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2 years ago

6

You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling a

lmost touches the floor and that its period is 15 s. The acceleration of gravity is 9.81 m/s2 .
How tall is the tower? Answer in units of m

1 answer:

expeople1 [14]2 years ago

7 0

Answer:

L=55.9m

Explanation:

The equation for the period of a simple pendulum is:

$T=2\pi\sqrt{\frac{L}{g}}$

In our case what we know is the period and the acceleration of gravity, and we need to know the length of the pendulum, so we can write:

$L=(\frac{T}{2\pi})^2g$

Which for our values is:

$L=(\frac{15s}{2\pi})^2(9.81m/s^2)=55.9m$

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A 42.0-cm diameter disk rotates with a constant angular acceleration of 2.70 rad/s2. It starts from rest at t = 0, and a line dr

Zielflug [23.3K]

Answer:

6.21 rad/s

1.3041 m/s, 0.567 m/s²

$106.4778\ ^{\circ}$

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration = 2.3 rad/s²

$\theta$ = Angle of rotation

t = Time taken = 2.3 s

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=0+2.7\times 2.3\\\Rightarrow \omega_f=6.21\ rad/s$

The angular speed is 6.21 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=0.21\times 6.21\\\Rightarrow v=1.3041\ m/s$

Linear velocity is 1.3041 m/s

Tangential acceleration is given by

$a_t=r\alpha\\\Rightarrow a_t=0.21\times 2.7\\\Rightarrow a_t=0.567\ m/s^2$

Tangential acceleration is 0.567 m/s²

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2.7\times 2.3^2\\\Rightarrow \theta=7.1415\ rad$

In degress the angle would be

$57.3+7.1415\times \dfrac{180}{\pi}=466.47780\ ^{\circ}$

From x axis it would be

$466.47780-360=106.4778\ ^{\circ}$

The angle is $106.4778\ ^{\circ}$ from x axis

7 0

3 years ago

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? ( g = 10m/s2 )

LekaFEV [45]

Answer:

Height is 11.25m

Explanation:

<u>Given the following data;</u>

Initial velocity, u = 0

Final velocity, v = 15m/s

Acceleration due to gravity, g = 10m/s²

To find the height, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement (height) measured in meters.

$V^{2} = U^{2} + 2aS$

<em>Making S the subject, we have;</em>

$S = \frac {V^{2} - U^{2}}{2a}$

But a = g = 10m/s²

<em>Substituting into the equation, we have;</em>

$S = \frac {15^{2} - 0^{2}}{2*10}$

$S = \frac {225 - 0}{20}$

$S = \frac {225}{20}$

S = 11.25m

<em>Therefore, the ball will reach a height of 11.25m before it begins to fall. </em>

4 0

2 years ago

A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then

Dominik [7]

Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

$v^{2}-u^{2}= 2as$

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$v^{2}-(0)^{2}= 2(1.44)(30)$

$v= 9.3 m^{-1}$

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

$s=ut+\frac{(1)}{2}at^2$

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$ .

$30=(0)t+\frac{(1)}{2}(1.44)t^2$

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

$T=6.45+\frac{s'}{v}$

Put s= 30 m, $v= 9.3 m^{-1}$ and s'= 100 m.

$T=6.45+\frac{(100)}{9.3}$

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

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3 years ago

an object is moving in a straight line with a constant acceleration initially it is traveling at 16 meters per second. 3 seconds

VLD [36.1K]

A)It moved 6 meters.

B)It would 3 meters.

I think this is the answer.

Have a good day.

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3 years ago

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Aneli [31]

when you solve an equation the quadratic of g(f) must be treated like any other mathematical quantity— they must be multiplied, divided, raised to powers, cancelled, etc in exactly the same way as the numbers to which they belong

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2 years ago