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2 years ago

13

Let's say you have two tuning forks which are supposed to produce the same frequency, 512 Hz. One is of good quality, but the ot

her is cheaply made and vibrates at 510 Hz. If you bonk them both how many beats per second will you hear

1 answer:

solong [7]2 years ago

7 0

Answer:

= 2 beats per seconds

Explanation:

From |f -f'| = modulus of the difference between the frequency given.

f = 510Hz and f' = 512Hz

Difference between the frequency will give us the number of beat per seconds.

i.e 2 beats per seconds

These also shows how to get the period of the tuning forks.

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"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front

cluponka [151]

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2 years ago

The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration

Veseljchak [2.6K]

Answer:

acceleration = v-u/ t

= 15-10/5

= 5/5

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Explanation:

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2 years ago

Read 2 more answers

Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), th

Zinaida [17]

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

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2 years ago

If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o

Scilla [17]

Answer:

Maximum acceleration will be equal to $3.43m/sec^2$

Explanation:

We have given coefficient of kinetic friction $\mu _k=0.7$

And coefficient of static friction $\mu _s=1$

Acceleration due to gravity $g=9.8m/sec^2$

When truck moves maximum force will be equal to $F=\mu _kmg$

It is given that half of the weight is supported by its drive wheels

So force required $=\frac{\mu _kmg}{2}$

From newtons law maximum acceleration will be equal to $a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2$

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2 years ago

Imagine that you drop an object of 10 kg, how much will be the acceleration and

Lelu [443]

If you do this on Earth, then the acceleration of the falling object is 9.8 m/s^2 ... NO MATTER what it's mass is.

If its mass is 10 kg, then the force pulling it down is 98.1 Newtons. Most people call that the object's "weight".

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2 years ago