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4 years ago

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if a bicyclist, with initial velocity of zero, steadily gain speed until reaching a final velocity of 26m/s, how far away did sh

e travel during the race (in the same amount of time?

1 answer:

Alja [10]4 years ago

5 0

Answer:

<h2><em>34.46m</em></h2>

Explanation:

Using one of the equation of motion to solve the question. According to the equation v² =u²+2as where;

v is the final velocity of the bicyclist = 26m/s

u is the initial velocity of the bicyclist = 0m/s

a is the acceleration due to gravity = 9.81m/s

s is the distance covered during travel

Substitute the given parameters into the formula above to get the distance traveled

26² = 0² + 2(9.81)s

676 = 19.62s

Divide both sides by 19.62

676/19.62 = 19.62s/19.62

s = 34.46m

<em>The distance traveled by the bicyclist during the race is 34.46m</em>

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The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate start

Yanka [14]

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction $\mu _s=0.24$

Coefficient of kinetic friction $\mu _k=0.22$

mass of crate $m=20\ kg$

Force applied $F=200\ N$

maximum static Friction $F_s=\mu _sN$

$N=mg$

$F_s=0.24\times 20\times 9.8$

$F_s=47.04\ N$

thus applied force is greater than Static friction therefore kinetic friction will come into play

$F_k=\mu _kN$

$F_k=0.22\times 20\times 9.8=43.12\ N$

net Force on crate $F-F_k=ma$

$a=\frac{200-43.12}{20}=7.84\ m/s^2$

Magnitude of velocity can be obtained by using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

$v^2-0=2\times 7.84\times 2$

$v=\sqrt{31.37}$

$v=5.60\ m/s$

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4 years ago

What skills are used to move the body from one place to another? *

Anuta_ua [19.1K]

Your muscles, bones, heart and other things in the body.

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3 years ago

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Suppose a tidal basin is 6 m above the ocean at low tide and that the area of the basin is 5×107 m2. Estimate the gravitational

Neko [114]

Answer:

Potential energy will be $176.58\times 10^{11}j$

Explanation:

We have given the height of the basin is h = 6 m

Area of the basin $A=5\times 10^7m^2$

Volume $V=area \times height=5\times 10^7\times 6=30\times 10^7m^3$

Density $\rho =1000kg/m^3$

We know that mass is given by $m=\rho V=1000\times 30\times 10^7=3\times 10^{11}kg$

We know that potential energy is given by $E=mgh=3\times 10^{11}\times 9.81\times 6=176.58\times 10^{11}j$

3 0

3 years ago

In a loading a lorry a man lifts boxes each weight 200N through a height of 3.5m.

vekshin1

The work done on the boxes and the energy spent is 700J while the power required to lift 4 boxes per minute is 2.8kW

This work done on an object is the force required to move that object from point A to point B.

Data Given;

Weight (mg) = 200N
Distance (s) = 3.5m

<h3>Work Done</h3>

This is the work done to move the boxes from point A to point B.

$w = F.s\\F = ma = mg = 200N\\w = 200 * 3.5\\w = 700J$

The work done to move the object is 700J

<h3>Energy</h3>

This is used to calculated the energy used to carry the work done and the formula is given as

$E = force * displacement\\E = F.s\\E = 200 * 3.5\\E = 700J$

The energy transferred when he lifts one box is 700J

<h3>Power</h3>

Power is the rate at which energy is used with respect to time

$P = E/t\\$

The energy required to lift 4 boxes in one minutes is

$700 * 4 = 2800$

Now we can calculate the power used

$P = Energy / time\\P = 2800 / 1 \\P = 2800W = 2.8kW$

The power required to move four boxes in 1 minutes is 2.8kW

From the calculations above, the following was gotten

work done = 700J
Energy = 700J
Power = 2800W or 2.8kW

Learn more on work done, energy and power here;

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3 years ago

The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by

Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

$m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}$

we have

a. for c = 20 cm

$m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,$

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

$\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.$

Evaluated at c = 20 cm, we have

$\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.$

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

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4 years ago