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aliina [53]
2 years ago
9

Which variable mentioned in Table 2-1 is kept constant? a. amount of time spent swimming b. type of swimming stroke c. number of

calories used d. the person swimming

Physics
1 answer:
Vladimir [108]2 years ago
7 0

Answer:

a. amount of time spent swimming

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Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
Read 2 more answers
Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area
cluponka [151]

Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r²           [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

4 0
2 years ago
Determine the value of the current in the solenoid so that the magnetic field at the center of the loop is zero tesla. Justify y
sleet_krkn [62]

Answer:

I will explain the concept of magnetic field and how it can be calculated.

Explanation:

The formula for magnetic field at the center of a loop is given as

B = μ_{o}I / 2R

where B is the magnetic field

R is the radius of the loop

I is the current

and μ_{o} is the magnetic permeability of free space which is a constant 4π × 10^{-7} newtons/ampere²

If the magnetic field at the center of the loop is 0, then μ_{o}I = 0

I = 0 which means there will be no current flow in the loop.

3 0
3 years ago
How long could you survive without electricity? What parts of your life would be affected by loss of electricity?
Law Incorporation [45]

Answer:

Depends on the person, if you are hooked to your phone which most people are these days, you might go a little crazy. But think cavemen or anybody long ago, they survived without electricity. People in Africa bathe in the lake or any water nearby, they also walk for miles just to get drinking water. So yes, if you really tried you could go without electricity for possibly your whole life. It sucks if you have homework though because you don't have Brainly :)

Explanation:

6 0
3 years ago
A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
valentinak56 [21]

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0
3 years ago
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