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katrin2010 [14]

3 years ago

10

A bicyclist maintains a constant speed of 4.00 m/s up a hill that is inclined at 10.0∘ with the horizontal. Calculate the work d

one by the person and the work done by gravity if the bicycle moves a distance of 18.0 m up the hill. The combined mass of the rider and the bike is 82.0 kg. 1) Calculate the work done by gravity. (Express your answer to three significant figures.)

1 answer:

Alex3 years ago

8 0

Answer:

the work done by the person $W_p=2.52*10^3J$

the work done by gravuty $W_g=-2.52*10^3J$

Explanation:

Because the velocity is constant the cyclist has to excert a force equal to the component of the weight.

$F_c-F_w=0$

$F_w=m*g*cos(\theta)\\F_w=82.0*9.81*sin(10^o)\\F_w=140N$

The work done by the person is given by:

$W_p=F_c*d\\W_p=140N*18.0\\W_p=2.52*10^3J$

and the work done by the gravity:

$W_g=F_g*d\\W_g=-82.0*9.81*sin(10)*18.0\\\\W_g=-2.52*10^3J$

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A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin

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Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
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In our problem, $p_1 = 1.5 atm$ , $V_1 =6.5 L$ and $V_2 =13.0 L$ , so the final pressure of the gas can be found by re-arranging eq.(1):
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3 years ago

What is the value for the kinetic energy for a n = 5 Bohr orbit electron in zeptoJoules?

Ne4ueva [31]

Answer:

The kinetic energy is 86.6 zepto joules.

Explanation:

Given that,

Number of orbit n =5

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Bohr's radius for hydrogen atom is

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Now, put the value of n in the formula of radius

$r=0.53\times10^{-10}\times5^2$

$r =1.33\times10^{-9}\ m$

We need to calculate the kinetic energy

Using formula of kinetic energy

$E_{k}=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{e^2}{2\times r_{s}}$

Put the value into the formula

$E_{k}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times1.33\times10^{-9}}$

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We know that,

$1\ zepto\ joule=1\times10^{-21}\ J$

The kinetic energy is

$E_{k}=\dfrac{8.66\times10^{-20}}{1\times10^{-21}}$

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3 years ago

If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55.5 m/s in the opposite directi

Lelu [443]

Answer:

ΔP = 14.5 Ns

I = 14.5 Ns

ΔF = 5.8 x 10³ N = 5.8 KN

Explanation:

The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:

ΔP = mv₂ - mv₁

ΔP = m(v₂ - v₁)

where,

ΔP = Change in Momentum = ?

m = mass of ball = 0.145 kg

v₂ = velocity of batted ball = 55.5 m/s

v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)

Therefore,

ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)

<u>ΔP = 14.5 Ns</u>

The impulse applied to a body is equal to the change in its momentum. Therefore,

Impulse = I = ΔP

<u>I = 14.5 Ns</u>

the average force can be found as:

I = ΔF*t

ΔF = I/t

where,

ΔF = Average Force = ?

t = time of contact = 2.5 ms = 2.5 x 10⁻³ s

Therefore,

ΔF = 14.5 N.s/(2.5 x 10⁻³ s)

<u>ΔF = 5.8 x 10³ N = 5.8 KN</u>

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