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3 years ago

Use the graph below to answer the following question: What is happening to the object's velocity?

Physics

1 answer:

hjlf3 years ago

4 0

The object's velocity is decreasing.

Explanation:

From graph is the attached image, we can clearly point that the velocity of this motion is decreasing with time.

Velocity is a vector quantity.

The y-axis represent displacement.
The x-axis depicts time
Using the graph, we know that the slope of the line on the graph gives us the velocity as it denotes the change of displacement with time.
When we find the slope, it will give us a negative value which shows that the body is slowing down and not increasing speed.

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

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Which data set has the largest standard deviation

AfilCa [17]

Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

$\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}$

Where $x_i$ is the value of each measurement, n is the number of elements in the set, and $\mu$ is the average or media of the values, defined as

$\displaystyle \mu=\frac{\sum x_i}{n}$

Let's analyze each set of data:

A.3,4,3,4,3,4,3

The average is

$\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}$

$\sigma=0.5$

B.1,6,3,15,4,12,8

The average is

$\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}$

$\sigma=4.7$

C. 20, 21,23,19,19,20,20

The average is

$\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}$

$\sigma=1.3$

D.12,14,13,14,12,13,12

The average is

$\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}$

$\sigma=0.8$

We can see the data set marked as B has the largest standard deviation

5 0

4 years ago

In running a ballistics test at the police department, Officer Rios fires a 6.1g bullet at 350 m/s into a container that stops i

Citrus2011 [14]

Answer:

–0.11 N.

Explanation:

Hope this helps

4 0

3 years ago

A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2

Tom [10]

the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

x = A cos (wt + Ф)

w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

w= $\sqrt{ \frac{200}{0.300} }$

w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

v= $\frac{dx}{dt}$

v= - A w sin (wt +Ф)

0 = -A w sin Ф

Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

4.00 = A cos 25.8t

Speed.

50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

Cos² 25.8t = $\frac{16}{A^2}$

sin² 25.8t = $\frac{3.756}{A^2 }$

1 = $\frac{1}{A^2} \ (16 + 3.756)$

A = $\sqrt{19.756}$

A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here: brainly.com/question/14311816

4 0

2 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

3 years ago

Read 2 more answers

Why is it impossible to create a perpetual motion machine?

enyata [817]

Answer:

The first law of thermodynamics is the law of conservation of energy. It states that energy is always conserved. ... To keep a machine moving, the energy applied should stay with the machine without any losses. Because of this fact alone, it is impossible to build perpetual motion machines.

Explanation:

8 0

3 years ago