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Mekhanik [1.2K]
3 years ago
12

An object moving at a constant velocity accelerates at a constant rate of 2.5 m/s2 for a distance of 30 m. At the end of the acc

eleration the object has a velocity of 90 m/s. What was its velocity before accelerating?
Physics
1 answer:
nydimaria [60]3 years ago
5 0

Answer:

The initial velocity is 89.16m/s

Attached is the solution

Explanation:

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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:

1)

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

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

3)

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

4)

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
Which of the following is NOT a layer of the Sun?
Alja [10]
It depends when you look at a pichture it could possibly help.
3 0
4 years ago
Any five chapter 9 institutions found in the constitution of south africa
navik [9.2K]
Five institutions from the chapter 9 found in the constitution of south Africa are : 
- The Public Protector
- The south African Human right commission
- The commission of Gender Equality
- The Auditor- General
- The Independent Electoral Commission

Hope this helps
7 0
3 years ago
An airplane travels 2800 km at a speed of 700 km/h, decreases its speed to 500 km/h for the next 1500 km and travels the last 10
Dafna11 [192]

Answer:

he average speed for the airplane is 558 km/hr

Explanati

6 0
3 years ago
A sprinter who is running a 200-m race travels the second 100 m in much less time than the first 100m because
Harrizon [31]

Answer:

He is warmed up now

Explanation:

His muscles are better and stretched now

3 0
3 years ago
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