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3 years ago

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An electric motor rotating a workshop grinding wheel at a rate of 151 rev/min is switched off. Assume constant angular decelerat

ion of magnitude 2.23 rad/s 2 . Through how many revolutions does the wheel turn before it finally comes to rest

1 answer:

tamaranim1 [39]3 years ago

8 0

Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

= 15.8 rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² - 2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99 rad

one revolution = 2π radian

55.99 radian = 55.99 / 2 π no of terns

= 9 approx .

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A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears Tower in Chicago. On the level grou

Sophie [7]

Answer:

442.36038 m or 1451.31362 ft

Explanation:

$P_1$ = Initial pressure = 30.15 inHg

$P_2$ = Final pressure = 28.607 inHg

$\rho$ = Density of air = 0.075 lb/ft³

$1\ lb/ft^3=16.0185\ kg/m^3$

$1\ in=0.0254\ m$

$1\ m=3.28084\ ft$

Density of mercury = 13560 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Difference in pressure is given by

$P_1-P_2=\rho gh\\\Rightarrow h=\frac{P_1-P_2}{\rho g}\\\Rightarrow h=\frac{(30.15-28.607)\times 13560\times 0.0254\times 9.81}{0.075\times 16.0185\times 9.81}\\\Rightarrow h=442.36038\ m\\\Rightarrow h=442.36038\times 3.28084\\\Rightarrow h=1451.31362\ ft$

The height of the building is 442.36038 m or 1451.31362 ft

4 0

3 years ago

For a maximum superelevation of 0.08 ft/ft and a degree of curve of 4o, calculate the maximum safe speed for the curve assuming

Mamont248 [21]

Answer:

Explanation:

Given that

Superelation= 0.08ft/ft

Given curve= u•

Curve junction factor= 0.13

DR= 5729.57795

R = 5729.57795/D

R = 5729.57795/4

R = 1432.4ft

c + f = V^2/gG

0.08 + 0.13 = V^2 / (32*1432.4)

V^2 = 9625.728 or V = 98 ft/sec

The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed

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3 years ago

Kevin jumps straight up in the air to a height of 1 meter.At the top of his jump, he has potential energy of 1,000 joules.Answer

Llana [10]

Gravitational potential energy can be given by the equation
PE = mgh
where m is the mass,
g is the gravitational constant 9.81 or 10 depending on rounding
and h is the height

well weight is a force equiavlent to
W= m*g

so comparing that to the potential energy equation, divide the potential energy by the height and you will get weight in Newtons

3 0

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

3 years ago

At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa

ivann1987 [24]

Answer:

(b) EAST

Explanation:

you can assume that the magnetic field points rightward, that is, in the positive x direction (NORTH). Furthermore, you can assume that the direction of the motion of the electron is in the positive y direction. Hence, you have:

$\vec{B}=B_o\hat{i}\\\\\vec{v}=v_o\hat{j}$

You use the Lorentz formula to known which is the direction of the magnetic force over the electron:

$F=qv\ X\ B$

which implies the cross product between the unitary vecors j and i, that is

$\hat{i} \ X\ \hat{j} = -\hat{k}$ (WEST)

However, the minus sign of the charge of the electron changes the direction 180°. Hence, the direction is k. That is, to the EAST

3 0

3 years ago

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