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3 years ago

6

How fast must a cyclist climb a 6.0º hill to maintain a power output of 0.25 hp? Neglect work done by friction, and assume the m

ass of cyclist plus bicycle is 68 kg.

1 answer:

N76 [4]3 years ago

4 0

Answer: 2.7m/s

Explanation:

cyclist fight only against the component of weight along the road (no friction) , that is:

fx =Wsin6° = mgsin6°=68*9.8*sin6°=69.6N downward

so the force that moves the bike over the hill will be equal to fx but upward

Now we now that the power P = 0.25hp= 186.5W is:

P=f*v so v = P/fx = 186.5/69.6 = 2.7m/s

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As the wavelength of light decreases,<br> What happens

stich3 [128]

Answer:

Waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. ... The wavelength decreases as the light enters the medium and the light wave changes direction.

Explanation:

As a wavelength increases in size, its frequency and energy (E) decrease. From these equations you may realize that as the frequency increases, the wavelength gets shorter. ... Mechanical and electromagnetic waves with long wavelengths contain less energy than waves with short wavelengths.

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3 years ago

Read 2 more answers

A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the

insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

$\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}$

Req =3.03Ω , R1 =12.18Ω

$\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}$

$\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}$

$\frac{1}{R2}=0.2479$

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0

3 years ago

Two metal spheres are suspended from strings. The

nikdorinn [45]

Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = -2 \times 10^2 C$

$q_2 = -4 \times 10^{-8} C$

r = 3.0 m

now we have

$F = \frac{(9 \times 10^9)(2 \times 10^2)(4\times 10^{-8})}{3^2}$

$F = 8000 N$

since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be

B. A repulsive force of 8.0*10^3 N.

3 0

2 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

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2 years ago

7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca

Misha Larkins [42]

Answer:

q = C V charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor

N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors

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2 years ago