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2 years ago

Help me with this please

Physics

2 answers:

NARA [144]2 years ago

5 0

I think 9 is function

12345 [234]2 years ago

4 0

Answer:

for this test or homework or whatever it is I will give you my opinion of what I know.

7. For number 7 I think it's D but I'm not that sure.

8. when they say all of the following is not a family except will that means that which of the following in the multiple choice is not a type of family, well as so far beginning (C) is not the type of family.

9. this question sees given love and security is an example of...........the family, well to answer is function because love and secure makes up the function of a family without love and secure a family cannot be function that good.

10. I believe the answer is A, but not sure.

11. As we all know, we should wear a rubber boots to prevent electric shock.

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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

2 years ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

pantera1 [17]

Answer:

The period is $T = 0.700 \ s$

Explanation:

From the question we are told that

The mass is $m = 0.350 \ kg$

The extension of the spring is $x = 12.0 \ cm = 0.12 \ m$

The spring constant for this is mathematically represented as

$k = \frac{F}{x}$

Where F is the force on the spring which is mathematically evaluated as

$F = mg = 0.350 * 9.8$

$F =3.43 \ N$

$k = \frac{3.43 }{ 0.12}$

$k = 28.583 \ N/m$

The period of oscillation is mathematically evaluated as

$T = 2 \pi \sqrt{\frac{m}{k} }$

substituting values

$T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }$

$T = 0.700 \ s$

7 0

3 years ago

Which law represents the thermodynamic statement of the conservation of energy of a system

Tresset [83]

The law of conservation of energy is a law of science that states that energy cannot be created or destroyed, but only changed from one form into another or transferred from one object to another.<span>

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3 0

2 years ago

When jogging outside you accidentaly bumb into a curb. Your feet stop but your body continues to movr foward and you end up on t

sattari [20]

B.) Newtons first law of motion

8 0

2 years ago

Read 2 more answers

What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s^2? (a) 104N (b) 338 N

Brilliant_brown [7]

Answer:

b) 338 N

Explanation: let m be the mass of the gymnast and a be the acceleration of the gymnast.

the force required to accelerate the gymnast is given by:

F = m×a

= (45.0)×(7.50)

= 337.5 N

Therefore, the force a trampoline has to apply is 138 N.

6 0

3 years ago

Help me with this please ​

Help me with this please