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Brilliant_brown [7]
3 years ago
9

Calculate the rate of loss of heat through a glass window of area 200 CM square and thickness 0.5 CM where temperature inside is

35 degree Celsius and outside is -5 degree Celsius conductivity of Glass is 2.2 into 10 to the power 3 cal per s per cm per k .
​
Physics
1 answer:
saw5 [17]3 years ago
7 0

Answer:

The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass

Explanation:

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A crate falls from an airplane flying horizontally at an altitude of 2,000 m.
Hatshy [7]

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

7 0
3 years ago
A plane starting from rest (vo = 0 m/s) when t0 = 0s. The plane accelerates down the runway, and at 29 seconds, its velocity is
IrinaK [193]

Answer:

Acceleration, a=2.48\ m/s^2

Explanation:

Given that,

The plane is at rest initially, u = 0

Final speed of the plane, v = 72.2 m/s

Time, t = 29 s

We need to find the average acceleration for the plane. It can be calculated as :

a=\dfrac{v-u}{t}

a=\dfrac{72.2}{29}

a=2.48\ m/s^2

So, the average acceleration for the plane is 2.48\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
QUESTION 3
Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately <u>2.937 m/s²</u>.

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, \mu_s = 0.5

Wood on wood kinetic friction, \mu_k = 0.3

Solution;

The force of friction, F_f, acting when the crate is moving is given as

follows;

F_f = m × g × \mu_k

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

F_f = 100 × 9.81 × 0.3 = 294.3

The force of friction, F_f = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

a = \dfrac{F}{m}

Therefore;

a = \dfrac{293.7 \ N}{100 \ kg} = 2.937

The acceleration of the crate, a ≈ <u>2.937 m/s²</u>.

Learn more about friction here:

brainly.com/question/94428

8 0
2 years ago
a lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block
Airida [17]

Answer:

577g

Explanation:

Given parameters:

Temperature change = 5.9°C

Amount of heat lost = 427J

Unknown:

Mass of the block = ?

Solution:

The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.

                H =  m c Ф

  H is the heat capacity

 m  is the mass of the block

  c is the specific heat capacity

   Ф is the temperature change

Specific heat capacity of lead is 0.126J/g°C

   m = H / m Ф

   m = \frac{427}{0.126 x 5.9}  = 577g

Mass of the lead block is 577g

7 0
2 years ago
An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
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