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3 years ago

Is my answer correct or no?

Physics

2 answers:

irinina [24]3 years ago

8 0

No your answer is incorrect. Would you like the correct one?

OLEGan [10]3 years ago

8 0

Your answer is incorrect would you like me to help you or do you think you can figure it out on your own

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Does wind have atoms and molecules in int

satela [25.4K]

All forms of matter are comprised of atoms. Wind in itself is the movement of particles(air). Wind does not have particles since it is an action and not a form of matter.

4 0

3 years ago

Convert an acceleration of 12m/s^2 to km/h^2

ozzi

Answer:

43.2

because to convert from m/sec to kmph we need to multiply by 3600/1000

4 0

3 years ago

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

Calculate the force of gravity on the 0.60- kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth

nignag [31]

The gravitational force between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the first object has a mass of $m_1=0.60 kg$ , while the second "object" is the Earth, with mass $m_2=5.97 \cdot 10^{24}kg$ . The distance of the object from the Earth's center is $r=1.3 \cdot 10^7 m$ ; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
$F=G \frac{m_1m_2}{r^2}=(6.67\cdot 10^{-11}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}= 1.41 N$

5 0

3 years ago

A school bus moves slower and slower. Using what you have learned about forces, explain why the bus moves slower and slower.

MariettaO [177]

Explanation:

the weight of the people inside the bus

4 0

3 years ago