Someone plz help me with this problem!!!

Katyanochek1 [597]

DDDDDDDDDDDDDDDDDDDDDDDDDDDDD

5 0

2 years ago

Which of the following is always true about the particles that make up matter? A. They can only be found in solid substances. B.

Neko [114]

B.They are too small to be seen with only our eyes

4 0

2 years ago

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What did Thomson contribute to the model of atom

Ganezh [65]

<h3>Answer;</h3>

Electrons 

<h3>Explanation;</h3>

Thomson contributed to the model of an atom by discovery of electrons and thus proving the existence of sub-atomic particles in an atom.
Thomson used cathode ray tube, and demonstrated that cathode rays were negatively charged. According to his model normally known as the plum pudding in which he stated that an atom is composed of electrons as subatomic particles that are surrounded by positive charges to balance the electrons.

5 0

2 years ago

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What is the magnitude of the acceleration vector which causes a particle to move from velocity −5i−2j m/s to −6i+ 7j m/s in 8 se

shusha [124]

Answer:

Acceleration, $a=\dfrac{1}{8}(-i+9j)\ m/s^2$

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

Final velocity of particle in vector form, v = (-6i + 7j) m/s

Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}$

$a=\dfrac{(-i+9j)}{8\ s}\ m/s^2$

or

$a=\dfrac{1}{8}(-i+9j)\ m/s^2$

Hence, the value of acceleration vector is solved.

4 0

2 years ago

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

2 years ago