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NISA [10]
3 years ago
12

Help !

Physics
1 answer:
Black_prince [1.1K]3 years ago
7 0
First off, you need to know the weight of the projectile, lift and drag coefficients something like a high Reynolds number is preferred, then use the gravitational constant of 9.8 meters per second squared those would be a good start to get closer to your goal
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According to Kepler, what do all bodies in orbit around another have in common?
dolphi86 [110]
D.
All follow an elliptical path that has two foci, rather than a circular path.

Kepler found paths are elliptical, not circular
3 0
3 years ago
A dog walks 12 meters to the east and then 16 meters back to the west for this motion what is the distance moved What is the mag
FinnZ [79.3K]

The distance is 28 meters, and the displacement is -4.

For the distance it would be 12 + 16 = 28.

For the displacement it would be 12 - 16 = -4.

would really appreciate a brainliest! Hope this helped!

6 0
3 years ago
An athlete runs a 100 m race in 10 seconds against a friction force of 100N. How do I work out his power output?
Rashid [163]
They seem to cancel each other out which is odd
6 0
3 years ago
A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate
qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg =  0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0
3 years ago
The average density of the body of a fish is 1080 kg/m3. To keep from sinking, the fish increases its volumeby inflating an inte
Dimas [21]

Answer:

f = 8 %

Explanation:

given,

density of body of fish = 1080 kg/m³

density of air = 1.2 Kg/m³

density of water = 1000 kg/m²

to protect the fish from sinking volume should increased by the factor f

density of fish + density of water x increase factor = volume changes in water                                                    

1080 +f x 1.2 =(1 + f ) x 1000                

1080 + f x 1.2 = 1000 + 1000 f      

998.8 f = 80                                  

f = 0.0800                            

f = 8 %                                        

the volume increase factor of fish will be equal to f = 8 %

7 0
3 years ago
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