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svetoff [14.1K]
3 years ago
15

Most of the Earth’s water is found in the rivers and lakes. t or d

Physics
2 answers:
salantis [7]3 years ago
5 0
False most of earths water is found in oceans .
Temka [501]3 years ago
4 0
False the majority of the water on earth is found in the oceans
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A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?
NemiM [27]

The average current is 0.10 A.

<h3>Current </h3>

Charge moving through a location on a circuit at a constant rate is called current. When numerous coulombs of charge pass over a wire's cross section in a circuit, it produces a large current. It is not necessary for a wire to be moving at a fast speed in order to have a high current if the charge carriers are tightly packed into the wire. To put it another way, many charge carriers traveling through the cross section is sufficient; they do not need to travel a great distance in a single instant. The amount of charges that flow through a cross section of wire on a circuit, as opposed to how far they travel in a second, is what determines current.

A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?

Learn more about current here:

brainly.com/question/2264542

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6 0
1 year ago
1. Briefly Explain why sliding friction is less than static friction.
stellarik [79]

Explanation:

No surface is perfectly smooth.  Everything has irregularities, like bumps and ridges.  Static friction is the friction between non-moving objects when their surfaces lock together.  When an object is sliding, its surface skips and jumps along the surface of the other object.  Because there's less locking between the surfaces, the sliding friction is less than the static friction.

5 0
3 years ago
A Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia In one set of experimen
balu736 [363]

Answer:

Coefficient of static friction = 0.37

Explanation:

At the point the the quoll slides, quoll attains its maximum velocity.

So Ne = (mv^2)/r ....equa 1

And N =mg....equ 2

Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s

Sub equ 2 into equ 1

Mge= (mv^2)/r ...equa3

Simplfy equ3

e = v^2/(gr)...equ 4

Sub figures above

e = 5.76/(9.8*1.6)

e = 0.37

7 0
3 years ago
To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be
Romashka-Z-Leto [24]
I think the answer is A.
7 0
2 years ago
Read 2 more answers
At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until
Sav [38]

Answer:

a) \Delta \theta = 617.604\,rad, b) \Delta t = 10.392\,s, c) \alpha = -14.128\,\frac{rad}{s^{2}}

Explanation:

a) The final angular speed at the end of the acceleration stage is:

\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)

\omega = 111.5\,\frac{rad}{s}

The angular deceleration is:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}

\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}

\alpha = -14.128\,\frac{rad}{s^{2}}

The change in angular position during the acceleration stage is:

\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}

\theta = 177.604\,rad

Finally, the total change in angular position is:

\Delta \theta = 440\,rad + 177.604\,rad

\Delta \theta = 617.604\,rad

b) The time interval of the deceleration interval is:

\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }

\Delta t = 7.892\,s

The time required for the grinding wheel to stop is:

\Delta t = 2.50\,s + 7.892\,s

\Delta t = 10.392\,s

c) The angular deceleration of the grinding wheel is:

\alpha = -14.128\,\frac{rad}{s^{2}}

4 0
3 years ago
Read 2 more answers
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