Ask question

Ask question

All categories

3 years ago

8

During which step of the technological design process would a scientist try to determine if anyone has already studied the probl

em they have identified?

1 answer:

Vadim26 [7]3 years ago

6 0

Background research.

You might be interested in

A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go? (b) how long is it in the air

Mariulka [41]

Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

Answer: The ball stays in the air for 4.5 s (nearest tenth)

3 0

4 years ago

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser

Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em></em>

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0

3 years ago

A 100 kg person rides in an elevator moving at a constant speed of

Airida [17]

Answer:

Explanation:

All this information only applies to the person. There is an extra tension force if we are talking about the elevator, but we are not. Dont forget to apply the units

Acceleration means change in speed or velocity. The elevator is moving at a constant speed of 3 meters. You wont even know you are moving because there is no change in acceleration. It equals 0

The forces ONLY acting on the person would be the force of gravity pulling them down, and the normal force that the elevator is reciprocating from the person standing on it.

Force = mass x acceleration. You have 100 kg and you are accelerating at 0 m/s. The force is 0. Which makes sense because the force of gravity and the net force completely cancel each other out.

8 0

3 years ago

What waves have high amplitudes

Olegator [25]

-- loud sounds

-- bright lights

-- strong radio signals

-- Slinkies that can pinch you painfully

-- a tsunami in the ocean

-- earthquakes above Richter 5 or 6

5 0

4 years ago

At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a

enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

6 0

3 years ago