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zysi [14]
3 years ago
8

In a Venn diagram, the separate circles contain characteristics unique to each item being compared and the intersection contains

characteristics that are common to both items being compared. Ernie is working on the Venn diagram below to compare the career pathways of Biotechnology Research and Development and Diagnostic Services.
What else could Ernie put in the common section?

Collecting data and analyzing results
Designing and implementing systems
Maintaining and using diagnostic equipment
Designing and using laboratory equipment
Mark this and return
Physics
2 answers:
kumpel [21]3 years ago
4 0

A- Collecting data and analyzing results

i just took the test

Vesna [10]3 years ago
3 0
<span>I think its- Collecting data and analyzing results
Thats what I put.</span>
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A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will
gtnhenbr [62]
1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
x(t)=v_0t
y(t)=h- \frac{1}{2}gt^2
where the motion on the x-axis is a uniform motion with constant speed v_0=18.1 m/s, while the motion on the y-axis is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2, and initial height h=1.85 m.

First of all, we can find the time t at which the ball reaches the ground by requiring y(t)=0:
0=h- \frac{1}{2}gt^2
t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }=  0.61 s

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one: 
v_x=18.1 m/s
On the y-axis, the velocity is given by
v_y(t)=gt
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s
And the speed of the ball is the magnitude of the resultant of the two components:
v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s
8 0
3 years ago
Read 2 more answers
n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad
Oksi-84 [34.3K]

Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

F1/A1 = F2/A2

From our question,

The radius of the smaller piston is 5 cm = 0.05 m

The radius of the larger piston is 15 cm = 0.15 m

The force of the larger piston is 13300 N

The force of the smaller piston is unknown = F

A1 = πr² = 3.142 * 0.05² = 0.007855 m²

A2 = πr² = 3.142 * 0.15² = 0.070695 m²

F1/0.007855 = 13300/0.070695

F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

5 0
3 years ago
A 4 kg block is pulled with 5.6 N of force to the right. The block experiences 1.1 N of friction. What is the acceleration of th
BARSIC [14]

Answer:

The acceleration of the box is 1.125 m/s² towards right.

Explanation:

Mass of the box, m=4 kg

Force acting towards right, F=5.6 N

Frictional force acting towards left, f=1.1 N

Let the acceleration be a m/s².

Now, net force acting on the box towards right is given as:

F_{net}=F-f=5.6-1.1=4.5\textrm{ N}

From Newton's second law of motion,

F_{net}=ma\\4.5=4a\\a=\frac{4.5}{4}=1.125\textrm{ }m/s^2

Therefore, the acceleration of the box is 1.125 m/s² towards right.

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In 10s a car accelerates 4m/s^2 to 60m/s. How fast was the car going before it accelerated?
slega [8]

v2 = v1 + at 60 = v1 + 4*10  v1 = 20 m/s

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