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2 years ago

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What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters

/second in a circle with a radius of 10.0 centimeters? A. 1.82 × 102 meters/second2 B. 3.61 × 103 meters/second2 C. 5.64 × 103 meters/second2 D. 2.49 × 103 meters/second2 E. 1.18 × 103 meters/second2

1 answer:

Artist 52 [7]2 years ago

8 0

Answer: $3.61(10)^{3} \frac{m}{s^{2}}$

The centripetal acceleration $a_{c}$ of an object moving in a uniform circular path is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$

Where:

$V=19m/s$ is the velocity

$r=10cm=0.1m$ is the radius of the circle

$a_{c}=\frac{(19m/s)^{2}}{0.1m}$

$a_{c}=3610m/s^{2}=3.61(10)^{3}m/s^{2}$

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We have that the values for F north, F east, F up are

$F_N=1.09090909*10^{-5}$
$F_E=5.18181818*10^{-6}$
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From the Question we are told that

electric force $F_1 = 1.2 x 10^{-3} N(N)$

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electric force , $F_3=2.2 x 10^{-4} N (U)$

charge on this ball one $q_1= 110 nC.$

charge on this ball two $q_2= -50 nC.$

Generally the equation for the F north is mathematically given as

$F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}$

$F_N=1.09090909*10^{-5}$

For F East

$F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}$

$F_E=5.18181818*10^{-6}$

For F UP

$F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}$

$F_E=2*10^{-6}$

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2 years ago

adelina 88 [10]

Answer:

I'm pretty sure that the answer is A

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2 years ago

Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are

Mars2501 [29]

Answer:

θ = 36°

Explanation:

given,

F₁ = 22.8 N

F₂ = 16.6 N

magnitude of force = ?

direction of force = ?

$F = \sqrt{F_1^2 + F_2^2}$

$F = \sqrt{22.8^2 + 16.6^2}$

$F = \sqrt{795.4}$

F = 28.20 N

direction

$\theta = tan^{-1}(\dfrac{F_2}{F_1})$

$\theta = tan^{-1}(\dfrac{16.6}{22.8})$

$\theta = tan^{-1}(0.728)$

θ = 36°

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2 years ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

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A car moves with the speed of 40km/hr for the first half distance and 60km/hr for second half distance.find average speed of car

slava [35]

Answer:

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average = (a + b) / 2

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