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professor190 [17]
3 years ago
5

Why does air resistance increase when a sky diver Opens his parachute?

Physics
2 answers:
Sonbull [250]3 years ago
7 0
Once the parachute is opened, the air resistance overwhelms the downward force of gravity.
UkoKoshka [18]3 years ago
4 0
When the parachutes open, the amount of space that is going against the air increases, so the resistance increases, and it overwhelms the force of gravity for a while until it reaches equilibrium
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garik1379 [7]

Potential energy is highest at the top of the loop, and kinetic energy is highest at the bottom of the loop.

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4 years ago
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Name the principle in which rocket work.
REY [17]
The rocket engine works on the basic principle proposed by Newton which is Newton’s Third Law.
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3 years ago
3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final
MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0
3 years ago
In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.
denis23 [38]
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
5 0
4 years ago
Calculate the strength and direction of the electric field 300 mm to the right of a -200 C electric
Norma-Jean [14]

Answer: -1.91*10^13 v/m

Explanation:

6 0
3 years ago
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