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3 years ago

14

A BALL IS THREWN WITH AN ANGULAR DIRECTION TOTHE HORIZONTAL AXIS .IT TAKES 4SECONDS TO MOVE BETWEEN TWO PILLARA WICH HAS A HEIGH

T OF 20 METERES.WHAT IS THE MAXIMUM HEIGHT IT CAN REACH.

1 answer:

hjlf3 years ago

6 0

Answer:

So -24.5 m merely means the depth of the base of the tower ie the ground from the top of the tower. So wrt the ground height of the tower = 24.5 m.

Explanation:

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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:

a)

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:

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

b)

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,

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

c)

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

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

d)

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 particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoi

Pavlova-9 [17]

Answer:

$v(1.5)=0.7648\ m/s$

Explanation:

<u>Dynamics</u>

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

$\displaystyle a=\frac{F}{m}$

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

$F=2cos1.1t$

The variable acceleration is calculated by:

$\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}$

$a=0.8439cos1.1t$

The instant velocity is the integral of the acceleration:

$\displaystyle v(t)=\int_{t_o}^{t_1}a.dt$

$\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt$

Integrating

$\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}$

$\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)$

$\boxed{v(1.5)=0.7648\ m/s}$

3 0

3 years ago

You drive 200 miles in 3 hours before stopping for lunch and gas. After lunch you

Paraphin [41]

Given :

You drive 200 miles in 3 hours before stopping for lunch and gas. After lunch you travel 250 miles in an hour and a half.

To Find :

Average speed.

Solution :

We know, average speed is given by :

$v=\dfrac{\text{Total distance covered}}{\text{Total time taken}}\\\\v=\dfrac{200+250}{3+1.5}\ miles/hr\\\\v=\dfrac{450}{4.5}\ miles/hr\\\\v=100\ miles/hr$

Therefore, average speed of the journey is 100 miles/hr.

Hence, this is the required solution.

4 0

4 years ago

A spinning top initially spins at 16rad/s but slows down to 12rad/s in 18s, due to friction. If the rotational inertia of the to

Alex Ar [27]

Answer:

The change in angular momentum is $\Delta L = 0.0016 \ kgm^2/s$

Explanation:

From the question we are told that

The initial angular velocity of the spinning top is $w_1 = 16 \ rad/s$

The angular velocity after it slow down is $w_2 = 12 \ rad/s$

The time for it to slow down is $t = 18 \ s$

The rotational inertia due to friction is $I = 0.0004 \ kg \cdot m^2$

Generally the change in the angular momentum is mathematically represented as

$\Delta L = I *(w_1 -w_2)$

substituting values

$\Delta L = 0.0004 *(16 -12)$

$\Delta L = 0.0016 \ kgm^2/s$

5 0

3 years ago

What aspects of his fitness program could keep him from achieving his goal?

Flura [38]

Idk maybe excersice everyday

4 0

4 years ago