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Inessa [10]
4 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.
Physics
1 answer:
N76 [4]4 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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Why do covalent and ionic bonds form?
nevsk [136]

Answer:

Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals. An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.

Explanation:

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5 0
3 years ago
What causes competition among organisms?
irga5000 [103]

Answer:

Whats up Bmw,

The organisms is caused when both the organisms or species are harmed, which is limited supply of at least one resource from food or even water used by both can be a factor.

Explanation:

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8 0
4 years ago
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A student rides a bicycle in a circle at a constant speed and constant radius. A force diagram for the student-bicycle system is
skelet666 [1.2K]

Solution:

The gravitational force that acts on the bicycle system is

$F_{g} = 500 \ N$

Now the force, that is the gravitational force is related to mass of the system and the acceleration due to gravity of the system, 'm' and 'g' respectively.

Therefore, we can write

$F_g=mg$

500  = m x 10    (since , g = 10 m/s-s)

∴ m = 50 kg

Now the net vertical force acting on the student bicycle system is 0. And the vertical acceleration of system is also 0. The total horizontal force acts to the right of the system. So by Newton's 2nd law of motion, we can write

$F_f = ma$

$a=\frac{F_f}{m}$

  $=\frac{250}{50}$

Therefore $a= 5 \ m/s^2$

Hence (C) is correct option.  

6 0
3 years ago
What’s 4.5 inches into centimeters?
marshall27 [118]

Answer:

11.43 cm

Explanation:

yes

3 0
4 years ago
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A flutist assembles her flute in a room where the speed of sound is 340 m/s. When she plays the note A, it is in perfect tune wi
Yuki888 [10]

Incomplete question as we have not told to find which quantity.The complete question is here

A flutist assembles her flute in a room where the speed of sound is 340 m/s. When she plays the note A, it is in perfect tune with a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 347 m/s .

(a)How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?

(b) How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork?

Answer:

(a)f_{beat}=9.06Hz

(b) The flutist extends to the turning joint of her flute by 1.363×10⁻⁵m

Explanation:

Part A

The wavelength of sound wave of the note A

λ=v/f

Where v speed of sound

f is frequency

Substitute the given values

So

λ=(340m/s)÷440Hz

λ=0.7727 m

The new frequency of note A when the air inside the flute has warmed

f¹=v¹/λ

f¹=(347m/s)/0.7727 m

f¹=449.05Hz

The Beat frequency of two waves is:

f_{beat}=|f_{1}-f_{2} |\\f_{beat}=|449.05Hz-440Hz | \\f_{beat}=9.06Hz

Part B

Frequencies of standing waves modes of an open tube of Length L:

f_{m}=m(v/2L)\\where\\m=1,2,3,4........

So

L=m(v/2f_{m} )

L=mλ/2

From part (a) The wavelength of sound of note A is 0.7727m and m=1

So

L=(0.7727/2)=0.38635m

When the air inside the flute has warmed.Then the new length is given below at same frequency of 440 Hz

So

L_{a}=v/2f\\L_{a}=340m/s/2*440Hz\\L_{a}=0.38636m

So the length the flutist extends to the turning joint for the flute to be:

ΔL=La-L

ΔL=0.38636m - 0.38635

ΔL=1.363×10⁻⁵m

The flutist extends to the turning joint of her flute by 1.363×10⁻⁵m

8 0
3 years ago
Read 2 more answers
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