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AleksandrR [38]
3 years ago
12

List two examples of when you would use light years as a measurement.

Physics
1 answer:
astra-53 [7]3 years ago
7 0
One is when you are measuring a distance in space! I don't know the other but hope you find another example!
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A shadow from a surgeon's hand obstructs your view while operating. Make suggestions for an alternative light source that avoids
Charra [1.4K]

Answer:

B...................

4 0
3 years ago
The difference between the observed points and the regression line points is equal to the:________
pshichka [43]

The difference between the observed points and the regression line points is equal to the correlation.

The strength and direction of a relationship between two or more variables are described by the statistical measure of correlation, which is given as a number. However, a correlation between two variables does not necessarily imply that a change in one variable is the reason for a change in the values of the other.

Regression expresses the relationship as an equation, whereas correlation assesses the strength of the linear link between two variables. The square of the correlation coefficient, also known as Pearson's r, between the observed and predicted values in a regression is sometimes referred to as R2.

Learn more about correlation here;

brainly.com/question/6563788

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8 0
2 years ago
. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your d
Scorpion4ik [409]

Answer:

Explanation:

The total distance is how far you walk from the starting point.

Distance through west = 18.0m

Distance through north = 25.0m

Total distance covered = 18.0+25.0m

Total distance covered = 43.0m

This means that I am 43.0m from the starting point

Displacement is the distance covered in a specified direction. The displacement will be gotten using the Pythagoras theorem as shown:

d^2 = 25^2 + 18^2\\d^2 = 625+324\\d^2 = 949\\d = \sqrt{949}\\ d = 30.81m

The direction of your displacement is 30.81m

Direction is gotten according to the formula;

\theta = tan ^{-1}{\frac{y}{x} }\\\theta = tan ^{-1}{\frac{25}{-18} }\\\theta = tan ^{-1}-1.3889}\\\theta = -60.27^0\\\theta = 180-60.27\\\theta = 119.7^0

<em>Note that the direction to the west is negative, that is why the x is -18.0m</em>

<em></em>

5 0
2 years ago
Which properties can be used to determine whether a substance is a liquid or a plasma? Check all that apply.
Annette [7]
A liquid is a phase of a substance that is characterized by being flowing and has no shape. The molecules of this substance are far apart but much closer than the particles in a gas. Plasma is a state of a substance that consists of ionized gas. So, it in the phase of a gas and has a charge. We can therefore distinguish the two by checking which substance has a charge.
5 0
2 years ago
A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on t
Phoenix [80]

Answer:

a.I=981.34 N*s

b.v_f=3.96 m/s

c.v_{f1}=3.63m/s

d.y_f=0.673m

Explanation:

Given: m=67kg, h=0.720m, 0

a.

I=\int\limits^{t_1}_{t_2} {F(t)} \, dt

F(t)=9200*t-11500t^2

I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt

I=4600*t^2-3833.3*t^3|(0.80,0)

I=2944-1962.66=981.35

I=981.34 N*s

b.

v_f^2=v_i^2+a*y'

Starting from the rest

v_f^2=0+2*9.8m/s^2*0.80s

v_f^2=15.68

v_f=\sqrt{15.68m^2/s^2}=3.96 m/s

c.

I_{total}=p_f

I_1-m*g*d=m*v_{f1}-m*v_f

981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)

Solve to vf

v_{f1}=3.63m/s

d.

v_f^2=v_i^2+2*a*y_f'

y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2

y_f=0.673m

7 0
3 years ago
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