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dusya [7]
2 years ago
6

40 Points!!!!!!!!!!!!!

Physics
1 answer:
Firlakuza [10]2 years ago
3 0
The answer is A for sure
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Please helppp
Y_Kistochka [10]

Answer:

F = 17.3 kN

Explanation:

The normal force must support the weight of the car plus provide for the needed centripetal acceleration.

F = m(g + v²/R ) = 1000(9.8 + 15²/30) = 17,300

6 0
2 years ago
A book with mass 2.3 kg sits on a table. What is the normal force on the
natita [175]

Explanation:

The table is level and there are no other forces on the book, so the normal force is equal to the weight.

N = mg

N = (2.3 kg) (9.8 m/s²)

N = 22.5 N

6 0
3 years ago
Read 2 more answers
Which use combustion to release thermal energy?
julia-pushkina [17]
Heat of combustion.<span> The calorific value is the total energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon or other organic molecule reacting with oxygen to form carbon dioxide and water and release heat.</span>
8 0
3 years ago
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Name the processes of going (a) from a solid to a gas and (b) from a gas to a solid.
marissa [1.9K]
The answer to the first one is sublimation.
8 0
3 years ago
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
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