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kicyunya [14]
3 years ago
7

If a creature cats Ferrari, with an initial speed of 10 m/s, accelerate at a rate of 50 m/s/s for 3 seconds, what will it's fina

l speed be?
Physics
2 answers:
Anton [14]3 years ago
7 0
Vf = x
Vi = 10 m/s
A = 50 m/s²
T = 3


Vf = Vi + at

x = 10 + 50(3)

x=160

Final Velocity is equal to 160 m/s
riadik2000 [5.3K]3 years ago
3 0


160 m/s because you start off with 10 and go up by 50*3 which is 150 then add that with the begging speed to achieve 160m/s.

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Which of the following best represents a chemical reaction?
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Answer:

The answer to your question should be D.

Explanation:

reactants are on the laft side of arrow and products are on right side of arrow

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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va
bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

8\times 10^{-12}\ F

Explanation:

E = Electric field = 4\times 10^6\ V/m

d = Gap = 2.5 mm

Q = Charge = 80 nC

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Potential difference is given by

V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V

The potential difference between the plates is 10000 V

Area is given by

A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2

The area of the plate is 0.00225988700565 m²

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F

The capacitance is 8\times 10^{-12}\ F

4 0
3 years ago
A 100-meter sprint is a race using only the straight side of a racetrack. A 400-meter sprint is a race that makes one complete l
Shalnov [3]
Speed uses distance and velocity uses displacement in its calculation.

For 100 m race, distance = displacement. Hence speed = velocity

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3 0
3 years ago
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How does the saturation of a solution affect crystal formation? In your own words please
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3 0
3 years ago
A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at
wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

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Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

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T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0
2 years ago
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