What are important physics points when bungee jumping :)

Light in vacuum travels at a speed of 3.00 x 10^8 m ^-1 s^-1 on average earth is 93,000,000 miles from the sun how many minutes

NemiM [27]

The correct unit for the speed of light is [ m s⁻¹ ].

Time = (distance) / (speed)

Time = (9.3 x 10^7 miles) x (1609 m/mile) / (3 x 10^8 m/s) = 498.8 seconds .

That would be <em>8.31 minutes</em>.

6 0

2 years ago

Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas

Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

$I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}$

For 4 particles:

$I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

$I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}$

c)

$I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

8 0

2 years ago

Batman chased the joker on his batbike for 20 minutes, traveling at the speed of 30 kilometers per minute. How far did Batman go

AURORKA [14]

Answer:

123 kilometer

Explanation:

none

8 0

2 years ago

A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car,

rusak2 [61]

The answer is D I’m not really sure yet

6 0

2 years ago

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity

$v=\sqrt{2.14^2+0.34^2}=2.17m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0$

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0

3 years ago