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Free_Kalibri [48]
3 years ago
6

A thick window pane 0.367 cm has an area of 1.247 square meters. The temperature difference between the inside and outside surfa

ces of the window is 33.2°C. What is the rate of heat flow through this window in J/s if the thermal conductivity for glass is 0.8 J/s/m/°C?A. 9.025B. 9025C. 902.5D. 90.25
Physics
1 answer:
____ [38]3 years ago
7 0

Answer:

B. 9025 J/s

Explanation:

thickness (d) = 0.367 cm = 0.00367 m

area (A) = 1.247 m^{2}

temperature difference (ΔT) =33.2 °C

thermal conductivity (K) = 0.8 J/s/m/°C

find the rate of heat flow

rate of heat flow = \frac{K.A.ΔT}{d}

rate of heat flow = \frac{0.8x1.247x33.2}{0.00367}

rate of heat flow = 9024.6 ≈ 9025 J/s

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DESCRIBE THE FORMATION OF THE SOLAR SYSTEM ACCORDING TO THE NEBULAR THEORY
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3 years ago
The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin
Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

K=730 N/m,m=0.053kg, h=1.90m.

v_f^2=v_i^2+2*g*h

v_i^2=2*g*h=2*9.8m/s^2*1.9m

v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}

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v_i=V*sin(31)

V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}

V=11.84 m/s

b).

K_k=\frac{1}{2}*K*x^2

No friction on the ball so:

x^2=\frac{2*K_k}{K}

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5 0
3 years ago
You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
blsea [12.9K]
<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
7 0
3 years ago
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marishachu [46]

Answer:

288.2 K

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T_f = Final temperature

v_0 = Original volume = 150 mL

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The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL

8 0
4 years ago
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