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3 years ago

Which data set has the largest range?

Physics

1 answer:

blsea [12.9K]3 years ago

3 0

Answer:

Line the numbers from smallest to largest the subtract the smallest from the largest numbers.

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A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0

2 years ago

Read 2 more answers

List three reasons that rain predictions made by meteorologists are sometimes incorrect. Explain your response.

Agata [3.3K]

1. the low pressure is moving slower than expected.
This make the meteorologist receive premature data which make them fail to interpret the data correctly and make the wronf prediction.
2. Sudden change in wind direction, which transfer the natural occurence into other region than where it initially predicted
3. We still haven't developed the methodology to 100% predict natural occurence

5 0

2 years ago

PLEASE HELP! I NEED TO ANSWER THIS QUICKLY!

rodikova [14]

Answer:

I'm not a genius ok?

Explanation:

1. Radar communication, Analysis of the molecular and atomic structure, telephone communication

2. c

4 0

2 years ago

Someone help this is a test and i need to finish this quick i will give brainliest of it lets me

tekilochka [14]

Sorry, I don't know but I think the correct answer is the first option.

7 0

2 years ago

Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

Given:

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

Assume:

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

4 0

3 years ago