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4 years ago

When an object is turning around, is it also at rest at the point?

Physics

2 answers:

Ganezh [65]4 years ago

8 0

Answer:

yes

Explanation:

Depolarization

AysviL [449]4 years ago

8 0

When an object changes direction, it’s velocity goes to 0 at that point. So the object will also be considered at rest.

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A hollow cubical box is 0.221 m on an edge. This box is floating in a lake with 1/4 of its height beneath the surface. The walls

yanalaym [24]

Answer:

3/4 filled with water

Explanation:

x = Fraction of box filled

g = Acceleration due to gravity = 9.81 m/s²

$\rho_w$ = Density of water

V = Volume of water

Weight of empty box is equal to the weight of the water displaced

$W_b=\frac{1}{4}V\times \rho_wg$

As all the forces are conserved

$W_b+xV\rho_wg=V\rho_wg\\\Rightarrow \frac{1}{4}V\times \rho_wg+xV\rho_wg=V\rho_wg\\\Rightarrow \frac{1}{4}+x=1\\\Rightarrow x=1-\frac{1}{4}\\\Rightarrow x=\frac{2}{3}$

So, the box starts to sink when the box is 3/4 filled with water

8 0

3 years ago

You have a new goal of saving at least $4,500 over the course of the next year.

expeople1 [14]

Answer:

You would need to save $300 every month to meet your savings goal.

Explanation:

4,500-900=3,600

3,600/12=300

3 0

3 years ago

Read 2 more answers

An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

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

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

The angular displacement of the wheel in a certain time interval t is given by

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

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

A major artery with a 1.7 cm2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area

Alex

Answer:

0.14

Explanation:

Flow rate is the volume flowing through a point at a particular time, in calcuing flow rate we have

Q= v*t

it in terms of Area, we have Q= A*v

Where A= area

v= velocity.

Solving the question , flow rate is constant then

A*v= constant

A(i) v(i)= A(f) v(f)

Where A(i)= initial area= 1.00cm^2

A(f)= final area= 0.400cm^2

V(i) and V(f) are the initial and final velocity respectively and the ratio of the two will gives us the factor

Substitute the values into the equation we have

1 V(i)= 4 V(f)

But we were told that the cross sectional area of 1.00cm^2 branches into 18 smaller arteries.

Then

1 V(i)=0.4 V(f)*(18)

1 V(i)=7.2V(f)

Then if we find the ratio of the velocity, we will get the factor.

V(f)/V(i)= 1/7.2

V(f)/V(i)=0.14

Hence, the factor of the average velocity of the blood reduced when it passes into these branches is 0.14

8 0

3 years ago

A rugby player passes the ball 7.88 m across the field, where it is caught at the same height as it left his hand. (Assume the p

damaskus [11]

Explanation:

Given

Range of ball =7.88 m

Initial speed=13.1 m/s

and we know

Range of Projectile is

$R=\frac{u^2sin2\theta }{g}$

$13.1=\frac{13.1^2\times sin2\theta }{9.8}$

$7.88\times 9.8=13.1^2\times sin2\theta$

$sin2\theta =0.4499$

$2\theta =26.74^{\circ}$

$\theta =13.37 ^{\circ}$

(b)other angle would be complementary of $\theta$

$=90-\theta =76.63 ^{\circ}$

8 0

3 years ago