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3 years ago

A mercury thermometer is constructed as

Physics

1 answer:

Tamiku [17]3 years ago

8 0

Answer:

The change in height of the mercury is approximately 2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

$\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C$

from which we solved for the change in volume $\Delta V$ due to a given change in temperature $\Delta T$

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

$V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3$

Therefore, the change in volume with a change in temperature of 36°C becomes:

$\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3$

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

$V_{cyl}=\pi r^2\,h\\h =\frac{V_{cyl}}{\pi r^2} \\h=\frac{0.0000474248\, cm^3}{\pi \, (0.00225\,cm)^2} \\h=2.98188 \,cm$

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A roller coaster travels 41.1 m at an angle of 40.0°

Iteru [2.4K]

Answer:

41.1 ÷ 40.0

Explanation:

Did you learn about Newton

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3 years ago

A skateboarder traveling with an initial velocity 9.0 meters per second,

meriva

Answer:

25m/s

Steps:

First, The equation v= u + a * t shows us what we need to find, (the finale velocity).

Second, we substitute the values given:

v= 9m/s + 4m/s2 * 4s

Last, We calculate the values:

Multiply 4m/s2 * 4s = 16m/s

Add 9m/s + 16m/s

Answer: 25m/s

Hope this helps :)

4 0

3 years ago

Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i

max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0

4 years ago

A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The

Liula [17]

What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be $\frac{F}{2}$ now:

a= $\frac{F}{2m}$

So the accelaration will also he halved (it's the original acceleratation divided by 2)

b) The object's mass is halved?
a= $\frac{F}{m/2}$ =a= $\frac{F2}{m}$

which is the original acceleration times two!! so it will double

c) The force and the object's mass are both halved?
now we have

a= $\frac{F/2}{m/2}$ =a= $\frac{2F}{2m}$ =a= $\frac{F}{m}$

so they will cancel each other out and the acceleration will stay the same!

5 0

3 years ago

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º

The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

Vectors are decomposed into a coordinate system

The components are added

The resulting vector is constructed

Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

v = $\frac{\Delta d}{t}$

Δd = v t

Where v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

t₁ = 2 min = 120 s

Δd₁ = v₁ t₁

Δd₁ = 600 120

Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s

time t₂ = 1 min = 60 s

Δd₂ = v₂ t₂

Δd₂ = 400 60

Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

cos (180 -35) = $\frac{x_1}{\Delta d_1}$

sin 145 = $\frac{y_1}{\Delta d1}$

x₁ = Δd₁ cos 125

y₁ = Δd₁ sin 125

x₁ = 72 103 are 145 = -58.98 103 m

y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

cos (90+ 30) = $\frac{x_2}{\Delta d_2}$

sin (120) = $\frac{y_2}{\Delta d_2}$

x₂ = Δd₂ cos 120

y₂ = Δd₂ sin 120

x₂ = 24 103 cos 120 = -12 10³ m

y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

Rₓ = x₁ + x₂

R_y = y₁ + y₂

Rx = - (58.98 + 12) 10³ = -70.98 10³ m

Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

R = $\sqrt{R_x^2 +R_y^2}$

R = $\sqrt{70.98^2 + 62.08^2}$ 10³

R = 94.30 10³ m

We use trigonometry for the angle

tan θ ’= $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{62.08}{70.98}$

θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

θ = 180 - θ'

θ = 180 -41.2

θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

v = $\frac{\Delta R}{t_3}$