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4 years ago

11

Calculate the average velocity of a dancer who moves 5 m toward the left of the stage over the course of 15 s. ** Velocity = dis

placement/time Question 1 options: A. 3 m/s B. 1/3 m/s C. 1/3 m/s west D. 3 m/s west

1 answer:

Alex73 [517]4 years ago

3 0

Answer:

B

Explanation:

Velocity=disp/time

V=5m/15s

V=1/3 m/s

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A standard baseball has a mass of 144.3 g. Determine the weight, in newtons, of a baseball.

alex41 [277]

Fg = ma
Fg = (0.1143 kg) (9.81 N/kg)
Fg = 1.12 N

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3 years ago

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w

Tatiana [17]

Answer:

$a. \ f_1=7.9057Hz\\\\b. \ f_2=15.8114Hz\\\\c. \ f_3=23.7171Hz$

Explanation:

a. The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

$f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}$ #where t=250N*10=2500N, $\mu=0.1kg$

#substitute for actual values for the lowest frequency.

$F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{1}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=7.9057Hz$ #n=1, lowest frequency

Hence, the lowest frequency for standing waves is 7.9057Hz

b.The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

$f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}$ #where t=250N*10=2500N, $\mu=0.1kg$

#The second lowest frequency happens at $n=2$ :

$F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{2}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=15.8114Hz$

Hence, the second lowest frequency is 15.8114Hz

c.Given that the wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

$f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}$ #where t=250N*10=2500N, $\mu=0.1kg$

The third lowest frequency happens at $n=3$

$F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{3}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=23.7171Hz$

Hence, the third lowest frequency is 23.7171Hz

6 0

3 years ago

A large, 34.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of m

Lorico [155]

Answer:

length L of the clapper rod for the bell to ring silently = 0.756m

Explanation:

We are given;

Mass of Bell;m_b = 34 kg

Distance of centre of mass from pivot;d = 0.7m

The bells moment of inertia about an axis at the pivot;I = 18 kg.m²

Mass of clapper;m_c = 1.8 kg

Length of slender rod is L

Now, the formula for period of physical pendulum having small amplitude is given as;

T_b = 2π√(I/mgd)

Where;

I is moment of inertia

m is mass

g is acceleration due to gravity = 9.8 m/s²

d is distance from rotation axis to centre of gravity

Plugging in the relevant values and using mass of bell, we have;

T_b = 2π√(18/(34*9.81*0.7)

T_b = 2π√(18/(34*9.81*0.7)

T_b = 1.745 s

Now, the formula for period for a simple pendulum which is essentially what the clapper rod is would be;

T_c = 2π√(L/g)

Now, we want to find length of clapper L.

Thus, let's make it the subject;

L = g(T_c/2π)²

Now, we are told that for the bell to ring silently, T_b = T_c.

Thus, T_c = 1.745 s.

So,

L = 9.8(1.745/2π)²

L = 0.756m

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3 years ago

WHAT HAPPENS TO THE VELOCITY AND ACCELERATION OF A BODY FALLING FREELY IN A VACUUM?

mixas84 [53]

both remains constant

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4 years ago

The diagram shows what happens to a system undergoing an adiabatic process.

posledela

The answer is:
B. <span>X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.</span>

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4 years ago

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