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Naddik [55]
4 years ago
13

Ondrea could drive a Jetson's flying car at a constant speed of 540.0 km/hr across oceans and space, approximately how long woul

d she take to drive from the Sun to Pluto in years? (Assume Pluto is at its average distance of 5.9 × 109 km from the Sun)
Physics
1 answer:
Tcecarenko [31]4 years ago
7 0

Answer:

The time taken in years is   x = 125 \  years

Explanation:

From the question we are told that

   The  speed is  v  = 540.00 \ km /hr  =  \frac{540 *1000}{3600} =  150\  m/s

    The distance from the sun to Pluto is  d =  5.9*10^{9} \  k m =  5.9*10^9 * 1000 =  5.9*10^{12} \  m

Generally the time  taken is mathematically represented as

     t =  \frac{d}{v}

=>   t = \frac{5.9*10^{11}}{150}

=>   t =  3.933*10^{9}

Converting to years

   1 year  \to  3.154*10^7 \  s

    x \  years  \to 3.933*10^{9}

=>  x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}

=>    x = 125 \  years

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Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a to
stepan [7]

Answer:

0.168°C

Explanation:

The frictional force between the palms enables rubbing and initiates the heat deposed.

The energy generated per rub is;

The force × distance

Hence we have; 71.3 N×0.0750m

Hence for 11 rubs we have; 11× 71.3 N×0.0750m= 58.82J

Now this heat acquired by the tissue is the same as mass of tissue × specific heat capacity of tissue × temperature rise;

Expressed mathematically as;

58.82=0.100 kg×3500 J/(kg · °C)×∆T

∆T= 58.82/ 0.100 ×3500 =0.168°C

6 0
3 years ago
a new car is advertised as having anti-noise technology. The manufacturer claims that inside the car any sound on negated. Evalu
nirvana33 [79]

Answer:

I guess you mean that any sound coming from outside is negated, first, this would mean that you can not hear some signals that may help you to avoid accidents, so you do not want this technology.

Now, let's see if it is possible.

Suppose you have a soundwave approaching the car, the car needs to cancel the soundwaves as the soundwaves approach to the surface, to do it, you need to create another wave that is equal in amplitude, but with a change of phase of pi.

in order to do this, the car must be able to "analyze" the coming soundwave and instantly create another one to cancel it. Suppose that the sound is canceled. now if the sound changes, the car must be able to also change the sound that the car is producing, this instantaneous change is one of the problems.

Another problem is that, as you know, the sound propagates as spherical waves, this means that the wavefront of a wave produced far away will have a bigger radius than the soundwave produced by the car, then we never will have a situation where the wavefront is canceled: This means that we only can cancellate the sound in some areas of the car, and we still will have some sound in other parts of the car.

Another way to isolate the car is with isolating panels, those panels absorb the sound and transmit a very little amount of it (those panels are used in recording studios, for example). With enough of those you can cancel almost all the sound coming from the outside, but to do this you will need a big car because the amount of isolating material needed is a lot.  

We can conclude that the manufacturers claim is not correct.

5 0
3 years ago
A Van de Graaff generator causes a total charge q to build up on a metal sphere of radius r. Which variable does not affect the
Maru [420]

Answer:

The radius r of the metal sphere.

Explanation:

From Gauss's law we know that for a spherical charge distribution with charge Q, the electrical field at distance R from the center of the sphere is given by

E=\frac{Q}{4\pi \epsilon_oR^2}

What is important to notice here is that the radius of the sphere does not matter because any test charge sitting at distance R feels the force as if all the charge Q were sitting at the center of the sphere.

This situation is analogous to the gravitational field. When calculating gravitational force due to a body like the sun or the earth, we take not of only the mass of the sun and the distance from it's center; the sun's radius does not matter because we assume all of its mass to be concentrated at the center.

6 0
3 years ago
Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo
kifflom [539]

Answer:

893 Newtons

Explanation:

m = Mass of Usain Bolt = 94 kg

a = Acceleration of Usain Bolt in the first 0.89 seconds = 9.5 m/s²

From Newton's Second law

Force

F=ma

\\\Rightarrow F=94\times 9.5

\\\Rightarrow F=893\ N

The average horizontal force exerted by Bolt against the ground during the first 0.890 s of the race is 893 Newtons

3 0
3 years ago
PLEASE HELP ASAP
Liula [17]

Answer:A skier could benefit from a sports-specific training program because it could help them become better at skiing. ... Other activities such as squats and lunges will also improve skiing skills since it builds up lower body strength which is necessary to ski tough terrain.

Explanation:

7 0
3 years ago
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