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natka813 [3]
3 years ago
13

How much kinetic energy a spring that streches 4m would be?

Physics
1 answer:
miv72 [106K]3 years ago
8 0

Answer:

Explanation:bjbhfcvvjkkknbnnnm

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What is the science principle that explains magnetic fields? I need this fast, please!
bekas [8.4K]

Answer:

I literally just learned this last week and if I remember correctly it is Faraday's Law of Induction.

Explanation: Hope this helps also I hope you have/had an amazing day today<3

3 0
2 years ago
In a mall, a shopper rides up an escalator between floors. At the top of the escalator, the shopper turns right and walks 6.40 m
Elden [556K]

Answer:

25 degrees above the horizontal.

5 0
2 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Vlada [557]

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M_S = 1.99 × 10³⁰ kg

Mass of the neutron star

M_N = 2( M_S )

M_N = 2( 1.99 × 10³⁰ kg )

M_N = ( 3.98 × 10³⁰ kg )

Radius of neutron star R_N = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω_N.

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM_N = / R_N² = mR_Nω_N²

ω_N² = GM_N = / R_N³

ω_N = √(GM_N = / R_N³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω_N = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω_N = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω_N = √ 120831133.3636777

ω_N = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

5 0
2 years ago
At which position is the northern hemisphere experiencing winter?
Leto [7]

The answer would be point A.

Hope this helped you.

6 0
3 years ago
A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches
tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

{v}^{2}  =  {z}^{2}  + 2ax

v = z + at

First let's find the final velocity the plane will have at the end of the runway using the first equation:

{v}^{2}  =  {0}^{2}  + 2(5)(1800)

v = 60 \sqrt{5}

Now we can plug this into the second equation to find t:

60 \sqrt{5}  = 0 + 5t

t = 12 \sqrt{5}

Then using 3 significant figures we round to 26.8 seconds

3 0
2 years ago
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