Give me 3 events of matter and energy that you can't explain yourself ( need some ideas)

A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

3 0

3 years ago

What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms

kirza4 [7]

Answer:

F' = (3/2)F

Explanation:

the formula for the electric field strength is given as follows:

E = F/q

where,

E = Electric Field Strength

F = Force due to the electric field

q = magnitude of charge experiencing the force

Therefore,

F = E q ---------------- equation (1)

Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:

F' = (E/2)(3 q)

F' = (3/2)(E q)

using equation (1)

3 0

3 years ago

Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris

telo118 [61]

Answer:

a

$F_A =425.42 \ N$

b

$F_A_H = 358.58 \ N$

Explanation:

From the question we are told that

The diameter of the Ferris wheel is $d = 80 \ ft = \frac{80}{3.281} = 24.383$

The period of the Ferris wheel is $T = 24 \ s$

The mass of the passenger is $m_g = 40 \ kg$

The apparent weight of the passenger at the lowest point is mathematically represented as

$F_A_L = F_c + W$

Where $F_c$ is the centripetal force on the passenger, which is mathematically represented as

$F_c =m * r * w^2$

Where $w$ is the angular velocity which is mathematically represented as

$w = \frac{2* \pi }{T}$

substituting values

$w = \frac{2* 3.142 }{24}$

$w = 0.2618 \ rad/s$

and r is the radius which is evaluated as $r = \frac{d}{2}$

substituting values

$r = \frac{24.383}{2}$

$r = 12.19 \ ft$

So

$F_c = 40 * 12.19* (0.2618)^2$

$F_c = 33.42 \ N$

W is the weight which is mathematically represented as

$W = 40 * 9.8$

$W = 392 \ N$

So

$F_A = 33.42 + 392$

$F_A =425.42 \ N$

The apparent weight of the passenger at the highest point is mathematically represented as

$F_A_H = W- F_c$

substituting values

$F_A_H = 392 - 33.42$

$F_A_H = 358.58 \ N$

4 0

3 years ago

A box with a mass of 18 kg is pushed across the floor. It has coefficient of friction of 0.39. Calculate the force of friction i

Taya2010 [7]

Answer:

68.8 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of box = 18 Kg

Coefficient of friction (μ) = 0.39

Force of friction (F) =?

Next, we shall determine the normal force of the box. This is illustrated below:

Mass (m) of object = 18 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal force (N) =?

N = mg

N = 18 × 9.8

N = 176.4 N

Finally, we shall determine the force of friction experienced by the object. This is illustrated below:

Coefficient of friction (μ) = 0.39

Normal force (N) = 176.4 N

Force of friction (F) =?

F = μN

F = 0.39 × 176.4

F = 68.796 ≈ 68.8 N

Thus, the box experience a frictional force of 68.8 N.

3 0

2 years ago

What is the mechanical advantage of the wheel and axle shown below?

BigorU [14]

The correct answer is A. 32.5

Mechanical advantage is the ratio of force that is input into a machine to the force output.

Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.

MA=R/r where R is the radius of the wheel and r is the radius of the axle.

Substituting for the values in the question gives:

MA=26cm/0.8cm

=32.5

6 0

3 years ago

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Give me 3 events of matter and energy that you can't explain yourself ( need some ideas)​

Give me 3 events of matter and energy that you can't explain yourself ( need some ideas)