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zvonat [6]
3 years ago
5

The electric field intensity at a particular location surrounding a Van de Graaff generator is 4.5x103 N/C. Determine the magnit

ude of the force that this field would exert upon ...
Physics
2 answers:
puteri [66]3 years ago
8 0

Answer:

Explanation:

Electric field, E = 4.5 x 10^3 N/C

charge on electron, e = 1.6 x 10^-19 C

The relation between the force and the electric field is given by

F = q E

F = 1.6 x 10^-19 x 4.5 x 10^3

F = 7.2 x 10^-16 N

Oksanka [162]3 years ago
3 0

Answer:

The magnitude of the force is 7.2\times10^{-16}\ N

Explanation:

Given that,

Electric field intensity E= 4.5\times10^3

Suppose, The electron when positioned at this location

We need to calculate the magnitude of the force

Using formula of electric field

E = \dfrac{F}{q}

F = Eq

Where, E = electric field

q = charge of electron

Put the value into the formula

F=4.5\times10^3\times1.6\times10^{-19}

F=7.2\times10^{-16}\ N

Hence, The magnitude of the force is 7.2\times10^{-16}\ N

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Answer:

Refracted

Explanation:

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3 years ago
A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd
PilotLPTM [1.2K]

The minimum coefficient of static friction  between the pavement and the tires is 0.69.

The given parameters;

  • <em>radius of the curve, r = 90 m</em>
  • <em>angle of inclination, θ = 10.8⁰</em>
  • <em>speed of the car, v = 75 km/h = 20.83 m/s</em>
  • <em>mass of the car, m = 1100 kg</em>

The normal force on the car is calculated as follows;

F_n = mgcos(\theta)

The frictional force between the car and the road is calculated as;

F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)

The net force on the car is calculated as follows;

mgsin(\theta) +  \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \  + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) +   \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) +  \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\

\mu_s = 0.69

Thus, the minimum coefficient of static friction  between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

8 0
3 years ago
If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

                         θ_1  = 90° -  ∅

                         θ_1  = 90° -  20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

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3 years ago
Which statement correctly lists the type of electromagnetic waves from highest to lowest energy? gamma rays, ultraviolet waves,
ahrayia [7]

Answer:

Thank you for posting your question on Brainly.

Explanation:

The waves from highest to lowest are :

* Gamma

* X- Rays

* UV

* Visible

* Infrared

* Microwaves

* Radio-waves

I really hope this was helpful. Have a great day : )

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Answer:

-you have to make yourse;f as light as possible so toss your bag, jacket, and shoes.

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-Keep your arms up and out of the quicksand.

-Try to reach for a branch or person's hand to pull yourself out.

-Take deep breaths.

-Move slowly and deliberately.

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