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Debora [2.8K]
4 years ago
10

stion 13 of 20 : Select the best answer for the question. 13. If a certain mass of mercury has a volume of 0.002 m3 at a tempera

ture of 20°C, what will be the volume at 50°C? A. 0.002021624 m3 B. 0.004021624 m3 C. 0.000010812 m3 D. 0.002010812 m3
Physics
1 answer:
Free_Kalibri [48]4 years ago
7 0

Answer : The correct option is, (A) 0.002010812m^3

Solution : Given,

Volume of mercury at temperature 20^oC = 0.002m^3

As we know that the mercury is a liquid substance. So, we have to apply the volume of expansion of the liquid.

Formula used for the volume expansion of liquid :

V_{T}=v_{1}[1+\gamma (T_{2}-T_{1})]

or,

V_{2}=V_{1}[1+\gamma (T_{2}-T_{1})]

where,

V_{T} = volume of liquid at temperature T^oC

V_{1} = volume of liquid at temperature 20^oC

V_{2} = volume of liquid at temperature 50^oC

\gamma = volume expansion coefficient of mercury at 20^oC is 0.00018 per centigrade    (Standard value)

Now put all the given values in the above formula, we get the volume of mercury at 50^oC.

V_{2}=0.002[1+0.00018(50-20)]=0.0020108m^3

Therefore, the volume of mercury at 50^oC is, 0.002010812m^3

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
How to calculate energy needed for a change of state ​
Alenkasestr [34]

Answer:

change in temperature = (100 - 25) = 75.0°C.

change in thermal energy = mass × specific heat capacity × change in temperature.

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vova2212 [387]

Answer:

t=240s

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x=1200m/s

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A: They maintain stable concentrations.

Explanation:

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