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Gala2k [10]
4 years ago
6

A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward

acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30m/s. How long did the burn phase last?
Physics
1 answer:
SVETLANKA909090 [29]4 years ago
3 0
The important thing to note here is the direction of motion of the test rocket. Since it mentions that the rocket travels vertically upwards, then this motion can be applied to rectilinear equations that are derived from Newton's Laws of Motions.These useful equations are:

y = v₁t + 1/2 at²
a = (v₂-v₁)/t

where
y is the vertical distance travelled
v₁ is the initial velocity
v₂ is the final velocity
t is the time 
a is the acceleration

When a test rocket is launched, there is an initial velocity in order to launch it to the sky. However, it would gradually reach terminal velocity in the solar system. At this point, the final velocity is equal to 0. So, v₂ = 0. Let's solve the second equation first.

a = (v₂-v₁)/t
a = (0-30)/t
a = -30/t

Let's substitute a to the first equation:
y = v₁t + 1/2 at²
49 = 30t + 1/2 (-30/t)t²
49 = 30t -15t
49 = 15 t
t = 49/15
t = 3.27 seconds

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A car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The diameter of a tire is 80.2 cm. Find the numb
BaLLatris [955]

By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.

ANGULAR MOTION

Since the car accelerate from rest, initial velocity will be zero.

Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters

  • Initial velocity U = 0
  • Final velocity V = 24.3 m/s
  • Time t = 9.1 s

If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.

First equation

V = U + at

Substitute all necessary parameters into the equation.

24.3 = 0 + 9.1a

a = 24.3/9.1

a = 2.67 m/s^{2}

Third Equation of motion

V^{2} = U^{2} + 2aS

Substitute all the necessary parameters

24.3^{2} = 0 + 2 x 2.67 x S

590.49 = 5.34S

S = 590.49 / 5.34

S = 110.58 m.

Given that the diameter of a tire is 80.2 cm,

the radius (r) will be 80.2/2 = 40.1 cm

convert it to meter

r = 40.1/100 = 0.401 m

The Circumference of the tire = 2\pir

Circumference = 2 x 3.143 x 0.401

Circumference = 2.52 m

Assuming no slipping, number of revolutions = 110.58/2.52

Number of revolutions = 43.89 rev.

Number of revolutions = 43 rev.

Therefore, the number of revolutions the tire makes during this motion is 43 rev.

Learn more about circular motion here: brainly.com/question/6860269

4 0
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you have dissolved 10g sodium oxide in 200 ml water .calculate the concentration of the solution .physical​
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Answer:

0.05

Explanation:

Divide the mass of the solute by the total volume of the solution. Write out the equation C = m/V, where m is the mass of the solute and V is the total volume of the solution. Plug in the values you found for the mass and volume, and divide them to find the concentration of your solution.

6 0
3 years ago
During the launch from a board, a diver's angular speed about her center of mass changes from zero to 8.10 rad/s in 240 ms. Her
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Answer:

Part a)

\alpha = 33.75 rad/s^2

Part b)

\tau = 391.5 Nm

Explanation:

Part a)

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so it is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{8.10 - 0}{240\times 10^{-3}}

\alpha = 33.75 rad/s^2

Part b)

average external torque is given as

\tau = I\alpha

here we know that

I = 11.6 kg m^2

\tau = 11.6 \times 33.75

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6 0
3 years ago
I know this ain't right but can someone please correct the mistakes I have it for a test​
nikklg [1K]

Answer:

NICE HANDWRITING. I've done this before and I'm sure everything is correct. Although I would re-check to see if you had any mistakes with your calculations. Good job!

Explanation:

8 0
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