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3 years ago

What is water that flows across eatlrtus surface

Physics

1 answer:

ira [324]3 years ago

4 0

Water that flows across the surface is called a;

Runoff

That's when rain has saturated the ground to the point it cant hold anymore and it runs over the surface.

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A 3.00 kg stone is dropped from a 39.2 m high building. when the stone has fallen 19.6 m, the magnitude of the impulse the earth

tatiyna

<span>The formulas are, Impulse = mv-mu ....... (1) v^2 = u^2 + 2as .......... (2) We know that, u=0 a=acceleration=gravity = 9.80665 m/s^2 = 9.81 m/s^2 s=19.6 sub (2) we get, v^2 = 0+ 2*9.81*19.6 v^2 = 2*9.81*19.6 v^2 = 384.552 v = 19.6099 v = 19.61 m/s Sub v=19.61 m/s in (1) we get, Impulse = mv - mu we know that u=0; v= 19.61 m/s; m= 3.00 kg Impulse = 3(19.61) - 3(0) Impulse = 58.83-0 Impulse = 58.83 Ns. Therefore the gravitational force exerted by the stone is 58.83 Ns.</span>

4 0

4 years ago

a force of 5.5 n is applied to an object. the moment arm for the force is 0.84 m. what is the torque produced by the force?

Kipish [7]

A force of 5.5N is applied to an object. The moment arm for the force is 0.84 m, the torque produced by the force is <u>4.62N-m</u>.

A force applied perpendicularly to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. We can find the torque from a force by taking the perpendicular component of that force and multiplying by the magnitude of the R vector where this R vector is the vector that points from the axis to the point where the force is applied.

To calculate torque produced by the force, we have-

Torque= forcexmoment arm

T = fxd

T =5.5Nx0.84m

T=4.62N-m.

To learn more about torque, visit---

brainly.com/question/20691242

#SPJ4

6 0

2 years ago

Evelynn is measuring the pitch of a piano note. What unit of measurement is she most likely recording her value

sweet [91]

Answer:

hertz

Explanation: play piano

5 0

2 years ago

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)