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3 years ago

When water changes into vapor this is called?

1 answer:

4 0

When water changes into vapor, it is called evaporation. BONUS: This is formed by the boiling point of water, which is 230°F (Fahrenheit) or 110°C (Celsius).

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A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the

svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength is given by

$\lambda = \dfrac{2L}{n}$

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

$f = n\dfrac{v}{2L}$

now, wavelength calculation

n = 1

$\lambda_1 = \dfrac{2\times 2}{1}$

λ₁ = 4 m

n = 2

$\lambda_2 = \dfrac{2\times 2}{2}$

λ₂ = 2 m

n =3

$\lambda_3 = \dfrac{2\times 2}{3}$

λ₃ = 1.333 m

now, frequency calculation

n = 1

$f = n\dfrac{v}{2L}$

$f_1 =1\times \dfrac{50}{2\times 2}$

f₁ = 12.5 Hz

n = 2

$f = n\dfrac{v}{2L}$

$f_2 =2\times \dfrac{50}{2\times 2}$

f₂= 25 Hz

n = 3

$f = n\dfrac{v}{2L}$

$f_3 =3\times \dfrac{50}{2\times 2}$

f₃ = 37.5 Hz

8 0

3 years ago

How does resistant affect the current in a cirtuit

Burka [1]

Resistance reduces the current. If there is more resistance, there is less current.

7 0

3 years ago

Electromagnetic waves are created by vibrating ________________ and ________________ fields.

BARSIC [14]

The correct answer is C

5 0

4 years ago

A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the

Len [333]

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t = 55 s

v2 = ?

We can use a kinematic equation to solve this problem. We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

7 0

3 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

3 years ago