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Maksim231197 [3]
3 years ago
9

Anyone knows how to do this? I think it’s 0 because it’s not moving————

Physics
1 answer:
dedylja [7]3 years ago
8 0

Answer:

1 m/s  

Explanation:

2 to 3 second interval would be zero.

3 to 4 is 2 m/s

so average is 1 m/s

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A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi
kati45 [8]

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well  nitrate per 15 \mathrm{m}^{3} of water

nitrate in distant reservoir =5 \mathrm{~g} / \mathrm{m}^{3}

Let the flow rate of well

flow rate of reservoir =y m^{3} / s

Drinking water requirement is 45 \mathrm{ppm} or 45 \mathrm{~g} / \mathrm{m}^{3}

therefore, the total flow of drinking water

 

8 0
2 years ago
What is the velocity of a 50kg skater if her momentum is 225kg. m/s?
Gnom [1K]

|Momentum| = (mass) x (speed)

225 kg-m/s =(50kg) x (speed)

Divide each side by (50kg):  Speed=(225 kg-m/s) / (50 kg) = 4.5 m/s .

Regarding the velocity, nothing can be said other than the speed, because
we have no information regarding the direction of the object's motion.

8 0
3 years ago
Read 2 more answers
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0
3 years ago
The amplitude of a paricular wave is 4.0 m. The crest to trough distance
kozerog [31]

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

6 0
3 years ago
a 4m long straight wir that carries acurrent of 0.5A is placed perpendicular to a uniform magnetic field. if the size of magneti
PolarNik [594]

Answer:

B=0.2T

Explanation:

given required solution

l=4m B=? <em>F</em><em>=</em><em>BIL</em>

i=0.5A B=F/IL

F=0.4N B=0.4N/0.5A*4m

B=0.4/2=0.2T

5 0
3 years ago
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