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Pepsi [2]
3 years ago
15

When we plot all the points that satisfy an equation or inequality we ___ it.

Physics
2 answers:
snow_tiger [21]3 years ago
6 0
We<span> use </span>inequalities<span> when there is a range of possible answers for a situation. ... </span>inequalities—inequalities<span> that can be written in the form of a linear </span>equation. ... the bounded region, and anypoint<span> within this region </span>satisfies<span> the </span>inequality<span> x ≥ -</span><span>2. ... </span>All<span> of the </span>points<span> under the line are shaded; this is the range of </span>points<span> where the ...</span>
Sergio039 [100]3 years ago
5 0

I just took this test!

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A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
3 years ago
A good way to de-magnetize something is to___.
Dominik [7]

HEYA MATE

YOUR ANSWER IS <em><u>D.PLACE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>IN</u></em><em><u> </u></em><em><u>AN</u></em><em><u> </u></em><em><u>ELECTRI</u></em><em><u>C</u></em><em><u> </u></em><em><u>FIELD</u></em>

<em><u>BE</u></em><em><u>CAUSE</u></em><em><u> </u></em><em><u>IT </u></em><em><u>makes</u></em><em><u> sense you can use alternating current to remove magnetism</u></em>

4 0
3 years ago
Read 2 more answers
Density is given in ____.<br> a. Pa/cm3<br> c. g/s2<br> b. N/m2<br> d. g/cm3
Ainat [17]
Density is defined as  [mass] / [volume] .

The only choice listed with those physical dimensions is 'd' .
3 0
3 years ago
What is the combined resistance for two resistors with resistances of 10 Ohms and 23.7 Ohms if they are connected in series
Agata [3.3K]

Answer: 33.7Ω

Explanation:

Since there are two resistors connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i.e Rtotal = R1 + R2

R1 = 10Ω

R2 = 23.7Ω

Hence, Rtotal = 10Ω + 23.7Ω

Rtotal = 33.7Ω

Thus, the combined resistance for two resistors is 33.7Ω

3 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
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