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kap26 [50]
3 years ago
8

According to the Holland Codes, which career is best suited to someone who scored highest in the investigative category?

Physics
1 answer:
Yuri [45]3 years ago
8 0
<span>Laboratory chemist working on cancer research.</span>
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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
Am arrow of mass 1000kg is shot into a wooden block of mass 5000lg lying at rest in a smooth surface.If the arrow travels 15m/s
Ber [7]

Answer:

Vf=3

Explanation:

you must first write your data

data before impact

M1=1000 M2=5000

V1=0 m/s V2 =0m/s

data after impact

M1=1000 M2=5000

V1=15m/s V2=?

M1V1 +M2V2=M1V1 +M2V2f

(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf

0=15000+5000Vf

- 15000÷5000=5000Vf÷5000

Vf= -3

Vf =3

6 0
2 years ago
A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.
Inessa [10]

Hi there!

We can use Newton's Second Law:
\Sigma F = ma

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}

6 0
2 years ago
Que cuerpos celestes observó laika durante su viaje
sergejj [24]
Do you want me to translate it?
4 0
3 years ago
Identify the wave with the shortest wave length.
Valentin [98]

Answer:

d) gamma

Explanation:

hope i helped

7 0
3 years ago
Read 2 more answers
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