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Licemer1 [7]
3 years ago
7

A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end

from the mass m, the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity as the rod swings through its lowest (vertical) position?
Physics
1 answer:
konstantin123 [22]3 years ago
3 0

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

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The lungs are located in the chest true or false
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8 0
2 years ago
Read 2 more answers
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
lisov135 [29]

Answer:

(a) 70cm³

(b) 805 grams

Explanation:

(a) V = L×B×H

= 7cm×5cm×2cm

= 35cm×2cm

= 70cm³

(b) Mass = Volume × Density

= 70cm³ × 11.5g/cm³

= 805 grams

3 0
3 years ago
Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th
Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0
3 years ago
Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).
MrRissso [65]

Answer:

You will hear the note E₆

Explanation:

We know that:

Your speed = 88m/s

Original frequency = 1,046 Hz

Sound speed = 340 m/s

The Doppler effect says that:

f' = \frac{v \pm v0 }{v \mp vs}*f

Where:

f = original frequency

f' = new frequency

v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz

This frequency is close to the note E₆ (1,318.5 Hz)

7 0
2 years ago
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