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Zarrin [17]
3 years ago
10

Which group of stars is represented by the line image

Physics
2 answers:
Lerok [7]3 years ago
8 0

Answer:

ursa Major

Explanation:

AURORKA [14]3 years ago
8 0

Answer: ursa major

Explanation:

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"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree
kenny6666 [7]

I do not agree with the statement.
The "substance" can be a compound.  It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.
 
6 0
2 years ago
In our solar system, most asteroids are found _____.A.In the kuiper beltb.Orbiting close to the sunc.Between mars and jupiterd.B
nydimaria [60]
Between Mars and Jupiter
7 0
3 years ago
Read 2 more answers
As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equat
DaniilM [7]

Answer:

the volume in terms of time t ⇒ \\V = e^{0.05t} + 399\\

Explanation:

Given that :

\frac{dv}{dt}= kv\\\\\int\limits \, \frac{dv}{v}=  \int\limits \, kdt\\In v = kt + C_1\\v = e^{kt} + C\\400 = e^{k*0} + C\\400 = 1 + C\\C = 400 -1\\C = 399

V = e^{kt} + 399

When v = 300 ;  \frac{dv}{dt}= - 15

then

\frac{dv}{dt}= kv\\\\-15 = 300*k\\\\k = \frac{-15}{300}\\\\\\k = \frac{-1}{20}\\\\k = -0.05

∴ \\V = e^{0.05t} + 399\\

Therefore, the volume in terms of time t ⇒ \\V = e^{0.05t} + 399\\

6 0
3 years ago
A 3.06kg stone is dropped from a height of 10.0m and strikes the ground with a velocity of 7.00m/s. What average force of air fr
xxTIMURxx [149]
<span>22.5 newtons. First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as: E = 0.5mv^2 where E = Energy m = mass v = velocity Substituting known values and solving gives: E = 0.5 3.06 kg (7 m/s)^2 E = 1.53 kg 49 m^2/s^2 E = 74.97 kg*m^2/s^2 Now ignoring air resistance, how much energy should the rock have had? We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So 3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2 So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is 299.88 J - 74.97 J = 224.91 J So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division. 224.91 J / 10 m = 224.91 kg*m^2/s^2 / 10 m = 22.491 kg*m/s^2 = 22.491 N Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
3 0
3 years ago
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
3 years ago
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