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3 years ago

Why do small bells have a high pitch

1 answer:

8 0

Small objects vibrate quickly and produce lots of waves each second, and as the length of the bells are short the wave length become short too,so small bells have a high frequency hence high pitch.

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A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s afte

Arturiano [62]

Answer:

a) a = 2.35 m/s^2

Explanation:

(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:

$x=v_ot+\frac{1}{2}at^2$ (1)

x: position of the ball after t seconds = 87 m

t: time = 8.6 s

a: acceleration of the ball = ?

vo: initial velocity of the ball = 0 m/s

You solve the equation (1) for a:

$x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}$

You replace the values of the parameters in the previous equation:

$a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}$

The acceleration of the ball is 2.35 m/s^2

6 0

3 years ago

2 characteristics of constant speed

svetoff [14.1K]

Answer:

it has no acceleration

Explanation:

8 0

3 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

3 0

2 years ago

A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b

lakkis [162]

Answer:

$T = 17.26 ^oC$

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

$Q_1 = Q_2$

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

so we will have

$32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)$

so we have

$(45.8 - T) = 15.1(T - 15.4)$

so we have

$16.1 T = 277.87$

$T = 17.26 ^oC$

6 0

3 years ago

A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100

Verizon [17]

Answer:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

Explanation:

Attached you will find the plot of position vs time of Ellie´s movement.

The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.

So, the velocity is calculated as follows:

v = Δx / Δt

where

Δx = final position - initial position

Δt = elapsed time

1) The average velocity of Ellie while running is:

v = 1000 m - 0 m / 200 s = 5 m/s

While resting:

v = 0 m - 0 m / 100 s = 0 m/s

And while walking back:

v = 0 m - 1000 m / 1000 s = - 1 m/s

Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.

Comparing with the graphic, the velocity is the slope of the function position(t).

Then:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

2) The speed is the distance traveled over time:

Running speed = 1000 m / 200 s = 5m /s

Resting speed = 0 m / 100 s = 0 m/s

Walking speed = 1000 m/ 1000 s = 1 m/s

Then:

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

4 0

3 years ago