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3 years ago

8

A construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity if he travels a distance o

f 50 m how much work is being done

1 answer:

Anastasy [175]3 years ago

8 0

Answer:

W = 0

Explanation:

We are given with, a construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity. He travels a distance of 50 m.

The work done by an object is given by :

$W=Fd$

F = ma

So,

$W=mad$

m is mass

a is acceleration

d is displacement

The worker is moving with constant velocity, its acceleration will be 0. So, the work done by the worker is 0.

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Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit

pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let $T_{0}$ = $70^{0}$ F be the temperature and T that of the body

Explanation:

Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = <em>kdt</em>

In [T-70]= <em>kt </em>+ $C_{1}$

T = 70 + $C_{2}$ $e^{kt}$

Finding he value for $C_{2}$ using the initial value of T (0)= 300, therefore we get:

300=70+ $C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

Finding the value for <em>k </em>using T (3) = 200, therefore we get

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

<em>K </em>= $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

T(t) = 70+230 $e^{-0.19018}$

4 0

3 years ago

A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition

Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

$\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}$

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

d = slit separation = 0.175 mm = 0.000175 m

Therefore,

$\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\$

<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>

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3 years ago

What is the technology used behind scanning probe microscopes?

Wittaler [7]

hola ya sabes la respuesta si si ,me la dices porfa

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4 years ago

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White raven [17]

Answer:

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3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

4 years ago