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olga2289 [7]
2 years ago
6

2. What do you understand by balanced and unbalanced force​

Physics
1 answer:
labwork [276]2 years ago
8 0

Answer:

forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

\\

hope helpful ~

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The fastest animal in the world is the peregrine falcon. It is capable of diving after its prey at an amazing 83.00 meters per s
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1 metre per second = 2.237 miles per hour

so 83 m/sec = 185.666 miles per hour !! ...answer !!

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4 years ago
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In order for water to condense on an object, the temperature of the object must be ______ the dewpoint temperature.
enyata [817]

Answer:

at   ( or below)

Explanation:

at the dewpoint.......water will condense out of the air onto the surface

7 0
2 years ago
You can determine the index of refraction of a substance by measuring its critical angle for total internal reflection. True or
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4 years ago
There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame
slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3

Similarly

V_2=\frac{4}{24}\pi d_2^3

Dividing the above two equations, we get

\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3

∴ The larger one hold 163.296 cm³ of paint

5 0
3 years ago
The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the elect
RideAnS [48]

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

E=\dfrac{\lambda}{2\pi \epsilon_o r}

It is clear that the electric field is inversely proportional to the distance. So,

\dfrac{E}{E'}=\dfrac{r'}{r}

E'=\dfrac{Er}{r'}

E'=\dfrac{125\times 3.5}{1.5}  

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

5 0
4 years ago
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