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3 years ago

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A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height

of 3.0 m. What is its angular momentum if its radius is 7.4 cm? (I know the answer is 9.6 kgm^2/s but I don't understand how to get it)

1 answer:

stira [4]3 years ago

4 0

Answer:

$9.5 kg m^2/s$

Explanation:

The angular momentum of an object is given by:

$L=mvr$

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

$L=(0.35)(9.0)(3.0)=9.5 kg m^2/s$

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adelina 88 [10]

Answer:

0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli

1.1m to 0.04125 m

Explanation:

T= 1/f=

if f= 300Hz then T = 1/300 =0.003333 s

if f= 8000 then T= 1/8000 = 0.000125s

now v=f×wave length

or wavelength = speed/ frequency

when f = 300 Hz

wavelength = 330/300=1.1 m

wavelength = 330/8000 = 0.04125m

note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s

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2 years ago

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex

Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, $d = 1.8 \ m$

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point $D$ , the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$$AD-BD = \frac{\lambda}{2}$

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

$f=340.9744$ Hz

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2 years ago

Use wave equation calculate the speed of sound in the air if frequency of 110 hz has a wave length of 3 m

kirill115 [55]

Answer = 330 m/s

The wave equation is as follows:

Wave speed = wavelength x frequency

The known values are:
Wavelength = 3m
Frequency = 110 Hz

Substitute the known values into the wave equation to find the wave speed.

Wave speed = 3 x 110

Wave speed = 330 m/s

8 0

1 year ago

What is the current in a 120V circuit if the resistance is 20Ω?

Kaylis [27]

We have: $I=\frac{U}{R}=\frac{120}{20}=6A$

ok done. Thank to me :>

6 0

2 years ago

A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

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3 years ago