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g100num [7]
2 years ago
15

A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at

0 degrees celsius. The aluminum ring ( a = 2.4 x 10^-5/degrees celsius) has an inside diamter of 3.994 cm at 0 degrees celsius. At what temperature will teh sphere fall through teh ring??Mulitple Choice Questiona, 461.5 degrees celsiusb. 208.0 degrees celsiusc. 115.7 degrees celsiusd. 57.7 degrees celsius
Physics
1 answer:
mote1985 [20]2 years ago
8 0

Answer:

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

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2 years ago
At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made u
Alex_Xolod [135]

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

P = \frac{E}{t} \rightarrow E= Energy, t = time

P = 62W = 62 \frac{J}{s}

The rate of energy loss per day would then be,

P = 62\frac{J}{s} (\frac{86400s}{1day})

P = 5356800 \frac{J}{day}

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that 1KCal = 4.184*10^{6}J

In this way the energy in Cal is,

E = 5356800J \frac{1KCal}{4.184*10^{6}J}

E = 1279.694 KCal

The number of kilocalories (food calories) must be 1279.694 KCal

4 0
2 years ago
A spring balance consists of a pan that hangs from a spring. A damping force Fd = −bv is applied to the balance so that when an
Citrus2011 [14]

Answer:

b ≈ 64 Kg/s

Explanation:

Given

Fd = −bv

m = 2.5 kg

y = 6.0 cm = 0.06 m

g = 9.81 m/s²

The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).

m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):

∑Fy = 0 (↑)

k*y - W = 0    ⇒   k*y - m*g = 0   ⇒   k = m*g / y

⇒   k = (2.5 kg)*(9.81 m/s²) / (0.06 m)

⇒   k = 408.75 N/m

Hence, if

b² = 4*k*m    ⇒     b = √(4*k*m) = 2*√(k*m)

⇒     b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)

⇒     b = 63.9335 Kg/s ≈ 64 Kg/s

5 0
2 years ago
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
2 years ago
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AlladinOne [14]

Answer:

True statements:

Power is the rate at which work is done.

Power is the rate at which energy is transformed.

The SI unit of power is watt.

A person is limited in the total work he or she can do by the rate at which energy can be transformed.

Explanation:

The above statements are true.

1. Power is defined as the rate at which work is done.

2. Power is the rate at which energy is transformed. Whenever energy changes its form, or moves from one place to another, the rate at which it makes those changes is described as power.

3. The SI Unit of power is watts.  If 1 joule of energy is converted by an object in 1 second then its power is called one watt.

So, the statement<em> 'the SI unit of power is the horsepower' is</em> false.

4. A person is limited in the total work he or she can do by the rate at which energy can be transformed i.e, his power. He is not only limited by the total energy required.

So, the statement <em>'A person is limited in the total work he or she can do only by the total energy required'</em> is not completely true.

5 0
2 years ago
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