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iren [92.7K]
4 years ago
4

What quantities determine how quickly a wave travels along a string? Choose all that apply..a)Mass of the stringb)Length of the

stringc)Tension in the stringd)Linear density of the string
Physics
1 answer:
Ray Of Light [21]4 years ago
7 0

Answer:

a)Mass of the string

b)Length of the string

c)Tension in the string

d)Linear density of the string

Explanation:

The speed of a wave on a string is given by:

v=\sqrt{\frac{T}{m/L}}

where

T is the tension in the string

m is the mass of the string

L is the length of the string

m/L is the linear density of the string

Looking at the formula, we notice that the speed of the wave, v, depends on all the following quantities:

a)Mass of the string

b)Length of the string

c)Tension in the string

d)Linear density of the string

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Which of the following is a common source of groundwater pollution?
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One of the common source is C. <span>Spraying crop fields with pesticides
After spraying the crops with pesticides, the waste will fall and absorbed by the ground.
This will contaminate all substance that exists (nutrients that needed by the plants)  in that ground and make that area became toxic for years.</span>
4 0
3 years ago
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Two long, straight wires are parallel and 10 cm apart.One carries a current of 2.0 A, the other a current of 5.0A. (a) If the tw
lubasha [3.4K]

Answer:

Explanation:

Given

Distance between two wires d=10 cm

Current value i_1=2 A

i_2=5 A

(a)If current flows in opposite direction

When current Flows in opposite direction the two wires will repel each other

Force due to current carrying wire F=\frac{\mu _0i_1i_2}{2\pi r}

F_{12}=F_{21}=\frac{\mu _0\times 2\times 5}{2\times \pi \times 0.1}

F_{12}=\frac{4\pi \times 10^{-7}\times 2\times 5}{2\pi \times 0.1}

F_{12}=2\times 10^{-5} N (Repulsive Force)

(b)If current is flowing in the same direction

When direction of current is same then force will be attractive in nature

F_{12}=F_{21}=\frac{\mu _0i_1i_2}{2\pi r}

F_{12}=\frac{4\pi \times 10^{-7}\times 2\times 5}{2\pi \times 0.1}

F_{12}=2\times 10^{-5} N (attractive Force)

5 0
4 years ago
What explains the key difference between a bomb calorimeter and a coffee cup calorimeter?.
Virty [35]

The key difference between a bomb calorimeter and a coffee cup calorimeter is high temperature.

<h3>What is bomb calorimeter?</h3>

A bomb calorimeter is an apparatus that can measure heats of combustion, used in various applications such as calculating the calorific value of foods and fuels.

<h3>What is coffee cup calorimeter?</h3>

A coffee cup calorimeter is a cup used to provide insulation when materials are mixed inside of it.

<h3>Difference between the two calorimeter</h3>
  • The coffee cup calorimeter can't be used for high-temperature reactions, either, because they would melt the cup.
  • A bomb calorimeter is used to measure heat flows for gases and ​high-temperature reactions

Learn more about calorimeter here: brainly.com/question/1407669

#SPJ1

3 0
2 years ago
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and
mina [271]

Answer:

\frac{dQ}{dt}= 4312 W

Explanation:

As we know that base of the slab is given as

A = 11 \times 8

A = 88 m^2

now we know that rate of heat transfer is given as

\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)

here we know that

k = 1.4 W/m k

Also we have

x =0.20

\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)

\frac{dQ}{dt}= 4312 W

7 0
3 years ago
Design a tension member and slip-critical splice to carry a factored load of 500 kips. Please use a wide-flange section for the
Svetlanka [38]

Answer:

Kindly check the explanation section.

Explanation:

For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.

From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.

Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.

Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).

If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.

The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500

The fracture strength = .75 × Ah × Fhb = 309 kips.

The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.

4 0
4 years ago
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