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SashulF [63]
3 years ago
7

PLEASE I NEED HELP ASAP!!!!!!!!!!

Physics
1 answer:
max2010maxim [7]3 years ago
3 0

tooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

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What is the displacement for a person that walks 3.0km north and 8.0 km south
Oduvanchick [21]

Answer:

5 km

Explanation:

x = |3 - 8| = 5 km

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What rocket color code represents the rocket motor?
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Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
Which of the following are likely to form a covalent bond?
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The answer is
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8 0
3 years ago
Read 2 more answers
A 1.20 × 104 kg railroad car moving at 7.70 m/s to the north collides with and sticks to another railroad car of the same mass t
saul85 [17]

Answer: 4.77m/s

Explanation:

According to the law of conservation of momentum which states that the sum total of momentum of bodies before collision is equal to the sum of their momentum after collision. Note that the two bodies will move at a common velocity after colliding.

Let m1 and m2 be the mass of the first and second railroad cars

u1 and u2 be the velocities of the railroad cars

v be the common velocity

Using the formula

m1u1 + m2u2 = (m1 +m2)

m1 = 1.20×10⁴kg

m2 = 1.20×10⁴kg (body of same mass)

u1 = 7.70m/s

u2 = 1.84m/s

v = ?

(1.20×10⁴×7.7) + (1.20×10⁴×1.84) = (1.20×10⁴ + 1.20× 10⁴)v

9.24×10⁴ + 2.21×10⁴ = 2.4×10⁴v

11.45×10⁴ = 2.4×10⁴v

v = 11.45×10⁴/2.4×10⁴

v = 4.77m/s

The velocity of the cars after collision will be 4.77m/s

5 0
3 years ago
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