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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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Julie is making a capacitor with an area of 2.5 × 10–5 m2. The capacitance is 5.5 pF. What is the distance between the plates?

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The answer is A. 40 µm

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3 years ago

What is the relationship between the angle of incidence and the angle of refraction when light passes from a medium where its sp

Bogdan [553]

Answer:

This is because the speed of a wave is determined by the medium through which it is passing. When light speeds up as it passes from one material to another, the angle of refraction is bigger than the angle of incidence. For example, this happens when light passes from water to air or from glass to water ❤

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2 years ago

The acceleration of an object would increase if there was an increase in the

xxMikexx [17]

As per Newton's II law we know that

$F_{net} = ma$

here we know that

$F_{net} = F_{ap} - F_f$

so here we will have

$a = \frac{F_{ap} - F_f}{m}$

so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.

So here correct answer will be

<em>A) force on the object.</em>

4 0

3 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$ (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$

7 0

3 years ago

A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will be its final velocity in m/s,

balu736 [363]

Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

initial velocity of the car, u = 120 km/h = 33.33 m/s

duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + 13 x 2

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

4 0

3 years ago