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Define the terms resultant and equilibriant of

hjlf

Answer:

1) Resultant and Equilibrant explained below

2) Resultant = 24.18 N in a direction of 11.93° NE

3)Equilibrant = 24.18 N in a direction of 11.93° SW

Explanation:

1) Resultant force: If two or more forces are acting at a certain point, their resultant force would be defined as the force that when applied at that same point, it will have the exact same effect like the two independent forces acting together.

Equilibrant of 2 forces: This is defined as the force exactly equal and has an opposite direction to the resultant

2)a.) we are told that 10 N acts in the direction N 30 East and 15 N acts in the easterly direction, if both forces

act at a point. Thus, coordinates are;

(10cos30,10sin30) and (15,0)

Therefore, resultant coordinate is;

R→ = (10cos30, 10sin30) + (15, 0)

R→ = (15 + 10cos30, 0 + 10 sin 30)

This gives;

R→ = (23.66, 5)

Thus the resultant is;

R = √(23.66² + 5²)

R = √584.7956

R = 24.18 N

b) direction is gotten from;

tan θ = 5/23.66

θ = tan^(-1) 5/23.66

θ = 11.93° NE

c) Equilibrant is equal force but opposite direction. Thus, Equilibrant = 24.18 N in a direction of 11.93° SW

6 0

3 years ago

The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but

Usimov [2.4K]

Answer:

a) 4.33 pC b) 5.44*10² N/C

Explanation:

a) The vertical deflecting plates of an oscilloscope form a parallel-plate capacitor.

The value of the capacitance, for a parallel-plate capacitor with air dielectric, can be found to be as follows, applying Gauss' law to the surface of one of the plates, and assuming a uniform surface charge density:

C = ε₀*A / d

where ε₀ = 8.85*10⁻¹² F/m, A = (0.03m)², and d = 0.046 m (we assume that the informed value of 4.6 m is a typo, as no oscilloscope exists with this separation between plates).

Replacing by these values, we find the equivalent capacitance of the plates, as follows:

$C = \frac{8.85e-12F/m*(0.03m)^{2} }{0.046m} =1.73e-13 F = 0.173 pF$

By definition, the capacitance of any capacitor can be expressed as follows:

$C =\frac{Q}{V}$

where Q= charge on any of the plates, and V= potential difference between them.

As we know C and V, we can find Q as follows:

Q = C*V = 0.173*10⁻¹² F * 25.0 V = 4.33*10⁻¹² C = 4.33 pC

b) We can find the electric field in several ways, but one very easy is applying Gauss' Law to a pillbox with a face outside one of the plates (paralllel to it) and the other inside the surface.

The total electric flux through the surface must be equal to the enclosed charge, divided by ε₀.

If we look to the flux crossin any face, we find that the only one that has a non-zero flux, is the one outside the surface.

As the electric crossing the boundary must be normal to the surface (in electrostatic conditions, no tangential field can exist on the surface) , and we assume that the surface charge density that creates it is constant across the surface, we can write the Gauss ' Law as follows:

E*A = Q / ε₀

where A = area of the plate = (.03m)² = 9*10⁻⁴ m², Q= charge on one of the plates = 4.33*10⁻¹² C (as we found in a)) and ε₀ = 8.85*10⁻¹² N/C.

Replacing by these values, and solving for E, we have:

$E = \frac{4.33e-12C}{(0.03m)^{2} 8.85e-12F/m} =5.44e2 N/C$

⇒ E = 5.44*10² N/C

5 0

3 years ago

Students are given the following information about a 2.0 kg, motorized toy boat on water. what net force is exerted on the boat

Natasha_Volkova [10]

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

The motorized toy boat moves at 8 m/s (u) at 4 s and at 8 m/s (v) at 8 s. We can calculate the acceleration (a) in that period using the following kinematic expression.

$a = \frac{v-u}{\Delta t} = \frac{8m/s-8m/s}{8s-4s} = 0 m/s^{2}$

The object with a mass (m) of 2.0 kg experiences an acceleration of 0 m/s². We can calculate the net force (F) in that period using Newton's second law of motion.

$F = m \times a = 2.0 kg \times 0 m/s^{2} = 0 N$