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sesenic [268]
4 years ago
13

Describe how the amount of heat released or absorbed related to specific heat capacity and amount of it? A) Amount of heat absor

bed or released is doubled if quantity is doubled. If a different substance with half the specific heat capacity is used, the amount of heat absorbed or released is doubled. B) Amount of heat absorbed or released is doubled if quantity is cut in half. If a different substance with half the specific heat capacity is used, the amount of heat absorbed or released is doubled. C) Amount of heat absorbed or released is doubled if quantity is doubled. If a different substance with half the specific heat capacity is used, the amount of heat absorbed or released is cut in half. D) Amount of heat absorbed or released is doubled if quantity is cut in half. If a different substance with half the specific heat capacity is used, the amount of heat absorbed or released is cut in half.
Physics
2 answers:
Fofino [41]4 years ago
7 0

Answer:

c

Explanation:

i took the usa test prep

Grace [21]4 years ago
3 0
The correct answer is
<span>C) Amount of heat absorbed or released is doubled if quantity is doubled. If a different substance with half the specific heat capacity is used, the amount of heat absorbed or released is cut in half. 

In fact, the amount of heat released or absorbed is given by:
</span>Q=m C_s \Delta T
<span>where 
m is the mass of the substance
Cs is the specific heat capacity of the material
</span>\Delta T is the change in temperature
<span>
We can see that Q is directly proportional to both m (mass, or quantity of the substance) and Cs (specific heat capacity). So if the quantity (m) is doubled, the amount of heat (Q) is doubled as well. Similarly, if the specific heat capacity (Cs) is cut in half, the amount of heat (Q) is cut in half as well.</span>
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The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the
xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First,  convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

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3 years ago
In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit
Minchanka [31]

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}

Putting, t = 1 s  and x = 4 m in above equation, we get :

4 = \dfrac{a(1)^2}{2}\\\\a = 8 \  m/s^2

Therefore, the acceleration of the cart is 8 m/s².

5 0
3 years ago
A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F
ivanzaharov [21]

Answer: 32.65\ m

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is \mu_s=0.4

The coefficient of kinetic friction is \mu_k=0.2

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction \mu_kg=0.2\times 9.8\ m/s^2

Using the equation of motion v^2-u^2=2as\\

insert the values

0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m

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answer: transverse and longitudinal

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Molecules can be made of atoms of the same element only. different elements only. the same or different elements.
Ira Lisetskai [31]

Both, there are two different types of molecules to distinguish that

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4 years ago
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