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Elena L [17]
3 years ago
11

What is the power of 10 when 0.00503 is written in scientific notation?

Physics
2 answers:
AnnZ [28]3 years ago
6 0

Answer:

-3

Explanation:

svetoff [14.1K]3 years ago
4 0

Answer:

Negative 3

Explanation:

Bc scientific notation is the zeros either ahead or behind the actual numbers

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a_sh-v [17]

Answer:

the shape how it involve into a picture

Explanation:

3 0
3 years ago
Describe a situation when you might travel at a high velocity, but with low acceleration
Alisiya [41]

Reading a book in your warm, comfy seat ... in Row-27 of a
passenger airliner cruising at 450 miles per hour.
 
5 0
3 years ago
An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele
kirza4 [7]

Answer:

W = 462.5 keV

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

F = qE

F = (1.6 \times 10^{-19})(1.85 \times 10^6)

F = 2.96 \times 10^{-13} N

now the distance moved by the electron is given as

d = 0.25 m

so we have

W = F.d

W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)

W = 7.4 \times 10^{-14} J

now we have to convert it into keV units

so we have

1 keV = 1.6 \times 10^{-16} J

W = 462.5 keV

5 0
3 years ago
A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary
Snezhnost [94]

Explanation:

Let N_p\ and\ N_s are the number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p}{N_s}=\dfrac{1}{3}

A resistor R connected to the secondary dissipates a power P_s=100\ W

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

V_s=(\dfrac{N_s}{N_p})V_p

V_s=3V_p...............(1)

The power dissipated through the secondary coil is :

P_s=\dfrac{V_s^2}{R}

100\ W=\dfrac{V_s^2}{R}

V_p^2=\dfrac{100R}{9}.............(2)

Let N_p'\ and\ N_s' are the new number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p'}{N_s'}=\dfrac{1}{24}

New voltage is :

V_s'=(\dfrac{N_s'}{N_p'})V_p'

V_s'=24V_p...............(3)

So, new power dissipated is P_s'

P_s'=\dfrac{V_s'^2}{R}

P_s'=\dfrac{(24V_p)^2}{R}

P_s'=24^2\times \dfrac{(V_p)^2}{R}

P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}

P_s'=6400\ Watts

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0
3 years ago
A rocket started from the rest and goes straight up with an acceleration of 4.0 m/s^2 for the first 100 seconds. Then it coasts
Dimas [21]

Answer:

idk

Explanation:

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3 years ago
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