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4 years ago

14

What is the biggest star in the milky way?

1 answer:

lora16 [44]4 years ago

5 0

The biggest star in our Milky Way is a hypergiant and is the one of the largest known stars. It's much larger than our Sun, probably more than 1,800+ times than our Sun, and since our Sun is acutally pretty big, that's saying a lot.

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You can also enter units that are combinations of other units. keep in mind that you have to indicate the multiplication of unit

omeli [17]

Watch this ! You've never seen anything like it !
No wait. I'm pretty sure you have.

Weight = (mass) · (gravity)

Weight = (10 kg) · (9.8 m/s²)

Weight = (10 · 9.8) kg·m/s²

Weight = 98 Newtons

7 0

4 years ago

he magnetic flux through a loop of wire decreases from 1.7 Wb to 0.3 Wb in a time of 0.4 s. What was the average value of the in

Pani-rosa [81]

Answer:

Induced emf through a loop of wire is 3.5 V.

Explanation:

It is given that,

Initial magnetic flux, $\phi_i=1.7\ Wb$

Final magnetic flux, $\phi_f=0.3\ Wb$

The magnetic flux through a loop of wire decreases in a time of 0.4 s, t = 0.4 s

We need to find the average value of the induced emf. It is equivalent to the rate of change of magnetic flux i.e.

$\epsilon=-\dfrac{\phi_f-\phi_i}{t}$

$\epsilon=-\dfrac{0.3\ Wb-1.7\ Wb}{0.4\ s}$

$\epsilon=3.5\ V$

So, the value of the induced emf through a loop of wire is 3.5 V.

5 0

4 years ago

Explain how the data collected and the calculations for the first and second resonance points in today's experiment would change

grandymaker [24]

Answer:

tssths

Explanation:

hgst

8 0

3 years ago

An electrolytic cell can be used to plate gold onto other metal surfaces. The plating reaction is: Au (aq) e- --> Au(s) Notic

Dmitriy789 [7]

Answer:

Explanation:

Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time

Q = I × t = n × F

Where

Q is charge

I is current = 0.2A

t is time = ?

n is number of mole

F is Faraday constant = 96500C/s

Mass of Au = 0.4 g = 0.0004kg

Molar mass of Au = 197 g/mol

Malar mass = mass / mole

n = 0.4 / 197

n = 2.03× 10^-3 mol

n = 0.00203 mol

t = n × F / I

t = 0.00203 × 96500 / 0.2

t = 979.7 seconds

t = 16.32 minutes.

4 0

3 years ago

A World-class sprinter can reach a top speed of about 11.5 m/s in the first 18.0 m of a race. What is the average acceleration o

irina [24]

Answer

a = 3.674 m / s ^ 2

t = 3.13 s

Using the kinematic equations for the movement we have:

$h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2$ (1)

$V_{f} = V_{0} + at$ (2)

Where:

$P_{0}$ = initial position

$V_{0}}$ = initial velocity

a = acceleration

t = time in seconds

$V_{f}$ = final speed

We know:

$P_{0}=0$

$V_{0}= 0$

h = 18 m

$V_{f} = 11.5\frac{m}{s}$

So:

From (2) we have that: $t =\frac{V_{f}}{a}$

$t =\frac{11.5}{a}$

From (1) we have to:

$h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2$

Then we clear "a" to find the acceleration.

$\frac{36}{t^2} = a\\a = \frac{36}{(\frac{11.5}{a})^2} \\\\a =\frac{11.5^2}{36}\\a = 3.674 m / s ^2$

Then, the time it takes to reach this speed is:

$t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s$

4 0

4 years ago

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