Which example shows an endothermic reaction?

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

Force = 27 Giga-Newtons (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 3,475 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

Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r

Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

$F_{c} = \dfrac{mv^{2}}{r}$

where, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0

2 years ago

A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel

aivan3 [116]

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

A. The applied force should be the same size as the friction force

7 0

2 years ago

A baseball m=.34kg is spun vertically on a massless string of length l=.52m. the string can only support a tension of tmax=9.9n

larisa86 [58]

4.5 m/s This is an exercise in centripetal force. The formula is F = mv^2/r where m = mass v = velocity r = radius Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop. Let's determine the force we get from gravity. 0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N Plug known values into formula. F = mv^2/r 13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m 6.88064 kg m^2/s^2 = 0.34 kg V^2 20.23717647 m^2/s^2 = V^2 4.498574938 m/s = V Rounding to 2 significant figures gives 4.5 m/s The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.

7 0

3 years ago

Read 2 more answers

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$