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madam [21]
3 years ago
10

Jonathan has six strings that are the same thickness and are all the same material. He cuts the strings to different lengths and

pulls so that they all have the same tightness. Which string will have the highest pitch when he plucks it with his finger?
Physics
2 answers:
harkovskaia [24]3 years ago
7 0
The shortest string will have the highest pitch.

tamaranim1 [39]3 years ago
4 0

Your answer is C.

The shortest string will have the highest pitch. This is because the pitch of a sound depends on the speed of its vibrations.

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A hollow conductor is positively charged. Asmall uncharged metal ball is lowered by a silk thread through asmall opening in the
dimulka [17.4K]

Answer:

Explanation:

According to the property of a conductor, the entire charge will reside on the outer surface of the conductor, there is no charge on the inner side of the conductor. As the uncharged metal ball touches the inner surface of the conductor, it does not attain any charge as the inner side of the conductor has no charge.

So option (c) is correct.

8 0
3 years ago
An enemy spaceship is moving toward your starfighter with a speed of 0.400 c , as measured in your reference frame. The enemy sh
Alexxandr [17]
The answer is (a) 0.859c (b) 31s
4 0
3 years ago
If a 160W light bulb consumes 1000J of electrical energy in a given time, then in the same time interval, how much energy will a
NNADVOKAT [17]
160w/4 = 40w
1000J/ 4= 250J
a 40w light bulb will consume 250J
7 0
2 years ago
Read 2 more answers
A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
3 years ago
Find acceleration given the following:
Lorico [155]

Answer:

Explanation:df=378

Di=0

Vi=0

T=6s

Formula

Df=Di+Vit+1/2(at)2

378=0+0+6+1/2×a×t2

378=6+1/2×a×36

378=6+18a

18a=378+6=384

a=385/18=21.33(cm/s)/s

5 0
3 years ago
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