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Sever21 [200]
3 years ago
13

Why do we use scatter plots for most data in Physical Science? Question 1 options: because the teacher said to use scatter plots

because scatter plots show 100% of the data like a pie because scatter plots show the relationship between the independent and dependent variables because scatter plots show data for categories that do not overlap
Physics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

scatter plots show the relationship between the independent and dependent variables

Explanation:

A scatter plot is a graph which shows two variables plotted along two axes (usually the x and y axes). Scatter plots are useful in establishing any form of correlation between the dependent and independent variables in any study.

Correlation simply means the degree of relationship between variables, that is, how much does one variable affect the other? When scatter plots are almost a straight line graph, there is a high correlation between the variables. When the points in a scatter plot are isolated, there is little (sometimes zero) correlation between the variables.

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Two objects of same material are travelling near you. Object A is a 1.9 kg mass traveling 8 m/s; object B is a 2 kg mass traveli
Alex73 [517]

To determine the object which could give the greatest impact we will apply the concept of momentum. The object that has the highest momentum will be the object that will impact the strongest. Our values are

Mass of Object A

m_A=1.9 kg

Velocity of object A

v_A=8ms

Mass of object B

m_B=2 kg

Velocity of object B

v_B=5ms

The general formula for momentum is the product between mass and velocity, then

p = mv

For each object we have then,

p_A=m_Av_A=1.9 kg(8ms)=15.2kg \cdot m/s

p_B=m_Bv_B=2 kg(5ms)=10kg \cdot m/s

Since the momentum of object A is greater than that of object B, then object A will make you feel force upon impact.

3 0
3 years ago
An object is at rest. There are several forces acting on the object, but the net force is zero. If all the
pantera1 [17]
It would not. Imagine four forces equal in magnitude but opposite in direction (e.g. north, east, south, and west). If these forces were to double in magnitude they would still have the same magnitude, meaning the net force is still equal to zero.
3 0
2 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in
Paraphin [41]

Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

v_{av}=\frac{d}{t}

where,

v_{av} = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

8 0
3 years ago
Read 2 more answers
Why metals have thermoconductivity higher than ceramic?
Pie

Answer:

Thermal Conductivity Easily Transmits Heat Among Fine Ceramics

4 0
2 years ago
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