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dangina [55]
3 years ago
12

An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is corr

ect at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What wull be the length of rod at 0 degree celcius?​
Physics
2 answers:
natulia [17]3 years ago
5 0
Given:
temperature at which the rod is measured= 25°
length of aluminium rod measured at 25°C= 1 meter
temperature at which the scale is correct= 0°C
Linear expansivity of steel= 12* 10^-6
linear expansivity of aluminum= 26* 10^-6
To find:
True length of rod at 25°C
Length of rod at 0°C
Solution:
Let the length of the aluminum and steel rod at 0°C be L0A and L0S respectively.
Length of the steel scale at 25°C , thermal expansion is:
LA25 = LA0(1+ (αS - αA)25)
LA0 = LA25/ (1+ (αS-αA)t)
LA0 = 100/ (1+ (12*10^-6 - 26*10^-6)25)
LA0 ≈ 100.05 cm at 0°C
2. Let the length at 25°C of the aluminium trod be LA25
The actual measure of aluminum rod at 25°C, will be
LA25 = 100 * (1+ 26* 10^-6*25)
LA25 ≈ 100.06 cm
marin [14]3 years ago
3 0

Answer:

An almunium rod when measured with a steel scale,both being at 25°C appears to be 1meter long. If the scale is correct at 0°C

Explanation:

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Explanation:

Given that,

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Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

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2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

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B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

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Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of
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Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r    where i₁ and i₂ are current in the two wires , r is distance between the two and  μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i²  = 3.98 x 10⁻²

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A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th
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The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

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  • A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

s=u_y t + \frac{1}{2}at^2

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

u_y is the initial vertical velocity of the ball, which is given by

u_y = u sin \theta

where

u = 23.5 m/s is the initial velocity

\theta=33.5^{\circ} is the angle of projection

Substituting,

u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity, downward

Substituting everything into the equation we get:

-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s

Therefore, the horizontal distance covered in a time t is

d=v_x t

And substituting t = 3.59 s, we find

d=(19.6)(3.59)=70.4 m

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

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