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3 years ago

Definition: In any chemical process, matter is neither lost nor gained.

Physics

1 answer:

kolbaska11 [484]3 years ago

7 0

Answer: Conversation of Matter

Explanation: According to Quizlet, Converstion of Matter is the answer.

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A force is acting on a 3-kg object during 5 seconds. If the object initially has a speed of 1.5m/s and after applying the force

uranmaximum [27]

Answer:

1.5 N

Explanation:

You've left us to guess what the question is. I will Assume it is what's the force?

Givens

m = 3 kg

vi = 1.5 m/s

vf = 4 m/s

t = 5 seconds

Formula

F = m * (vf - vi)/t

Solution

F = 3 * (4 - 1.5) / 5

F = 1.5 N

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3 years ago

A green light has a wavelength of 485 nm.<br> What is its frequency? (1 nanometer is 10-9 m)

Lynna [10]

Answer:

Explanation:

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3 years ago

An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 si

spayn [35]

Answer:

Explanation:

(r) = <10 cos 6t, 10 Sin 6t>

The distance traveled by the object is the magnitude of vector r.

The magnitude of vector r is given by

$r = \sqrt{(10 Cos 6t)^{2}+(10 Sin 6t)^{2}}$

$r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)$

r = 10

5 0

3 years ago

A boat is moving at the rate of 15.0 meters per second. It’s speed is then decreased uniformly to 3.0 meters per second. It take

yuradex [85]

Answer:

-3 m/s²

Explanation:

The formula for deceleration is given as the change in speed over a period of time

Given that,

Initial speed = 15 m/s

Final speed = 3 m/s

change in speed= 3-15= -12 m/s

Time taken= 4 sec

Deceleration = -12/4 = -3 m/s²

7 0

3 years ago

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$

Explanation:

From the question we are told that

The current on the copper is $I = 20 \ A$

The cross-sectional area is $A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2$

The number of copper atom in the wire is mathematically evaluated

$n = \frac{\rho * N_a}Z}$

Where $\rho$ is the density of copper with a value $\rho = 8.93 \ g/m^3$

$N_a$ is the Avogadro's number with a value $N_a = 6.02 *10^{23}\ atom/mol$

Z is the molar mass of copper with a value $Z = 63.55 \ g/mol$

$n = \frac{8.93 * 6.02 *10^{23}}{63.55}$

$n = 8.46 * 10^{28} \ atoms /m^3$

Given the 1 atom is equivalent to 1 free electron then the number of free electron is

$N = 8.46 * 10^{28} \ electrons$

The current through the wire is mathematically represented as

$I = N * e * v * A$

substituting values

$20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}$

=> $v = 0.0002808 \ m/s$

8 0

3 years ago