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3 years ago

In order for the eye to see an object, ______________ from the object must be reflected to your eye.

Physics

2 answers:

vovikov84 [41]3 years ago

6 0

Answer:

In order for the eye to see an object, light from the object must be reflected to your eye.

ipn [44]3 years ago

3 0

Answer:

light from the object

Explanation:

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A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

4 years ago

A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate

babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$

3 0

3 years ago

A 62 kg box is lifted 12 meters off the ground. How much work was done?

OleMash [197]

7291.2 I hope this is right

6 0

4 years ago

Read 2 more answers

Galileo's observational contributionsGalileo Galilei was the first scientist to perform experiments in order to test his ideas.

ozzi

<h2>Answer: </h2><h2>- Jupiter has orbiting moons.</h2><h2>- The Sun has sunspots and rotates on its axis.</h2><h2>- The Moon has mountains, valleys, and craters.</h2><h2>- Venus goes through a full set of phases.</h2>

Explanation:

In 1609 Galileo built a telescope, with which he observed mountains and craters on the Moon, discovered Jupiter’s major satellites and the next year he published these discoveries in his book <em>The Sidereal Messenger</em>.

In addition, Galileo observed that Venus presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that Venus rotates around the Sun and not around Earth. This is because <u>Venus presented its smaller size when it was in full phase and the largest size when it was in the new one, when it is between the Sun and the Earth. </u>

<u />

On the other hand, <u>although Galileo was not the first to observe sunspots</u>, he gave the correct explanation of their existence, which supported the idea that planets revolve around the Sun.

These observations and discoveries were presented by Galileo to the Catholic Church (which supported the geocentric theory at that time) as a proof that completely refuted Ptolemy's geocentric system and affirmed Copernicus' heliocentric theory.

4 0

4 years ago

PLS ANSWER FAST TIMED TEST WILL GIVE BRAINLY!!!!!!!!!

lutik1710 [3]

The answer would be true

7 0

3 years ago

Read 2 more answers