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nydimaria [60]
3 years ago
15

9. A wave on Beaver Dam Lake passes by two docks that are 40.0 m apart.

Physics
1 answer:
Kobotan [32]3 years ago
4 0

Answers:

a) 10 m

b) time=1.6 s, frquency=0.625 Hz

c) 6.25 m/s

Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength \lambda is the distance between to crests; this means we have 4\lambda in 40 m:

40 m=4\lambda

Clearing \lambda:

\lambda=\frac{40 m}{4}

\lambda=10 m

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- T

Then:

T=\frac{(1 wave)(16 s)}{10 waves}

T=1.6 s This is the period of the wave

On the other hand, the frequency f of the wave has an inverse relation with its period T:

f=\frac{1}{T}

f=\frac{1}{1.6 s}

f=0.625 Hz This is the frequency of the wave

c) The speed v of a wave is given by the following equation:

v=\frac{\lambda}{T}

v=\frac{10 m}{1.6 s}

Finally:

v=6.25 m/s

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The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
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Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

5 0
3 years ago
The work done to compress a spring with a force constant of 290.0 N/m a total of 12.3 mm is: a) 3.57 J b) 1.78 J c) 0.0219 J d)
iren2701 [21]

Answer:

Work done, W = 0.0219 J

Explanation:

Given that,

Force constant of the spring, k = 290 N/m

Compression in the spring, x = 12.3 mm = 0.0123 m

We need to find the work done to compress a spring. The work done in this way is given by :

W=\dfrac{1}{2}kx^2

W=\dfrac{1}{2}\times 290\times (0.0123)^2

W = 0.0219 J

So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.

7 0
3 years ago
The density of blood is 2kg/m.convert to g/cm
sertanlavr [38]

Explanation:

if we convert it into g/cm it'll be

2000 grams

6 0
2 years ago
Can anyone please help me with answering this question? : An accelerating (speeding up) body has a Net Force acting in the direc
Katarina [22]
That's true.

Netwon's second law states that the resultant of the forces F acting on a body is equal to the product between its mass m and its acceleration a:
\sum F = ma
This means that if the net force acting on an object is different from zero (term on the left), than the acceleration of the object (term on the right) must be different from zero as well, and therefore the body is accelerating.

In particular, both F and a in the equation are vectors: this means that if the acceleration is positive, F and a have the same direction. In this problem, the acceleration is positive (because the object is speeding up), therefore the force and the acceleration have same direction.
7 0
3 years ago
An Arrow (0.5 kg) travels with velocity 20 m/s to the right when it pierces an apple (2 kg) which is initially at rest. After th
sattari [20]

Answer:

4 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')...................... Equation 1

Where m = mass of the arrow, u = initial velocity of the arrow, m' = mass of the apple, u' = initial velocity of the apple, V = Final velocity of the apple and the arrow after collision.

make V the subject of the equation

V = (mu+m'u')/(m+m').................... Equation 2

Given: m = 0.5 kg, m' = 2 kg, u = 20 m/s, u' = 0 m/s(initially at rest)

Substitute into equation 2

V = (0.5×20+2×0)/(2+0.5)

V = 10/2.5

V = 4 m/s.

Hence the final velocity of the apple and the arrow after the collision = 4 m/s

6 0
3 years ago
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