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3 years ago

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A rocket initially at rest accelerates at a rate of 99.0 meters/second2. calculate the distance covered by the rocket if it atta

ins a final velocity of 445 meters/second after 4.50 seconds.

2 answers:

Paraphin [41]3 years ago

7 0

Initial velocity(u) = 0m/s. Acceleration(a)= 99m/s^2. Final velocity(v)= 445m/s. Time taken by rocket to cover s distance = 4.5s. Using kinematic equation v^2 - u^2 = 2as, (445)^2 - 0^2 = 2x99xs. s = 445^2 / 2x99 = 1000.12 m. Hope it helps.

saveliy_v [14]3 years ago

6 0

Something must be wrong in the data you have, since this is basic using of linear motion's formulas.
vf=v0+at. Where vf= final velocity; v0= initial velocity, a=acceleration and t=time.
If the rocket is initially at rest, v0=0. Therefore vf=at. Plugging numbers in gives 445=99*4.5, However
445≠445.5.

Check it and then calculate the distance from x=a*t^2.

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2 years ago

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A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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2 years ago

A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft

erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

$SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)$

Now, differentiating both sides with respect to time 't', we get:

$2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}$

Now, we are given speed as 25 mph. So, $\frac{de}{dt}=25\ mph$

Also, when $e=91\ mi$ , we can find 'x' using equation (1). This gives,

$x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi$

Now, plug in the values of 'e' and 'x' and solve for $\frac{dx}{dt}$ . This gives,

$\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph$

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

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3 years ago

Complete the passage to describe the relationship between kinetic energy, internal energy, thermal energy, and

butalik [34]

Answer:

Increases

Increases

Increases

Explanation:

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2 years ago

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A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil

o-na [289]

Answer:

a) U = 735 J , b) U = 125.7 J , c) U = 0 J

Explanation:

The gravitational power energy is

U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

y = 2.10 m

w = mg = 350 N

U = 350 2.10

U = 735 J

b) when the angle is 34º

y = L - L cos 34

y = L (1- cos34)

y = 2.10 (1- cos 34)

y = 0.359 m

U = 350 0.359

U = 125.7 J

c) in this case this point coincides with the reference system

y = 0

U = 0 J

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3 years ago