All categories

3 years ago

The frequency of a sound wave is 457 Hz. What is the period?

Physics

1 answer:

kondor19780726 [428]3 years ago

7 0

Period: 1/Frequency
1/457 = 0.002188 seconds
Round it to 0.0022 seconds

The answer is D

You might be interested in

CAN PEOPLE PLEASE HELP, I NEED TO GET TO THE NEXT LEVEL OR WHATEVER ITS CALLED BUT I NEED 5 BRAINLIEST IN 2 DAYS, I ALREADY HAVE

lapo4ka [179]

Answer:

answer the question i have asked go to my account

Explanation:

can i get brainliest

6 0

3 years ago

Read 2 more answers

the strength s of a rectangular wooden beam is proportional to its width and the cube of is septh. If the beam is to be cut from

babymother [125]

Answer:

Can see it in the pic

Explanation:

Can see it in the pic

4 0

3 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago

The grounded conductor and the grounding conductor are one and the same conductor and perform the same function.

STALIN [3.7K]

Answer:

Grounded used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.

As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.

Explanation:

A grounded conductor is referred to as one of the wire that needed in an electric circuit. it is basically a neutral conductor. It used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.

As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.

5 0

3 years ago

Find the frequency if the period is 15 us.

Sav [38]

Answer:

The frequency is $0.667$ × $10^{6}$ $Hz$ .

Explanation:

Period is defined as the time taken by any element to attempt one complete oscillatory motion in one second.
Frequency is the number of complete oscillations made in one second.

Frequency and period are inversely proportional to each other.

In mathematical term: $frequency$ = $\frac{1}{period}$ .
As we are given, $period$ = $15$ × $10^{-6}$ $s$ .
We can easily calculate frequency as: $f\\$ = $\frac{1}{15. 10^{-6} }$
So frequency is $f$ = $0.667$ × $10^{6}$ $Hz$ .

6 0

3 years ago