What is a successful result of science

a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of

Papessa [141]

The initial speed of the bolt is not 58.86 m/s.

Let a be the acceleration of the rocket.

During the 4 sec lift off, the rocket has reached a height of

h = (1/2)*a*t^2

with t=4,

h = (1/2)*a^16

h = 8*a

Its velocity at 4 sec is

v = t*a

v = 4*a

The initial velocity of the bolt is thus 4*a.

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,

h = (1/2)*g*t^2 + V0*t

Substituting h0=8*a, t=6 and V0=-4*a into it,

8*a = (1/2)*g*36 - 4*a*6

Solving for a

a = 5.52 m/s^2

6 0

2 years ago

Which component is found only in electric generators and not in electric motors?

liraira [26]

The answer is D. Slip rings.

4 0

3 years ago

Read 2 more answers

if an object is moving at a constant speed in a changing direction, What is the acceleration of the object (zero or not zero)

Andreas93 [3]

We know, acceleration is the change of velocity by time.
Velocity is the speed of an object which also indicates the direction.
Hence, acceleration is both dependant upon the speed as well as the direction.
So, if an object is moving at a constant speed in a changing direction, the acceleration will also change. It will not be zero.
An example is that of uniform circular motion.

Answer:

if an object is moving at a constant speed in a changing direction, the acceleration of the object will not be zero.

4 0

2 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):