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Serjik [45]
3 years ago
13

An 8.38 gram sample of water lost 870 J of heat. What is the change in temperature

Physics
1 answer:
Virty [35]3 years ago
6 0

Answer:

Find the mass of a sample of water if its temperature dropped 24.8 degrees Celsius when it lost 870 J of. heat. (specific heat of water is 4.18 J/g x degrees Celsius) 8.38 grams. How many degrees would the temperature of a 450 g piece of iron increase if 7600 J of.

Explanation:

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Ann is driving down a street at 61 km/h.
Wittaler [7]

Answer:

12.267 seconds approximately.

Explanation:

The units can be simplified into m/s, in which case you would have 61000/3600. Simplify that to 16 and 17/18. This is your meters per second, so multiply that by .724 to get the answer.

4 0
3 years ago
Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse
777dan777 [17]

Answer:

      Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

      d sin θ = (m + ½) λ

As the observation is on a distant screen

     tan θ = y / x

     tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

          tan θ = sin θ

Let's replace

     d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance  

     y₁ = (m + ½) λ x / d

     m = 1

      y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

             m = 2

             y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

           Δy = λ x / d (5/2 -3/2)

           Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

           Δy= 5,075 10⁻⁶ m

8 0
3 years ago
How much of the electromagnetic spectrum is visible light?
Margarita [4]
I believe the answer is D, only a small part of it
8 0
3 years ago
Read 2 more answers
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground
Blababa [14]

Answer:

KE_2 = 3.48J

Explanation:

Conservation of Energy

E_1 = E_2

PE_1+KE_1 = PE_2+KE_2

m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²

(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2

4.10J+0.914J = 1.53J + KE_2

5.01J = 1.53J + KE_2

KE_2 = 3.48J

3 0
3 years ago
Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t
GenaCL600 [577]
B) 14.0 N    

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is: 

 E = 0.5 M V^2 

 where 

 E = Energy 

 M = Mass 

 V = velocity   

 Substituting the known values, let's calculate the before and after energy. 

 Before: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (2.25 m/s)^2 

 E = 6.75 kg 5.0625 m^2/s^2 

 E = 34.17188 kg*m^2/s^2 = 34.17188 joules   

 After: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (1.2 m/s)^2 

 E = 6.75 kg 1.44 m^2/s^2 

 E = 9.72 kg*m^2/s^2 = 9.72 Joules   

 So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.   

 24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N   

 Rounding to 1 decimal place gives 14.0 N which matches option "B".
8 0
3 years ago
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