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Vlad1618 [11]
4 years ago
14

determine whether the following equations are dimensionally correct if not how can you make them dimensionally correct 1 /2 mv2

=(mgh) 2​
Physics
1 answer:
serg [7]4 years ago
6 0

Answer:

1 /2 mv^{2} =mgh

Explanation:

The correct equation follows the law of conservation of energy where kinetic energy is all transformed to potential energy, since we know that kinetic energy is expressed as

1 /2 mv^{2} while potential energy is mgh where m is the mass of the object, v is the velocity or speed of the object, g is acceleration due to gravity and h is the vertical height. Therefore, relating the two equations we should have 1 /2 mv^{2} =mgh

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A triangle ∆P QR has vertices P(3, 2, −4), Q(1, 0, −4), R(2, 1, 1). Use the distance formula to decide which one of the followin
777dan777 [17]

Answer:

a. FALSE

b.TRUE

C. FALSE

Explanation:

The formula fot the distance between two points is given as

d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2} +(z_{2}-z_{1})^{2}}\\

hence we determine the distances between all the points

a.P(3,2,-4), Q(1,0,-4), R(2,1,1)

PQ=\sqrt{(1-3)^{2} +(0-2)^{2} +(-4-(-4))^{2}}\\PQ=\sqrt{4+4+0}\\ PQ=\sqrt{8}

For point PR

we have

PR=\sqrt{(2-3)^{2} +(1-2)^{2} +(1-(-4))^{2}}\\PR=\sqrt{1+1+9}\\ PR=\sqrt{11}\\

|PQ|\neq |PR|

B. For point RP

RP=\sqrt{(3-2)^{2} +(2-1)^{2} +(-4-1)^{2}}\\RP=\sqrt{1+1+25}\\ RP=\sqrt{27}

for point RQ  we have

RQ=\sqrt{(1-2)^{2} +(0-1)^{2} +(-4-1)^{2}}\\RQ=\sqrt{1+1+25}\\ RQ=\sqrt{27}

|RP|=|R Q|

C.

QP=\sqrt{(3-1)^{2} +(2-0)^{2} +(-4+4)^{2}}\\QP=\sqrt{4+4+0}\\ QP=\sqrt{8}

For point Q R

QR=\sqrt{(2-1)^{2} +(1-0)^{2} +(1-(-4))^{2}}\\QR=\sqrt{1+1+9}\\ QR=\sqrt{11}\\

QP\neq QR

6 0
3 years ago
To pull a box up a rough slope,the force required will be least when it is applied :
kaheart [24]

Answer:

The correct option is;

d) Parallel to the plane

Explanation:

The forces acting on the box of mass, m are;

Weight of the box acting an angle, θ, equal to the inclination of the plane to the perpendicular of the plane

Weight of the box acting along the plane = m×g×sin(θ)

The force of friction along the plane = μ×m×g×cos(θ)

The total force acting downward along the plane F_{down}, = m×g×sin(θ) + μ×m×g×cos(θ)

The Force needed to pull the box up along the plane F = The total force acting downward along the plane

F = m×g×sin(θ) + μ×m×g×cos(θ) = m×g×(sin(θ) + μ×cos(θ))

When the force, Fₐ is applied vertically, the force acting along the plane = Fₐ×cos(θ)

When the force is applied perpendicular, the force acting along the plane = Fₐ×sin(θ)

When the force is applied horizontally, the force acting along the plane = Fₐ×cos(θ)

When the force is applied parallel to the plane, the force acting along the plane = Fₐ

Therefore, since Fₐ > Fₐ×cos(θ) and Fₐ > Fₐ×sin(θ), for acute angles, we have that the least force is required when the force is acting parallel to the plane.

6 0
3 years ago
Calculate the electric field on the surface of a charged sphere with a radius of 5cm and charge 10μC. Thanks
victus00 [196]

As you approach the surface of the sphere very closely, the electric field should resemble more and more the electric field from an infinite plane of charge.

If you check Gauss's law (recalling that the field in the conductor is zero) you will see that if the surface charge density is σ=Q/4πR2, then indeed the field at the surface is σ/ϵ0 as in the infinite charge of plane case.

Such a field is constant, the field lines are parallel and non-diverging, and the infinities associated with the field due to point charge do not arise.

Explanation:

4 0
2 years ago
Does dispersive power of the material of a prism depends upon the shape, size and angle of prism? Explain.
schepotkina [342]

The dispersive power of the material of the prism is independent of the shape, angle of prism or size of the prism. It depends only on the refractive index {(\mu)} of the material of the prism.

5 0
2 years ago
CONVERSIONS
Sauron [17]

Im srry i dont know the answer hehe

4 0
3 years ago
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