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3 years ago

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