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34kurt
2 years ago
12

A windowpane is half a centimeter thick and has an area of 1.0 m2. The temperature difference between the inside and outside sur

faces of the pane is 15° C.
What is the rate of heat flow through this window? (Thermal conductivity for glass is 0.84 J/s⋅m⋅°C.)
a. 50 000 J/sb. 2 500 J/sc. 1 300 J/sd. 630 J/s
Physics
1 answer:
polet [3.4K]2 years ago
3 0

To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

Where

k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)

A = 1m^2 Area

L = 5*10^{-3}m Length

T_H= Temperature of the "hot"reservoir

T_C= Temperature of the "cold"reservoir

Replacing with our values we have that,

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

\frac{Q}{t} = \frac{(0.84)(1)}{0.005} (15)

\frac{Q}{t} = 2520J/s

Therefore the correct answer is B.

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A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts
tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

v = 70 m/s

now we have

v_x = 70 cos25

v_x = 63.44 m/s

v_y = 70 sin25

v_y = 29.6 m/s

now in Y direction we can use kinematics

v_y = v_i + at

v_y = 29.6 - (9.81 \times 5)

v_y = -19.5 m/s

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{63.44^2 + 19.5^2}

v = 66.4 m/s

4 0
2 years ago
A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
Calculate the mass of gold that occupies 5.0 × 10−3 cm3 . the density of gold is 19.3 g/cm3
ANEK [815]

Answer;

= 0.0965 g

Explanation;

Mass is given by multiplying density by volume

Mass = density * volume  

Mass = 19.3 g/cm³×0.005 cm³  

Mass = 0.0965 g

7 0
2 years ago
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from
Alenkasestr [34]

Answer:

V₁  = 5.6 m/s

V₂ = 7.2 m/s

V₃ = 8.8 m/s

Explanation:

Average velocity: Average velocity can be defined as the ratio of the total  displacement to the total time taken. The S.I unit of Average velocity is m/s.

For the first 2 s,

V₁ = Δd₁/t

Where V₁  = Average velocity for the first 2 s

Where Δd₁= distance, t = time

Δd₁ = 25.6-14.4 = 11.2 m t = 2 s

V₁ = 11.2/2

V₁ = 5.6 m/s

For the second 2 s,

V₂ =Δd₂/t

Where V₂ = average velocity for the second 2 s.

Δd₂= 40-25.6 = 14.4 m, t= 2 s

V₂ = 14.4/2

V₂ = 7.2 m/s

For the last 2 seconds,

V₃ =Δd₃/t

Where V₃ = average velocity for the last 2 s

where Δd₃ = 57.6- 40 = 17.6 m, t = 2 s

V₃ = 17.6/2

V₃ = 8.8 m/s.

8 0
2 years ago
A small pebble and one large boulder start at the same height and begin rolling down the side of a mountain. Which object would
vitfil [10]

Answer:

The small pebble

Explanation:

Since the potential energy, P.E lost equals kinetic energy, K.E gained,

P.E = K.E

P.E = mgh = K.E

So, K.E = mgh where g = acceleration due to gravity and h = height of drop

Since h and g are constant

K.E ∝ m

So, the kinetic energy of the object is directly proportional to its mass. Thus, the object with the smaller mass has the lesser kinetic energy.

Since the object with the smaller mass is the small pebble, so the small pebble would have less kinetic energy as it crashes on the road at the bottom of the mountain.

8 0
2 years ago
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