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murzikaleks [220]
2 years ago
12

Suggest two ways in which the boy could balance the beam?

Physics
2 answers:
Oduvanchick [21]2 years ago
7 0
He can balance it on his hand or look for something that can handle the weight of it
postnew [5]2 years ago
5 0

Answer:

Hope this is what you were looking for!

Explanation:

1) So, if we are talking about balancing a straight stick (beam), he could put a finger from each hand on either side of it and move them together. As a result of friction his fingers will meet at the balance point.

2) Method of trial and error.

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A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1
nasty-shy [4]

(a) 1.72\cdot 10^{-5} \Omega m

The resistance of the rod is given by:

R=\rho \frac{L}{A} (1)

where

\rho is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m, so the cross-sectional area is

A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega

And now we can re-arrange the eq.(1) to solve for the resistivity:

\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m

(b) 8.57\cdot 10^{-4} /{\circ}C

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega

The equation that gives the change in resistance as a function of the temperature is

R(T)=R_0 (1+\alpha(T-T_0))

where

R(T)=0.86 \Omega is the resistance at the new temperature (92.0°C)

R_0=0.81 \Omega is the resistance at the original temperature (20.0°C)

\alpha is the temperature coefficient of resistivity

T=92^{\circ}C

T_0 = 20^{\circ}

Solving the formula for \alpha, we find

\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C

5 0
3 years ago
A lemon with mass 0.3 kg falls out of a tree from a height of 1.8 m. How much mechanical energy does the lemon have just before
Zina [86]
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<span>PE=mass∗gravity∗height=0.3kg∗9.8m/s2∗1.8m=?

</span>The answer will typically be given in joules:

1J=kg∗m2s2 Could be wrong...  But I believe it is 5.3...? as a final product.
3 0
3 years ago
Read 2 more answers
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 1
nevsk [136]

Answer:

A. 2.2*10^-2m

Explanation:

Using

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So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

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A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

5 0
3 years ago
Two groups of scientists got different results while studying about the evolution of the universe. Which of these statements bes
aliya0001 [1]
The one that would explain why the two groups of scientists got different results is : 
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Probably one analyze it with x theory and the other use y theory

hope this helps
3 0
3 years ago
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How might a <br> Theory relate to a model
Alexxx [7]
A theory can help create a model
3 0
3 years ago
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