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2 years ago

A 60 kg bicyclist going 2 m/s increased his work output by 1,800 J. What was his final velocity?

Physics

2 answers:

Alinara [238K]2 years ago

6 0

8 was the correct answer, thanks

Bas_tet [7]2 years ago

3 0

Bicyclist initial kinetic energy is Ek=(1/2)*m*v² where m is his mass and v is his speed and that is equal to:

Ek=(1/2)*60*2²=120 J.

When we add the increased work output, we get the total kinetic energy:

Ek(total)=Ek+W= 120 J + 1800 J= 1920 J

So Ek(total)=1920 J = (1/2)*m*V² where V is the speed after the bicyclist increased his work output. So lets solve for V:

(1/2)*60*V²=1920

30*V²=1920, we divide by 30,

V²=64, and take the square root of both sides,

V=8 m/s.

So the speed of the bicyclist after the increased work output is V=8 m/s.

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A simple arrangement by means of which e.m.f,s. are compared is known

strojnjashka [21]

Answer:

A simple arrangement by means of which e.m.f,s. are compared is known as?

(a)Voltmeter

(b)Potentiometer

(c)Ammeter

(d)None of the above

Explanation:

5 0

2 years ago

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha

gregori [183]

The answers are:

a) $Work=125,923.61J$

b) $Power=1.46watt$

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

$1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}$

Also, we need to find the mass given the density of the blood.

$1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg$

Now, calculating how much work does the heart do in a day, we have:

$Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J$

Then, calculating what is the power output and its horsepower, we have:

$Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt$

Have a nice day!

7 0

2 years ago

Two sinusoidal waves travel along the same string. They have the same wavelength and frequency. Their amplitudes are ym1 = 2.5 m

Nimfa-mama [501]

Answer:

0.5 m

Explanation:

Givens:

ym1 = 2.5 mm

ym2 = 4.5 mm

Ф_1=π / 4

Ф_2=π / 2

We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is

Ym = (ym1 + ym2)cos(Ф_2/2)

By substitution we have

Ym= (0.025 + 0.045)cos(π/4) = 0.496 m

The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore

Ym^2=(ym1^2+ym2^2)

So we have Ym=√0.025^2+0.045^2

= 0.5 m

7 0

2 years ago

How high above the earth's surface is g reduced to 8.80m/^2?

Sladkaya [172]

Gravity changes as the altitude change. The gravitational force is proportional to 1/R2, where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.

g1/(g2) = R2^2/(R1)^2

so we set the constant values to R1 and the unknown distance as x

(9.8)/(8.80) = (6400-x)2/(6400)^2

solving for x we will get

x = 345.85 km above the earths surface

Hope my answer would be a great help for you.
If you have more questions feel free to ask here at Brainly.

6 0

2 years ago

Read 2 more answers

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a

fgiga [73]

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

$sin \theta = \frac{n \lambda}{a}$

where

$\theta$ is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

$\lambda$ is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because $\theta$ increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) $10.8^{\circ}$

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

$sin \theta = \frac{(1) \lambda}{a}$