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Marrrta [24]
3 years ago
14

Sort the statements about heat transfer into the correct columns.

Physics
2 answers:
drek231 [11]3 years ago
6 0

Answer:

here are the answers

Hoochie [10]3 years ago
5 0

Answer: Conduction- Touch transfer heat and Earth warms air

Convection- liquid/gas transfers heat and warm air rises

Radiation- Sun heats Earth and Waves transfer heat

Explanation:

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What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having
faltersainse [42]

Answer:

The wavelength of light is 4.53\times10^{-7}\ m

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

d \sin\theta=n\lambda.....(I)

The separation of the slits

d = \dfrac{1}{N}

d=\dfrac{1}{5000}

d=2\times10^{-6}\ m

Now put the value in equation (I)

2\times10^{-6}\sin13.1^{\circ}=\lambda

Here, n = 1

\lambda=4.53\times10^{-7}\ m

Hence, The wavelength of light is 4.53\times10^{-7}\ m

7 0
3 years ago
How much heat is released when 6.00 grams of ammonia and 5.00 grams of oxygen?
snow_tiger [21]
11 grams were released.think it help
7 0
3 years ago
what is the average speed of (a) a car that travels 400m in 20s. and (b) an athlete who runs 1500m in 4 minutes​
zmey [24]

Answer:

a) 20 m/s

b) 37.5 m)s

Explanation:

Average speed = total distance ÷ total time

=> (a) average speed of a car that travels 400m in 20s

= 400/20 = 20 m/s

& (b) average speed of an athlete who runs 1500m in 4 minutes (or 4×60=240 seconds)

= 1500/240 = 37.5 m/s

5 0
3 years ago
Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_{1} +q_{2} =-q_{3} -----eqution 1

where,

q_{1} is the heat absorbed by the solid at 0⁰C

q_{2} is the heat absorbed by the liquid at 0⁰C

q_{3} the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_{2} = -q_{3}

<u>Step 4:</u> calculate the final temperature of the water

79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}

Substitute in the values; we will have,

79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})

79.13 kJ + 990.66J* (T_{f}-218}) = -1463J*(T_{f}-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_{f}-218}) = -1.463KJ*(T_{f}-100})

79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3

collect like terms,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0
3 years ago
What kind of energy is in a rock at the edge of a cliff.
soldi70 [24.7K]

Answer:

kinetic energy

Explanation:

8 0
2 years ago
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