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3 years ago

What do scientists study to figure out how long it takes the Sun to make one rotation on its axis?

Physics

2 answers:

Alinara [238K]3 years ago

6 0

C. Sunspots there is your answer

lilavasa [31]3 years ago

3 0

B auroras because it is used to figure out how long it takes the Sun to make one rotation on it axis

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

m1 = 1.29 kg

m2 = 0.3 kg

x1 = -0.75 m

x2 = -0.2 m

g = 9.8 m/s^2

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

3 0

3 years ago

A long, thin solenoid has 390 turns per meter and a radius of 1.20 cm. The current in the solenoid is increasing at a uniform ra

gtnhenbr [62]

To solve this problem it is necessary to apply the concepts related to Faraday's law and the induced emf.

By definition the induced electromotive force is defined as

$\int E dl = -\frac{d\phi}{dt}$

$\int E dl = -(\frac{dB}{dt})A$

Where,

$\phi =$ Electric field

B = Magnetic Field

A = Area

At the theory the magnetic field is defined as,

$B = \mu_0 NI$

Where,

N = Number of loops

I = current

$\mu_0 =$ Permeability constant

We know also that the cross sectional area, is the area from a circle, and the length is equal to the perimeter then

A = \pi r^2

l = 2\pi r

Replacing at the previous equation we have that

$E (2\pi r) = \mu_0 n (\frac{di}{dt})(\pi R^2)$

Where,

R = Radius of the solenoid

r = The distance from the axis

Re-arrange to find the current in function of time,

$\frac{di}{dt} = \frac{Er}{\mu_0 NR^2}$

Replacing our values we have

$\frac{di}{dt} = \frac{(8.00*10^{-6})(0.0348)}{(4\pi*10^{-7})(390)(1.2*10^-2)^2}$

$\frac{di}{dt} = 3.94487A/s$

8 0

3 years ago

A satellite is to be launched into an orbit of radius,r Show that v = 2gr, where V is the

Snezhnost [94]

Explanation:

We start by using the conservation law of energy:

$\Delta{K} + \Delta{U} = 0$

$\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0$

Simplifying the above equation, we get

$v^2 = 2G\dfrac{M}{r}$

We can rewrite this as

$v^2 = 2\left(G\dfrac{M}{r^2}\right)r$

Note that the expression inside the parenthesis is simply the acceleration due to gravity $g$ so we can write

$v^2 = 2gr$

where $v$ is the launch velocity.

6 0

3 years ago

In figure 1, charge q2 experiences no net electric force. What is q1?

lukranit [14]

By using Coulomb's law, we want to find the value of q₁ given that q₂ experiences no net electric force. We will find that q₁ = 8nC

<h3>Working with Coulomb's law.</h3>

Coulomb's law says that for two charges q₁ and q₂ separated by a distance r, the force that each one experiences is:

$F = k\frac{q_1*q_2}{r^2}$

Where k is a constant

Here we can see that q₂ interacts with two charges, then the total force on q₂ will be:

$F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2}$

And we know that it must be equal to zero, so we can write it as:

$F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2} = 0\\\\k*q_2*(\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2}) = 0\\$

The parenthesis must be equal to zero, so we can write:

$\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2} = 0$

And now we can solve this for q₁ to get:

$q_1 = 2nC*(\frac{(20cm)^2}{(10cm)^2} ) = 8nC$

If you want to learn more about Coulomb's law, you can read:

brainly.com/question/24743340

3 0

3 years ago