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2 years ago

15

The equation below can be used to calculate a change in gravitational potential energy. What units must be used for h? Give the

full name, not the abbreviation.
e=m x g x h

1 answer:

Anni [7]2 years ago

8 0

Answer:

h = change in vertical position (height)

has units of distance.

Explanation:

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Seesaw unbalanced force explain

wolverine [178]

Like a seesaw, it shows that the forces aren’t equal because if it was the seesaw would stay put

3 0

3 years ago

How much of Earth's surface is dry, ice-free land? *

AleksandrR [38]

Answer:

Less than a quarter of Earth's surfaces are dry, ice free lands

8 0

2 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t

KatRina [158]

Answer:

T₂ = 123.9 N, θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

T₁ₓ - $T_{2x}$ = 0

T_{2x}= T₁ₓ

Y axis y

T_{1y} + T_{2y} - 200N = 0

T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

sin 60 = T_{1y} / T₁

cos 60 = t₁ₓ / T₁

T_{1y} = T₁ sin 60

T1x = T₁ cos 60

T_{1y}y = 100 sin 60 = 86.6 N

T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

T_{2y} = T₂ sin θ

T₂ₓ = T₂ cos θ

we substitute

T₂ sin θ= 200 - 86.6 = 113.4

T₂ cos θ = 50 (1)

to solve the system we divide the two equations

tan θ = 113.4 / 50

θ = tan⁻¹ 2,268

θ = 66.2º

we caption in equation 1

T₂ cos 66.2 = 50

T₂ = 50 / cos 66.2

T₂ = 123.9 N

8 0

3 years ago

In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor

VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

$I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596$ kg-m2

$I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2$

by perpendicular axis theorem

$I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2$

4 0

3 years ago