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2 years ago

What is the maximum negative displacement a dog could have if it started its motion at +3 m?

Physics

1 answer:

raketka [301]2 years ago

7 0

Answer:

- 3 meter

Explanation:

A dog has started motion from +3 meter. ...(Given)

∴ maximum positive distance = + 3 meter

Magnitude of distance = 3

Maximum negative distance = (-) (magnitude of distance)

Maximum negative distance = (-) (3)

Maximum negative distance = -3 meter

Hence, maximum negative distance is -3 meter.

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The rotational speeds of four generators are listed in RPM (revolutions per minute). Arrange the generators in order based on th

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with the same generator, so the only factor for producing the slectric field is only the speed. The faster the rotational speed of the generator the greater it produce electric field. So the sequence is 3000 rpm < 3200 rpm < 3400 rpm < 3600 rpm

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2 years ago

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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

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Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

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2 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me

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1) Current in the wire: 0.0875 A

The current in the wire is given by:

$I=\frac{Q}{t}$

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

$I=\frac{420 C}{4800 s}=0.0875 A$

2) Drift velocity of the electrons: $1.78\cdot 10^{-6} m/s$

The drift velocity of the electrons in the wire is given by:

$u = \frac{I}{nAq}$

where

I = 0.0875 A is the current

$n=5.8\cdot 10^{28}$ is the number of free electrons per cubic meter

A is the cross-sectional area

$q=1.6\cdot 10^{-19} C$ is the charge of one electron

The radius of the wire is

$r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m$

So the cross-sectional area is

$A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2$

So, the drift velocity is

$u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s$

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2 years ago

An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h

marin [14]

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

$K=\frac{1}{2}mv^{2}$

For the object thrown in the air:

$K=\frac{1}{2}.2.[v(t)]^{2}$

$K=(-9.8t+24)^{2}$

$K=96.04t^{2}-470.4t+576$

Kinetic energy of the object as a function of time: $K=96.04t^{2}-470.4t+576$

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

$U=mgh$

For the object thrown in the air:

$U=9.8.2.h(t)$

$U=9.8.2.(-4.9t^{2}+24t+60)$

$U=-96.04t^{2}+470.4t+1176$

Potential energy as function of time: $U=-96.04t^{2}+470.4t+1176$

Total kinetic and potential energy, also known as mechanical energy is

TME = $96.04t^{2}-470.4t+576$ + ( $-96.04t^{2}+470.4t+1176$ )

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

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1 year ago

5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

Anna007 [38]

Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

3 0

2 years ago