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xz_007 [3.2K]
3 years ago
7

A. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.

Physics
1 answer:
never [62]3 years ago
3 0

Answer:

a) F₁ = F₂,  F₃ = F₄,  b)  the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

      F₁ > F₃

therefore the correct answer is 3

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A 1,300 kg wrecking ball hits the building at 1.07 m/s2.
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3 years ago
An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi
stiv31 [10]

Answer:

Magnetic field, B=2\times 10^{-4}\ T

Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

It enters  a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, E=10^4\ V/m

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, F_m=qvB\ sin\theta.......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

qvB\ sin\theta=qE

B=\dfrac{E}{v}

B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}

B = 0.0002 T

or

B=2\times 10^{-4}\ T

Hence, this is the required solution.

3 0
3 years ago
What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

7 0
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