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2 years ago

Some mechanical advantages is lost to friction true or false

Physics

2 answers:

cupoosta [38]2 years ago

6 0

Answer:

I THINK it would be False.

Talja [164]2 years ago

3 0

Answer:

las respuestas es falso..... Mmm

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the average speed of a runner in a 483. meter race is 3.0 meters per second. How long me runner to complete the race? Dont inclu

lara [203]

Answer:

161

Explanation:

$v=\frac{d}{t}$ slove for t

$t=\frac{d}{v}$

Insert values of d and v

$t=\frac{483}{3} \\$

t=161

3 0

3 years ago

1. Boyle's law relates the pressure of a gas to its

KIM [24]

Answer:

volume is the correct answer

Explanation:

7 0

2 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

3 years ago

5. A man has a weight of 100 Newtons. How much work is done if he climbs 4 meters up a ladder? Plug numbers under the equation.

Virty [35]

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 100 × 4

We have the final answer as

Hope this helps you

5 0

2 years ago

Find the magnitude of the electric force between the charges 0.12 C and 0.33 C at a separation of 2.5 m. k=8.99×109N⋅m2/C2.

makkiz [27]

Answer:

F= k Q1 Q2 / r^2

plug in the numbers

8 0

2 years ago