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Ludmilka [50]
3 years ago
7

Which of the following are examples of etiquette? Check all of the boxes that apply.

Physics
1 answer:
Irina18 [472]3 years ago
6 0

Answer:

C& D

Explanation:

ON EDG2020

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How did she change the circuit?
klio [65]

Answer:

She changed a closed circuit to an open circuit

Explanation:

3 0
3 years ago
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Which physical phenomena is responsible for the earth’s sky appearing blue? scattering reflection dispersion refraction
klio [65]

Answer:  Scattering reflection

Sunlight reaches earth's atmosphere and is scattered in all directions by all the gasses and particles in the air. Blue light is seen more than others because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

Explanation:

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3 years ago
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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
3 0
3 years ago
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The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu
abruzzese [7]

Answer:

\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 6.8 \times  {5}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 3.4 \times 25

\sf \implies KE =3.4 \times 25

\sf \implies KE = 85 \: J

8 0
3 years ago
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How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),
fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice =  37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat  absorbed during this phase change be Q₂ .

Let heat  absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0
3 years ago
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