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valentinak56 [21]
3 years ago
15

In astronomy, the term perihelion refers to the point in the orbit of an asteroid, comet, or planet when it is closest to the su

n. Aphelion refers to the point when the object is farthest away from the sun. Research facts about the dwarf planet Ceres. Calculate the average orbital speed of Ceres in miles/hour when it is closer to the sun and when it is further away.
Physics
1 answer:
Ivenika [448]3 years ago
3 0
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Drag the positive or negative feedback loop on the left to each process on the right. terms may be used once, more than once, or
slamgirl [31]

The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

<h3>What is a feedback loop?</h3>

A system component known as a feedback loop is one in which all or a portion of the output is used as input for subsequent actions. A minimum of four phases comprise each feedback loop. Input is produced in the initial phase. Input is recorded and stored in the subsequent stage. Input is examined in the third stage, and during the fourth, decisions are made using the knowledge from the examination.

Both negative and positive feedback loops are possible. Insofar as they stay within predetermined bounds, negative feedback loops are self-regulating and helpful for sustaining an ideal condition. One of the most well-known examples of a self-regulating negative feedback loop is an old-fashioned home thermostat that turns on or off a furnace using bang-bang control.

To learn more about feedback loop, visit:

brainly.com/question/11312580

#SPJ4

5 0
2 years ago
. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass
alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

3 0
2 years ago
A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe
KIM [24]

Answer: V = 15 m/s

Explanation:

As  stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately.

8 0
2 years ago
A circus act involves a trapeze artist and a baby elephant. They are going to balance on a teeter-totter that is 10 meters long
elena-s [515]

Answer:

Explanation:

Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman

weight x distance from pivot

= 500x 5

= 2500 Nm

torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance

= 2500x d

for equilibrium

2500 d = 2500

d = 1 m

So elephant will have to walk up to 1 m close to pivot or middle point.

6 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
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