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xz_007 [3.2K]
4 years ago
11

What is surface water or runoff?

Physics
1 answer:
Rudik [331]4 years ago
7 0
Water that gathers on the surface of the ground is surface water
movement of landwater to the oceans is runoff
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Suppose that the emf from a rod moving in a magnetic field was used to supply the current to illuminate a light bulb in a circui
faltersainse [42]

Explanation:

The rod moving in a magnetic field and induces an emf which is used to illuminate the bulb. But if the bulb is removed form the circuit, the circuit is opened.

For an open circuit, no current is passes through the moving rod. If there is no current along the rod, then no magnetic field developed around the rod. Because moving charges nothing but current produces the magnetic field around the rod.

The formula for the magnetic force on the rod is,

F_{\mathrm{B}} &=I(l \times B)

=I l B \sin \theta

The current along the rod is zero because the bulb is removed and the magnetic field around the rod is zero because no current is passes through the rod.

Then calculate the magnetic force on the rod as follows:

\F_{\mathrm{B}} &=I l B \sin \theta

=(0)(l)(0) \sin \theta =0 \mathrm{N}

Thus, no force is needed because there is no longer magnetic field developed around the rod.

8 0
3 years ago
how much energy would microraptor gui have to expend to fly with a speed of 10 m/sm/s for 1.0 minute?
Ray Of Light [21]

Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.

The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.

P = E/t

Where:

P = power (W)

T = time (s)

Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is

P = E/t

E = P x t

E = 8.1 x 60

E = 486 Joules

Thus, the energy that Microraptor must expend to fly at 10 m/s for 1.0 minutes is the 486 J.

Learn more about Microraptor gui here brainly.com/question/1200755

#SPJ4

3 0
1 year ago
A cylinder measuring wide and high is filled with gas. The piston is pushed down with a steady force measured to be . Calculate
jekas [21]

Explanation:

Let us assume that the given cylinder is 2.6 cm wide and its height is 3.1 cm. And, when piston is pushed down then the steady force is equal to 15 N.

Now, radius of the cylinder will be as follows.

            r = \frac{diameter}{2}

              = \frac{2.6 cm}{2}

              = 1.3 cm

or,           = 0.013 m         (as 1 m = 100 cm)

As, area of cylinder = \pi \times r^{2}

                                = 3.414 \times (0.013 m)^{2}

                                = 5.77 \times 10^{-4} m^{2}

Relation between pressure and force is as follows.

             Pressure = \frac{Force}{Area}

                         = \frac{15 N}{5.77 \times 10^{-4} m^{2}}

                         = 25996 N/m^{2}

Since, 1 N/m^{2} = 1 Pa           (as 1 kPa = 1000 Pa)

Therefore,   P = 25996 N/m^{2}  

                      = 25.99 kPa

                      = 26 kPa (approx)

Thus, we can conclude that pressure of the gas inside the cylinder is 26 kPa.

7 0
3 years ago
Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 12
ikadub [295]

Answer:

The unknown mass of the bullet is  m1=3.751x10^{-3} kg

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass  <u><em>m</em></u> , it results in change in momentum of the body and is given by:

F=\frac{P}{dt}

Where:

P

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

m_{2}=5.79x10^{-3}kg  \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of  

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\

-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg

3 0
3 years ago
Please help me this is a test and it needs to be done in a few​
Amanda [17]

Answer:

I think A because astroids are stronger than a crash. Hope I helped! :)

Explanation:

7 0
3 years ago
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