Examples of drawing packages

A rocket ship is accelerating at 200 m/s2, its mas is 135,000,000 kg. What is the force generated by this acceleration?

Rina8888 [55]

Acceleration does NOT "generate" force. Acceleration NEEDS force to make it happen. Without force ... provided by something else ... acceleration can't happen.

The force NEEDED to accelerate a mass with a certain acceleration is

Force needed = (mass) times (acceleration)

For the rocket ship in the question,

Force = (135,000,000 kg) times (200 m/s²)

Force = (135,000,000 x 200) kg-m/s²

<em>Force = 27 Giga-Newtons </em>(27,000,000,000 Newtons)

The gas-generator cycle F-1 rocket engine, developed in the US by Rocketdyne in the late 1950s, was used in the Saturn V rocket, the main launch vehicle of NASA's Apollo moon lander program . Five F-1 engines were used in the first stage of each Saturn V.

==> The thrust of each F-1 engine at full throttle is 7,770 kilo-Newtons.

It would take <em>3,475 </em>of these F-1 rocket engines, running full-throttle, to provide the force calculated in the answer to this question. If you didn't have 3,475 F-1 rocket engines, then you couldn't accelerate 135,000,000 kg at 200 m/s².

(And by the way ... the mass of each F-1 engine is 8,400 kg. So 3,475 engines alone account for 22% of the mass you're trying to accelerate. And don't even get me started about the mass of the FUEL you'd need to carry.)

5 0

3 years ago

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

2 years ago

A ______ is a closed loop containing a source of electrical energy and a load.

gavmur [86]

Answer:

d. circuit

Explanation:

the parts of the circuit consists of a load or resistance

7 0

3 years ago

A 6.9 kg object is suspended by a string from the ceiling of an elevator. The acceleration of gravity is 9.8 m/s 2 . Determine t

lina2011 [118]

Answer:

Tension, T = 87.63 N

Explanation:

Given that,

Mass of the object, m = 6.9 kg

The string is acting in the upward direction, a = 2.9 m/s²

Acceleration due to gravity, g = 9.8 m/s²

As the lift is accelerating upwards, it means the net force acting on it is given by :

T = m(a+g)

= 6.9 (2.9+9.8)

= 6.9(2.7)

= 87.63 N

So, the tension in the string is 87.63 N.

6 0

3 years ago

Which symbol is used for an alpha particle? a. β c. γ b. Δ d. α

xenn [34]

It is the very last answer

6 0

3 years ago