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Sever21 [200]
3 years ago
13

What rule should be used to transform a table of data to represent the

Physics
1 answer:
adoni [48]3 years ago
4 0

Answer:

Multiply the x values with -1.

Explanation:

By multiplying the numbers by one, you are changing them to be the opposite of their original state.

You multiply the numbers that are in the x-value column because you are reflecting the image over the y-axis.

<em>Hope this helped and good luck!</em>

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Question 7 of 10
Dahasolnce [82]
Kinetic friction (also referred to as dynamic friction) is the force that resists the relative movement of the surfaces once they're in motion.
https://www.khanacademy.org › stat...
Static and kinetic friction example (video) | Khan Academy

Answer a would be static friction
Answer b is fluid friction
(Air resistance is fluid friction. Fluid friction is the friction experienced by objects which are moving in a fluid and the air is a fluid.)
Answer c is static friction
ANSWER D IS KINETIC FRICTION

Hope this helps :D



4 0
3 years ago
Two pipes of equal length are each open at one end. Each has a fundamental frequency of 470 Hz at 310 K. In one pipe the air tem
Tresset [83]

Answer:

Given:

Fundamental frequency: 470Hz

T1:310k,T2:315k

Calculating velocity

Recall v=(331m/s)✓[T1/273k)

V=331✓(310/273)

V1=331*(1.0656)=352.72m/s

V2=331✓(315/273)=355.5m/s

Fundamental frequency=4L

F2=F1(V2/V1)

F2=470(355.5/352.72)=474.4Hz

Beat=[F2-F1]=474.4-470=4.4Hz

Explanation:

7 0
3 years ago
A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration
bazaltina [42]

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

3 0
3 years ago
Which one of the risk factors that influence you learn how may be most difficult to control as a young teen
nalin [4]
I hope my answer helped u under stand better
8 0
3 years ago
A mass m is attached to a string connected to a force sensor on a rotating platform. The platform’s angular velocity, ω, can be
makkiz [27]

Answer:

The mass m is 0.332 kg or 332 gm

Explanation:

Given

The platform is rotating with angular speed , \omega =8.5\, \frac{rad}{sec}

Mass m is moving on platform in a circle with radius , r=0.20\, m

Force sensor reading to which spring is attached , F=4.8\, N

Now for the mass m to move in circle the required centripetal force is given by F=m\omega ^{2}r

=>4.8=m\times 8.5 ^{2}\times 0.20

=>m=0.332\, kg

Thus the mass m is 0.332 kg or 332 gm

7 0
3 years ago
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