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goldenfox [79]
2 years ago
14

The efficiency of a modern bicycle is 95% if you exert 200 Joules of input work pedaling a bicycle on level ground what is the u

seful output work?
Physics
1 answer:
Delvig [45]2 years ago
8 0

Answer:

95 J

Explanation:

You can calculate efficiency by dividing useful output by total input, then multiplying it to 100.

So the foumula goes like:

Efficiency= (Useful output/Total input)x100

In this question,

Efficiency= 95%

Useful output= x

Total input= 200

Therefore;

95=(x/200)x100

0.95=x/100

x=0.95x100

x=95 Joules

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An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of
Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

3\times \frac{\lambda}{2} = 0.39

\lambda = 0.26 m

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

d = \frac{\lambda}{4}

d = \frac{0.26}{4} = 0.065 m = 6.5 cm

So the distance will be 6.5 cm

3 0
3 years ago
What is the answer for this question (4x³)²​
Delvig [45]
16x to the power of 6
6 0
3 years ago
Read 2 more answers
In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring cons
Reptile [31]

Answer:

x = 5.29 m

Explanation:

given,

weight of stretch = 37 N

left-hand spring constant (k₁)= 2.7 N/cm

right hand spring constant(k₂)= 4.3 N/ cm

spring are connected in parallel

F = F₁ + F₂

F = k₁x + k₂x

F = (k₁+ k₂)x

37= (4.3+ 2.7)x

7 x = 37

x = 5.29 m

3 0
3 years ago
What is the resultant of a pair of one pound forces at right angles to each other?
OleMash [197]

Answer:

F_R=\sqrt{2} \ pound.force

Explanation:

Given that there are two force of 1 pound each at right angles to each other.

The from the vector law of addition:

F_R=\sqrt{F_1^2+F_2^2}

where:

F_R= resultant force

F_1\ \&\ F_2 be the two of the forces to be added.

F_R=\sqrt{1^2+1^2}

F_R=\sqrt{2} \ pound.force

8 0
3 years ago
One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d
stira [4]

Answer:

a) Diffusion  coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

^{2} = 2Dt..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

^{2} = 2Dt

0.1^{2}  = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0
3 years ago
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