All categories

3 years ago

25.5 m/s in 5.75 s. What is the acceleration?

1 answer:

7 0

Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)

follow me on instagram : imrajsingh :):) thanks

You might be interested in

I NEED HELP ASAP<br>please please please

raketka [301]

Answer:

B AND D

HOPE ITS HELF FUL .........

6 0

3 years ago

Read 2 more answers

One clue we can observe that means a chemical reaction has occurred is the formation of light. What is one chemical reaction you

dimaraw [331]

Answer:

There are five signs of a chemical change:

Color Change.

Production of an odor.

Change of Temperature.

Evolution of a gas (formation of bubbles)

Precipitate (formation of a solid)

Explanation:

I just went ahead and gave you the five signs of chemical change hoped it helped

8 0

3 years ago

The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

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

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

PLEASE HELP ASAP...TIMED TEST. PLEASE ANSWER WITH AT LEAST A PARAGRAPH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

charle [14.2K]

Answer: I am pretty sure that you should pick radio waves.

Explanation: The scientist should use radio waves. I think this because you can use the radio waves to analyze the signals from outer space. This will work much better than anything there, to analyze it the best possible.

The best I could do.

8 0

3 years ago

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

3 years ago