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3 years ago

What term was used to describe the final digit

Physics

1 answer:

xxMikexx [17]3 years ago

7 0

Answer:

Significant digits (also called significant figures or “sig figs” for short) indicate the precision of a measurement. A number with more significant digits is more precise. For example, 8.00 cm is more precise than 8.0 cm.

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Afina-wow [57]

Answer:

2.76×10⁻¹⁰ C

Explanation:

Applying,

V = W/q................... Equation 1

Where V = Electric Potential, E = Electric potential energy, q = charge.

make q the subject of the equation

q = W/V................ Equation 2

From the question,

Given: W = 4.26×10⁻⁸ J, V = 154.5 V

Substitute these values into equation 2

q = 4.26×10⁻⁸/154.5

q = 2.76×10⁻¹⁰ C

3 0

3 years ago

"how long would it take for a ball dropped from the top of a 400-foot building to hit the ground? round your answer to two decim

Lelechka [254]

Hello

Let's convert first the height from foot to meters:
$S=400~ft=121.92~m$
This is the space covered by the ball in its accelerated motion, where the acceleration is the gravitational acceleration:
$a=g=9.81~m/s^2$
The law of motion is given by
$S= \frac{1}{2}at^2$
Re-arranging, we can get the total time of the motion:
$t= \sqrt{ \frac{2S}{a} }=4.99~s
$
which is the time the ball takes to hit the ground.

7 0

3 years ago

Read 2 more answers

Suppose you were marooned on a tropical island and had to use seawater (density 1.10 g·cm3) to make a primitive barometer. what

Paul [167]

We know that P = hρg

Where:

P - pressure Pa,

h - height in meter,

ρ – would be the density in kg / m^3; and

g - acceleration due to gravity is m / s^2

p = hρg if h = 0.735 meter, ρ = 13600 kg / meter^3, g = 10 meter/ sec^2

p = 0.735*13600*10 = 99960 Pa or

P = 1 x 10^5 Pa

Now with sea water if we have to make a barometer:

ρ = 1100 kg / meter^3 (given)

P = hρg if we put the value of P calculated and the value of ρ = 1100 kg / meter^3 given, we will

get, 1 x 10^5 = h*1100*10

therefore, h = 9.09 meter or 29.82 feet of water.

3 0

4 years ago

A man is running with a tennis ball in his hand. On his left is a bull's-eye target painted

san4es73 [151]

Answer: Horizontal force.

5 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

3 years ago