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3 years ago

Make the following conversion

Physics

1 answer:

ValentinkaMS [17]3 years ago

4 0

Answer:

The value of 60.2 mg in to g is 0.0602.

Explanation:

We need to calculate the 60.2 mg in to gram.

Conversion mg to g :

1 g = 1000 mg

Using this conversion,

$1 mg = 10^{-3}g$

So, $60.2\ mg = 60.2\times10^{-3}\ g$

$60.2\times10^{-3}\ g$

Hence, The value of 60.2 mg in to g is 0.0602.

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The amplitude, or magnitude, of a sinusoidal source is the maximum value of the source. What is the amplitude of the voltage sou

liraira [26]

Answer:

Explanation:

<u>Sinusoidal Functions</u>

A sinusoid or sinusoidal function is a sine or cosine which general equation is

$F(x)=A.sin(wt-\phi)$

Or also

$F(x)=A.cos(wt-\phi)$

Where A is the amplitude or maximum value, w is the angular frequency, t is the time and $\phi$ is the phase shift.

Comparing the given expression with the general formula

$v(t)=50cos(2000t-45^o) mV$

We can establish that A=50 mV = 0.05 V

$A = 0.05\ V$

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3 years ago

Which transition by an electron will release the greatest amount of energy? hurry!

Sphinxa [80]

highest energy level to the ground state.

Explanation:

The transition from the highest energy level to the ground state.

An electron has a discrete amount of energy accrued to it in any energy level it belongs to.

Electrons can transition between one energy level or the other.

When electrons change state, they either release or absorb energy.
When an atom absorbs energy, they move from their ground to final state which is consistent with the energy of the final state.
When electrons release energy, they move from excited state to their ground state.
Electrons will release the greatest amount of energy when they move from the highest energy level to the ground state.

Learn more:

Neil Bohr brainly.com/question/4986277

#learnwithBrainly

5 0

3 years ago

Read 2 more answers

A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.

mars1129 [50]

Answer:

a) v = 31.67 cm / s , b) v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

w = √ K / m

w = √ 23.2 / 0.37

w = 7,918 rad / s

a) the expression against the movement is

x = A cos (wt + Ф)

Speed is

v = dx / dt = - A w sin (wt + Ф)

The maximum speed occurs for cos = ± 1

v = A w

v = 4.0 7,918

v = 31.67 cm / s

b) as the object is released from rest

0 = -A w sin (0+ Фi)

sin Ф = 0

Ф = 0

The equation is

x = 4.0 cos (7,918 t)

v = -4.0 7,918 sin (7,918 t)

v = - 31.67 sin (7.918t)

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

7,918 t = cos⁻¹ 1.5 / 4.0

t = 1,186 / 7,918

t = 0.1498 s

We look for speed

v = -31.67 sin (7,918 0.1498)

v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

31.67 / 2 = 31.67 sin ( 7,918 t)

7.918t = sin⁻¹ 0.5

t = 0.5236 / 7.918

t = 0.06613

With this time let's look for displacement

x = 4.0 cos (7,918 0.06613)

x = 3.46 cm

6 0

3 years ago

The acceleration of a train moving from rest and its speed reaches 36m/sec in 9 sec

jenyasd209 [6]

Answer:

4 m/s^2

Explanation:

The acceleration is defined as: Δv/Δt (the difference of the velocity over a time period in which happens that difference).

Remember that a difference is calculated by subtracting the initial value of a physical quantity from its final value.

In our case:

Δv = Vfinal - Vinitial = 36m/s - 0 m/s = 36m/s

Δt = 9s

a = Δv/Δt = 36m/s / 9s = 4m/s^2

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2 years ago

A +0.2 µC charge is in an electric field. What happens if that charge is replaced by a +0.4 µC charge?

AleksandrR [38]

Explanation:

It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.

So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.

U = qV

Since, the electric potential is directly proportional to the charge. Hence, when 0.2 $\mu C$ tends to replaced by 0.4 $\mu C$ then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.

Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.

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3 years ago