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3 years ago

A guitar string transmits waves at

Physics

1 answer:

vichka [17]3 years ago

5 0

Answer:

The wavelength of the waves on the string is found using v = λf:

λ = v/f = 315/370 = 0.85135m

The first harmonic (fundamental) standing wave formed on the string has nodes only at the ends (see diagram in link) so:

String length = λ/2 = 0.85135/2 = 0.426m (to 3 sig. figs.)

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The volume of the nucleus as compared to the volume of the atom is

velikii [3]

The nucleus of an atom is about 10-15 m in size; this means it is about 10-5 (or 1/100,000) of the size of the whole atom. A good comparison of the nucleus to the atom is like a pea in the middle of a racetrack. (10-15 m is typical for the smaller nuclei; larger ones go up to about 10 times that.)

4 0

3 years ago

A canoe floats in a lake. Inside the canoe is a 25 kg steel solid ball. If the ball is thrown into the lake, does the level of t

Rus_ich [418]

Answer:

Explanation:

volume of ball bearing = 4/3 π r³

= 4/3 x 3.14 x 1.5³

= 14.13 cc.

if D be the density of steel , weight of the ball bearing

= 14.13 x D x g

For the first case , water will be displaced to keep it floating

weight of displaced water will be equal to weight of steel

weight of displaced water = 14.13 Dg

mass of displaced water = 14.13 D

volume of displaced water = mass / density of water

= 14.13 D / d ; (where d is density of water) .

Now when the steel ball bearing is dipped in water , it will displace water equal to its volume only and it will be drowned

its volume = 14 .13 cc

14.13 D / d > 14.13 ( because D/d is more than one , since D > d )

volume of water displaced in first case is greater

water level will go up higher in first case .

Hence in the second case water level will go down .

Same will happen in case of 25 kg steel .

4 0

3 years ago

The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i

Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

$T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1$

Block 2

$T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2$

Block 3

$- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3$

Solving Eq1,2,3 simultaneously

Divide 1 and 2

$\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4$