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Aleonysh [2.5K]
2 years ago
6

An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th

is process, in joules?
Physics
1 answer:
pogonyaev2 years ago
5 0

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, c=345J/^oC

Initial temperature, T_i=88^{\circ}C

Final temperature, T_f=45^{\circ}C

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

Q=mc\Delta T

Let the mass of the object is 10 g or 0.01 kg

So,

Q=0.01\times 345\times (88-45)

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

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If the light strikes the first mirror at an angle θ1, what is the reflected angle θ2? express your answer in terms of θ1.
Alexus [3.1K]

Answer:

θ₂ = 90° - θ₁

Explanation:

When the light falls on a mirror it bounces back. This is know as reflection. The incident angle is equal to the angle of reflection.

Here, the light strikes the mirror at an angle = θ₁

To find the angle of reflection we first need to understand angle of incidence. The angle of incidence is the angle made between the incident ray and normal. Normal is an imaginary line drawn perpendicular line on the boundary of the mirror.

Since the light strikes the mirror at angle of θ₁, which is the angle between light ray and the mirror.

Angle of incidence = 90° - θ₁.

Thus, angle of reflection, θ₂ = 90° - θ₁

3 0
2 years ago
How does the sun transfer energy to Earth?
aleksley [76]

Answer:

By electromagnetic waves.

Explanation:

The sun transfers heat to earth via electromagnetic waves  in twomajor  ways:

Radiation- this is the transfer of energy by invisible electromagnetic ways.

Convection-The radiant sun energy warms the atmosphere and becomes heat energy. This transfer of heat through movement of fluids or usually air is called convection.

4 0
2 years ago
Read 2 more answers
A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa
gavmur [86]

Answer : The correct option is, (d) 535^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of copper = 0.10cal/g^oC

c_2 = specific heat of water = 1.00cal/g^oC

m_1 = mass of copper = 120 g

m_2 = mass of water = 300 g

T_f = final temperature of mixture = 35^oC

T_1 = initial temperature of copper = ?

T_2 = initial temperature of water = 15^oC  

Now put all the given values in the above formula, we get:

120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC

T_1=535^oC

Therefore, the temperature of the kiln was, 535^oC

7 0
2 years ago
Read 2 more answers
A blue puck has a velocity of 0i – 3j m/s and a mass of 4 kg. A gold puck has a velocity of 12i – 5j m/s and a mass of 6 kg. Wha
Mnenie [13.5K]
By definition, the kinetic energy is given by:
 K = (1/2) * m * v ^ 2
 where
 m = mass
 v = speed
 We must then find the speed of both objects:
 blue puck
 v = root ((0) ^ 2 + (- 3) ^ 2) = 3
 gold puck
 v = root ((12) ^ 2 + (- 5) ^ 2) = 13
 Then, the kinetic energy of the system will be:
 K = (1/2) * m1 * v1 ^ 2 + (1/2) * m2 * v2 ^ 2
 K = (1/2) * (4) * (3 ^ 2) + (1/2) * (6) * (13 ^ 2)
 K = <span> 525</span> J
 answer
 The kinetic energy of the system is<span> <span>525 </span></span>J
6 0
2 years ago
A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco
34kurt

Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform  = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

gh = 1/2v²

v² = 2gh

v = √(2gh)

v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s

So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

= -310.1 kgm/s ÷ 0.02 s

= -15,505 N

So, the force transmitted to her bones is 15,505 N

3 0
2 years ago
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