All categories

3 years ago

Convert the unit of 0.0063 milliseconds into microseconds. (Answer in scientific notation)

Physics

2 answers:

liq [111]3 years ago

8 0

Answer:

6.3 x $10^{-3}$

Explanation:

1 millisecond = 1000 microseconds

0.0063 millisecond = 6.3 microseconds

djyliett [7]3 years ago

6 0

Answer:

microsecond = 1 × 10-6 seconds

1 second = 1 × 100 seconds

1 microsecond = (1 / 1) × 10-6 × 10-0 seconds

1 microsecond = (1) × 10-6-0 seconds

1 microsecond = (1) × 10-6 seconds

1 microsecond = 1 × 1.0E-6 seconds

1 microsecond = 1.0E-6 seconds,just do like it

You might be interested in

These waves are most harmful for living things.

scZoUnD [109]

Explanation:

G. gamma rays because they are produced by the hottest and most energetic objects in the universe, such as neutron stars and pulsars, supernova explosions, and regions around black holes.

hope this helps you

have a nice day:)

5 0

3 years ago

Two extremely large nonconducting horizontal sheets each carry uniform charge density on the surfaces facing each other. The upp

sashaice [31]

This problem refers to a parallel plate capacitor. There is an electric field between the two plates. The working equation to be used is the Gauss’s Law which is

Electric field = Surface charge density / ε0

The answer is -2.52 μC/m2.

8 0

3 years ago

A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr

NISA [10]

Answer:

The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If $\rho$ be the volume charge density,

Then, the charge will be,

$q=\rho\times\dfrac{4}{3}\pi R^3$ .....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

$\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}$

$E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}$

Put the value from equation (I)

$E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}$

$E=\dfrac{qr}{4\pi\epsilon_{0}R^3}$

Hence, The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

4 0

3 years ago

a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin

Sveta_85 [38]

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

6 0

4 years ago

Whats happening in this diagram?

yaroslaw [1]

The ship sends waves to the bottom of the ocean so if this waves hit any solid particle it reflect back to the main ship , and the ship send the divers to explore

6 0

3 years ago