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olganol [36]
2 years ago
10

A very tall building has a height H0 on a cool spring day when the temperature is T0. You decide to use the building as a sort o

f giant thermometer on a hot summer day by carefully measuring its height. Suppose you do this and discover that the building is a length h taller than it was on the cool spring day where h is much much less than H0. Assume the entire frame of the building is made of steel, which has a coefficient of linear expansion αsteel.
Required:
What is the temperature, assuming that the building is in thermal equilibrium with the air and that its entire frame is made of steel?
Physics
1 answer:
Vlada [557]2 years ago
5 0

Answer:

The temperature is   T  =  \frac{h}{H_O \alpha_{steel} }  + T_O

Explanation:

From the question we are told that

      The height on a cool spring day is H_O

      The temperature on a cool spring day is  T_O

      The difference in height between a cool spring day and a summer day  is     h

     The coefficient of static friction is \alpha _{steel}

The mathematical relation for the linear expansion of the steel buiding is represented as

               h  =  H_o \alpha_{steel}  [T-T_O]

Where T is the temperature of the steel during summer

Now making T the subject we have

                T  =  \frac{h}{H_O \alpha_{steel} }  + T_O

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A mover pushes a 255 kg piano
faust18 [17]

Answer:

0.495 ms^{-2}

Explanation:

According to the newton's second law of motion we can apply F=ma hear

Force = mass * acceleration

(assume the piano is moving left side )

←F = ma

F_(pull)+ F_(push)= M*a\\77.5 + 48.7 = 255 *a\\a = 0.495 ms^{-2}

7 0
2 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
2 years ago
Based on the free-body diagram, the net force acting on this wheelbarrow is {blank} N.
Semenov [28]

Answer:

diagram?

Explanation:

5 0
2 years ago
Read 2 more answers
A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when
choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

F_t - mg = ma

here since speed is constant or it is at rest

so we will have

a = 0

F_t = mg

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

F_t - mg = ma

F_t = mg + ma

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

F_t - mg = -ma

F_t = mg - ma

so tension force is less than the weight of crate if it is accelerating downwards

4 0
2 years ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0
2 years ago
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