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lbvjy [14]
3 years ago
9

Why do planents revolve around the sun ?​

Physics
1 answer:
Natali [406]3 years ago
4 0

Answer:

Sun's Gravitational Pull

Explanation:

Planets revolve around the sun because of the Sun's gravitational pull pulling all the planets to revolve around it.

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A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t
Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

3 0
2 years ago
Where does matter come from?
Pie
The Big Bang theory is matter and energy in the universe exploded out from one point. As the explosion occurred, energy and matter spread outward and formed the universe. The matter from the Big Bang formed clouds of gas.
8 0
2 years ago
a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction  30° east of north
liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement D_{1} = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey D_{2} = 20 km (red arrow) is at an angle with reference to y-axis.

So, we first find the components of the red arrow along X and Y.

Component of D_{2} along X-axis is given by  D_{2x}  = D_{2} Sin 30 = 10 km

Component of D_{2} along Y-axis is given by  D_{2y}  = D_{2} Cos 30 = 17.32 km

We now add all the vectors along X and along Y separately.

Net Displacement along X  D_{netX} = 10 km

Net Displacement along Y D_{netY} = 30 + 17.32 = 47.32 km

Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the D_{net}

D_{net}  = \sqrt{D_{netX} ^{2} + D_{netY} ^{2}}

Therefore, [tex]D_{net} = 48.37 km.

Hence, the distance between the ports P and R is 48 km.

6 0
3 years ago
A 10 m long steel beam is accidentally dropped by a construction crane from a height of 4.89 m. The horizontal component of the
Otrada [13]

Answer:

e = 1.21 mV

Explanation:

given,                                

length of rod = 10 m                

height of drop = 4.89 m          

Earth’s magnetic field =  12.4 µT

acceleration of gravity = 9.8 m/s²

velocity of the beam                      

v = \sqrt{2gh}

v = \sqrt{2\times 9.8 \times 4.89}

v = 9.79 m/s                        

emf of the beam

e = B l v                              

e = 12.4 x 10⁻⁶ x 9.79 x 10

e = 1.21 x 10⁻³ V

e = 1.21 mV

4 0
3 years ago
A bugle can be thought of as an open-end pipe. If a bugle were straightened out, it would be 2.65 m long. a. If the speed of sou
vitfil [10]

Answer

given,

Length of pipe, L = 2.65 m

speed of sound, v = 343 m/s

Pipe is open end Pipe

a) Lowest frequency

  condition, for open end pipe

     \lambda_1 = 2 L

     \lambda_1 = 2\times 2.65  

     \lambda_1= 5.30 m

   For lowest frequency

    f_1 = n\dfrac{v}{\lambda_1}

     n = 1

    f_1 = \dfrac{343}{5.30}

           f₁ = 64.72 Hz

b) For second frequency in bugle

        n = 2

   f_2 =2\times \dfrac{343}{5.30}

          f₂ = 129.43 Hz

   for n = 3

   f_3 =3\times \dfrac{343}{5.30}

          f₃ = 194.16 Hz

5 0
3 years ago
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