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Ray Of Light [21]

3 years ago

11

The transmission of heat requiring the movement of a liquid or a gas is

1 answer:

AysviL [449]3 years ago

7 0

Answer:

convection

Explanation:

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Four complete waves pass the duck in one second the frequency of this wave is

galben [10]

The frequency of a wave is the number of complete oscillations passing a given point per second.

In this case, assuming the duck is stationary, we have 4 complete waves passing the duck in one second: therefore, the frequency of the wave is
$f= \frac{4}{1 s}=4 Hz$

3 0

3 years ago

Match Newton's laws with their descriptions. 1. Every applied force is opposed by an equal force. 2 .A force must be applied to

Marina CMI [18]

A would be number 2. Newton's First Law states that an object at rest, will stay at rest and an object in motion, will stay in motion, unless acted upon by an unbalanced force. B would be number 3. His Second Law states that <span>the sum of the forces acting on a body is equal to the product of the mass of the body and the acceleration produced by the forces. And, C would be number 1. His Third Law states that for every action, there is an equal and opposite reaction. Hope this helps!</span>

6 0

3 years ago

Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8

sergejj [24]

Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

$T_2 = 39^oC$

$T_3 = 303 \ K$

Generally converting $T_3$ to Fahrenheit

$T_3' = (T_3 -273 ) * \frac{9}{5} + 32$

=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

=> $T_3' = 86 F$

Converting $T_2$ to Fahrenheit

$T_2' = T_2 * \frac{9}{5} + 32$

=> $T_2' = 39 * \frac{9}{5} + 32$

=> $T_2' =102.2 F$

Now comparing the temperature in Fahrenheit we see that $T_2$ is the hottest

3 0

3 years ago

What is the distance from the crest to the equilibrium of a wave called? A. amplitude B. period C. frequency D. phase

Goryan [66]

its the first one. A.

8 0

3 years ago

Read 2 more answers

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago