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4 years ago

What is the experimental group(s)?

1 answer:

5 0

<h3>I think it B The group(s) that gets the special treatment.</h3><h3 /><h3>I hope this is correct.</h3><h3 /><h3 /><h3 />

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PLEASE HELP!!!!

Goryan [66]

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let $u$ denote the initial velocity of the vehicle ( $20\; \text{mph}$ or $60\; \text{mph}$ ) and let $v$ denote the velocity of the vehicle after braking ( $0\; \text{mph}$ ). Let $x$ denote the braking distance.

Assume that the acceleration during braking are both constantly $a$ in both scenarios. The SUVAT equations would apply. In particular:

$\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}$ .

Since $v = 0$ <em> </em>(the vehicle has completely stopped), the equation becomes $x = (-u^{2}) / (2\, a)$ .

Assuming that $a$ (braking acceleration) stays the same, the braking distance $x$ would be proportional to $u^{2}$ , the square of the initial velocity.

Hence, increasing the initial speed from $20\; \text{mph}$ to $60\; \text{mph}$ would increase the braking distance by a factor of $3^{2} = 9$ .

7 0

1 year ago

Read 2 more answers

A potter's wheel has the shape of a solid uniform disk of mass 13.0 kg and radius 1.25 m. It spins about an axis perpendicular t

Fittoniya [83]

Answer: $10.82 kg-m^2$

Explanation:

Given

Mass of solid uniform disk $M=13 kg$

radius of disk $r=1.25 m$

mass of lump $m=1.7 kg$

distance of lump from axis $r_0=0.63$

Moment of inertia is the distribution of mass from the axis of rotation

Initial moment of inertia of disk $I_1=\frac{Mr^2}{2}$

$I_1=\frac{13\times 1.25^2}{2}=10.15 kg-m^2$

Final moment of inertia $I_f$ =Moment of inertia of disk+moment of inertia of lump about axis

$I_f=\frac{Mr^2}{2}+mr_0^2$

$I_f=10.15+1.7\times 0.63^2$

$I_f=10.15+0.674$

$I_f=10.82 kg-m^2$

3 0

4 years ago

5. Find the volume of the composite space figure to the nearest whole number.

Dmitry [639]

The shape is missing but let's consider it a semi-cylinder attached to the rectangular prism.
Given:
radius = 4.5 mm
<span>Height = 11 mm </span>

<span>Volume of cylinder = (1/2)(pi)(4.5)^2(11) (the shape is divided into half)
V = 349.89 mm cubed
Volume of prism = L x W x H
= 9 x 11 x 6
= 594 mm cubed
Total volume of the composite shape = 111.375 + 594
= 943.89 mm cubed
Rounded answer = 944 mm cubed.</span>

6 0

4 years ago

Jenni wants to start a print shop making specialty papers and cards. Which economic question has she answered?

Artemon [7]

I think it is Letter D, because she already said what to make

5 0

3 years ago

Read 2 more answers

Line segment gj is a diameter of circle l. angle k measures (4x 6)°. circle l is inscribed with triangle g j k. line segment g j

choli [55]

The value of x in the given right triangle in a semicircle is determined as 21.

<h3>What is the measure of a triangle in a semicircle?</h3>

The triangle in a semicircle is always a right angle triangle.

From the figure shown, we can say that the triangle G J K is right triangle and m<K = 90degrees.

Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:

4x + 6 = 90

4x = 90 - 6

4x = 84

x = 21

Thus, the value of x in the given right triangle in a semicircle is determined as 21.

Learn more about right angle here: brainly.com/question/64787

#SPJ4

8 0

2 years ago