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3 years ago

In Which figure below is the trend line drawn correctly

Physics

1 answer:

saul85 [17]3 years ago

5 0

Answer:

The first graph is the trend line.

Explanation:

Trend Line

It's a line indicating the tendency of something being measured or evaluated. The trend line is meant to represent an approximation of a set of points of a measured magnitude.

We have three possible 'trend lines' to choose from. The second is not a line, so it's not the correct option. The third is not a line either, it's just a graph who joins all the measured points. It doesn't mark a trend.

The first graph is a trend line because it's a single line and can be used to represent the whole set of values .

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2. Above right: The four positions of the thrown ball with no gravity are

padilas [110]

The characteristics of the projectile launch allows to find the results for the questions about the movement of the ball are:

In the attached we see the vertical speed decreases with height and the curve is of the parabolic type.

In the movement is several dimensions, each one is independent of the others, the movement in the x axis does not affect the movement in the y axis.

Kinematics studies the motion of bodies looking for relationships between position, velocity and acceleration. In the case of vertical and projectiles launch the acceleration on the vertical axis is the acceleration of gravity directed downward.

In the attachment we can see the position of the ball for two distances in the case of projectile launching.

We can see that the speed of the ball decreases with height according to the relation

y = go t - ½ g t²

Where y is the height, g is the initial vertical velocity, g is the acceleration of gravity and t is time.

In all movements in various dimensions we assume that each movement in an . axis is independent

In the case of projectile launching, on the vertical axis there is an acceleration of gravity and on the horizontal axis there is no acceleration, the only parameter that this gives the two movements is the time, which is a scalar.

In conclusion, using the characteristics of the projectile launch, we can find the results for the questions about the movement of the ball are:

In the movement is several dimensions, each one is independent of the others, the movement in the x axis does not affect the movement in the y axis.

In the attached we see the vertical speed decreases with height and the curve is of the parabolic type.

Learn more about projectile launch here: brainly.com/question/24888457

4 0

2 years ago

A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.

Elza [17]

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant $k=4.45\times 10^3 N/m$

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

$P_i=\frac{kx^2}{2}$

$P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J$

$P_f=mgh=0.25\times 9.8\times h$

$P_i=P_f$

$14.24 =0.25\times 9.8\times h$

$h=\frac{14.24}{0.25\times 9.8}=5.81 m$

6 0

2 years ago

A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

2 years ago

Why are SI units used for scientific works ?

sineoko [7]

Energy can be one answer! There are many, but energy is a main one.

5 0

3 years ago

Read 2 more answers

If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com

Lelu [443]

Answer:

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = $100*10^{-6} R$

for dry skin, voltage will be

V = IR = $100*10^{-6}*200R = 0.02R$

Comparing both voltages

$0.02R$ ÷ $100*10^{-6} R$ = 200

this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.

4 0

3 years ago