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3 years ago

How many significant digits are in the measurement 50.003010 nm?

Physics

2 answers:

masha68 [24]3 years ago

8 0

Answer:

There are 15 significant digits

Explanation:

50.00301 nm = 0.0000000500301

Counting the figures makes it 15

Lynna [10]3 years ago

5 0

Answer:

Significant digits = 8

Explanation:

Significant digits in a number which is more than 1 and if the number if having decimal in it then as per the rule all the digits present in the number must be significant digit

for an example we will say

1.00 = 3 significant figures

so in this way for all such kind of numbers we have to count all digits in it

so here given number is more than 1 and it contains the decimal in it

so we will say

50.003010 nm = 8 significant figures

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AIR MASSES!!

777dan777 [17]

Answer:

A : hot and moist, maritime tropical

B: cold and dry, maritime polar

C: hot and moist , maritime tropical

D: cold and dry, continental polar

E: hot and moist , maritime tropical

F: cold and dry , maritime polar

Explanation:

Cold air is denser than warm air. The more water vapor that is in the air, the less dense the air becomes. That is why cold, dry air is much heavier than warm, humid air.

Maritime polar (mP) air masses are cool, moist, and unstable. Some maritime polar air masses originate as continental polar air masses over Asia and move westward over the Pacific, collecting warmth and moisture from the ocean.

Maritime tropical (mT) air masses are warm, moist, and usually unstable.

5 0

3 years ago

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago

Light travels faster in warmer air. On a sunny day, the sun can heat a road and create a layer of hot air above it. Let's model

AysviL [449]

Answer:

Explanation:

If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.

n = c / v

If light travels faster in warmer air, it will have a lower refractive index

nh < nc

Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:

n1 * sin(θ1) = n2 * sin(θ2)

sin(θ2) = sin(θ1) * n1/n2

If blue light from the sky passing through the hot air will cross to the cold air, then

n1 = nh

n2 = nc

Then:

n1 < n2

So:

n1/n2 < 1

The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.

6 0

3 years ago

In an experiment, a variable, position-dependent force FC) is exerted on a block of mass 1.0 kg that is moving on a horizontal s

marshall27 [118]

Answer:

C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Fr.

Explanation:

Yo want to prove the following equation:

$W_N=\Delta K\\\\$

That is, the net force exerted on an object is equal to the change in the kinetic energy of the object.

The previous equation is also equal to:

$F(x)x-F_f=\frac{1}{2}m(v_f^2-v_o^2)$ (1)

m: mass of the block

vf: final velocity

v_o: initial velocity

Ff: friction force

F(x): Force

x: distance

You know the values of vf, m and x.

In order to prove the equation (1) it is necessary that you have C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F. Thus you can calculate experimentally both sides of the equation.

8 0

2 years ago

What is the movement of a stationary object?

galben [10]

If it is stationary, its not moving. there is no movement

8 0

3 years ago