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3 years ago

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as earth rotates about it’s axis, it takes three hours for the united states to pass beneath a point above earth that is station

ary relative to the sun. What is wrong with the following scheme? to travel from washington, D.C. to san francisco shoving very little fuel, simply ascend in a helicopter high over washington, D.C. and wait three hours until san francisco passes below

1 answer:

pickupchik [31]3 years ago

6 0

interesting question.

how much fuel to hover over one place ?

The Foucault pendulum is a v v v long pendulum which can show the eart's rotation over time if the pend bob motion is tracked and recorded

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If you use a 1500 watt sound system for 5 hours a day , calculate the amount of electricity used

Strike441 [17]

Answer:

Our appliance and electronic energy use calculator allows you to estimate your ... using the formulas provided below; Installing a whole house energy monitoring system. ... energy consumption a few watt-hours, and you can use a monitor to estimate those too. ... Estimate the number of hours per day an appliance runs.

Explanation:

6 0

2 years ago

. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110

7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x $10^{5}$ J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, $v_{0}$ = 110 km/h = 110 x 1000/3600 = 30.6 m/s $\frac{30.6^{2} }{2x9.8}$

initial height, $h_{0}$ = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = $\frac{1}{2} mv_{0} ^{2}$

gh = $\frac{1}{2} v_{0} ^{2}$

h = $\frac{v_{0} ^{2} }{2g}$

= 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

= mg (h - $h_{0}$ )

= 750 x 9.8 x (47.6 - 22)

= 188160 Joule

= 1.88 x $10^{5}$ J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

= 1.88 x $10^{5}$ /(22/sin 2.5°)

= 373 N

8 0

3 years ago

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a

vampirchik [111]

Answer:

$F = 1.5 \times 10^{-16} N$

this force is $1.68 \times 10^{13}$ times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

$KE = 1 keV$

$KE = 1 \times 10^3 (1.6 \times 10^{-19}) J$

$KE = 1.6 \times 10^{-16} J$

now the speed of electron is given as

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

now we have

$v = \sqrt{\frac{2 KE}{m}}$

$v = 1.87 \times 10^7 m/s$

now the maximum force due to magnetic field is given as

$F = qvB$

$F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})$

$F = 1.5 \times 10^{-16} N$

Now if this force is compared by the gravitational force on the electron then it is

$\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}$

$\frac{F}{F_g} = 1.68 \times 10^{13}$

so this force is $1.68 \times 10^{13}$ times more than the gravitational force

4 0

3 years ago

For each of the problems below, you will need to draw a graph to find the solution.

Gelneren [198K]

Answer:

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

Explanation:

3 0

2 years ago

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

3 years ago