\times 10^{18}\ inches

We need to convert 4.218 light-years barley corns.  

Since, 1 barleycorn = 1/3 inch  

1\ inch=3\ barleycorn

1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn

So, the nearest star to the Earth is at a distance of 4.68\times 10^{18}\ barleycorn. Hence, this is the required solution.

3 0
3 years ago
With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a
Aleks04 [339]

Answer:

The correct option is (b).

Explanation:

Given that,

Electric field, E=2\times 10^2\ N/C

We need to find the magnitude of the force on the electron as a result of the electric field.

We know that, the electric force is given by :

F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N

So, the required force on the electron is equal to 3.2\times 10^{-17}\ N.

5 0
3 years ago
5- A 2500g object is pushed with 55N for 12m in 11s, there was a force of friction of 30N.
Assoli18 [71]

Answer:

1kg =1000g

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55-30

average speed

final - initial

divide by time t(s)

3 0
2 years ago
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