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3 years ago

A pendulum takes 0.5 second to complete one cycle. What is the pendulum’s period and frequency?

1 answer:

8 0

Period: Time it takes to complete a cycle, it takes 0.5s to complete a cycle
Frequency: Number of cycles in a second. Since 0.5s = 1 cycle, 1s=2 cycles
Period is 0.5s, frequency is 2

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3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a

inessss [21]

Answer:

The average angular acceleration is $\alpha =125.487 rad /s^2$

Explanation:

From the question we are told that

The length of the bat is $l = 0.85m$ \

The initial linear velocity is $u = 0 m/s$

The time is $t = 0.15s$

The velocity at t is $v = 16 m/s$

Generally average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_o}{t}$

Where $w_f$ is the finial angular velocity which is mathematically evaluated as

$w_f = \frac{v}{l}$

$w_f = \frac{16}{0.85}$

$= 18.823 rad/s$

and $w_o$ is the initial angular velocity which is zero since initial linear velocity is zero

$\alpha = \frac{18.823 - 0}{0.15}$

$\alpha =125.487 rad /s^2$

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3 years ago

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri

omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if f = fs, i.e.

fs = μmg = m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N

7 0

3 years ago

Read 2 more answers

If the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amper

Elza [17]

Emf = d (phi-B) / dt
B dA/dt, where dA/dt is the area swept out by the wire per unit time. 
0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so 
L = (0.88 J/C) / [ (0.075 N s/C m)(4.2 m/s) ] = about 3 meters

6 0

3 years ago

Read 2 more answers

A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What

iogann1982 [59]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

$\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s$

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

$a=\omega^2 r$

At the instant t = 0,

$\omega=3.93 rad/s$

So the centripetal acceleration of the tip of the blades is

$a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2$

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3 years ago

What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e

Zinaida [17]

Answer:

$F=1.38*10^{-9}N$

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

Here k is the Coulomb constant. In this case, we have $q_1=-e$ , $q_2=e$ and $d=4.09*10^-10m$ . Replacing the values:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N$

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

$F=1.38*10^{-9}N$

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3 years ago