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Svetach [21]
3 years ago
8

A student carries a backpack for one mile , another student carries the same back pack for two miles . Compared to the first stu

dent , how much work did the second student do ?
Physics
2 answers:
NemiM [27]3 years ago
8 0

Answer:

he did twice as much since the first student carried the backpack for 1 mile whilst the second one carried it for 2 miles

Explanation:

Oxana [17]3 years ago
8 0

Answer:

2miles-1mile

the second person carried the for one mile longer.

So as your sentence you would sat

Compared to the first person the second person carried the backpack one mile longer.

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A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre
pantera1 [17]

Answer:

Part a)

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

T = \frac{m}{L}xg + Mg

so the speed of the wave at that position is given as

v = \sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{m}{L}

now we have

v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

time taken by the wave to reach the top is given as

t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}

t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})

t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})

t = 12 s

4 0
3 years ago
How high can a 40 N force move a load, when 395 J of work is done?
Juliette [100K]

Answer:

9.875

Explanation:

w=f×s

395=40×s

make s the subject of the formula

s=395/40

=9.875

7 0
3 years ago
If the force applied to an object is not greater than the starting friction, what will happen to the object? Which answer is cor
AleksandrR [38]

Answer:

The object will move in the opposite direction of the force applied. - 2.

8 0
3 years ago
Which of the waves has the smallest amplitude?
balandron [24]

Answer:

I think its the Blue wave, im not sure so dont take my word for it.

Explanation:

3 0
3 years ago
Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of
natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00  m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

r'=\sqrt{d^2+r^2}

Where, d = distance between two identical loudspeakers

r = distance between speakers and listener

Put the value into the formula

r'=\sqrt{(1.00)^2+(4.00)^2}

r'=\sqrt{1.00+16.00}

r'=4.12\ m

We need to calculate the path difference

Using formula of path difference

|r'-r|=4.12-4.00

|r'-r|=0.12\ m

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{v}{f}

Where, v = speed of sound

f = frequency

Put the value into the formula

\lambda=\dfrac{343}{300}

\lambda=1.143\ m

We need to calculate the phase difference

Using formula of phase difference

\phi=\dfrac{2\pi\times|r'-r| }{\lambda}

\phi=\dfrac{2\pi\times0.12}{1.143}

\phi=0.659\ rad

Hence, The phase difference is 0.659 rad.

7 0
3 years ago
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