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Oxana [17]
2 years ago
8

An object is dropped from 600 m. If it is initially at rest and achieves a top speed of 45 m/s just as it hits the ground, what

is its acceleration?​
Physics
2 answers:
marshall27 [118]2 years ago
4 0

Answer:

  1. i don't know answer
  2. sorry
  3. mistake
Brums [2.3K]2 years ago
3 0

Answer:

1.68m/s^2

Explanation:

using V^2=U^2+2×a×s formula.

If,

(Final Velocity)V=45

(Initial Velocity)U=0 because it starts from rest

(Distance)S=600m

(acceleration)a= ?

now,

using formula,

45^2=0^2+2×a×600

2025= 1200a

a=2025÷1200

a = 1.68m/s^2(The unit of acceleration is m/s^2)

Therefore the acceleration is 1.68m/s^2

Hope it works !!!

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gayaneshka [121]

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0
3 years ago
Water is boiled at 1 atm pressure in a 25-cm-internal- diameter stainless steel pan on an electric range. if it is observed that
patriot [66]
<span>3933 watts At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3 The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams. Rounding to 4 significant figures gives me 4705 grams of water. The heat of vaporization for water is 2257 J/g. So the total energy applied is 2257 J/g * 4705 g = 10619185 J Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds. Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So 10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3) = 3933 watts</span>
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A parallel-plate capacitor has a plate area of 0.2 m^2 and a plate separation of 0.1 mm. If the charge on each plate has a magni
barxatty [35]

Answer:

The potential difference across the plates is 226 V.

Explanation:

Given;

area of the capacitor plate, A = 0.2 m²

separation, d = 0.1 mm = 0.1 x 10⁻³ m

charge on each plate, Q = 4 x 10⁻⁶ C

Charge on the capacitor is given by;

Q = CV

Where;

C is the capacitance of the capacitor, given as;

C = ε₀A / d

Then, the potential difference across the plates is given by;

V = \frac{Q}{C} = \frac{Qd}{\epsilon_o A} = \frac{(4*10^{-6})(0.1*10^{-3})}{(8.85*10^{-12})(0.2)}\\\\V = 226 \ V

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