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3 years ago

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Use the information from the graph to answer the

1 answer:

Mumz [18]3 years ago

4 0

Answer:

-2.5 m/s

Explanation:

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Like the moon's orbit, Earth's orbit is not a perfect circle, but an oval How do you think this might affect Earth's tides?

ExtremeBDS [4]

During the parts in the orbit where the moon is farthest away from the earth the tides will be low. Whereas during the parts where the moon is closer to the earth the tides will be higher.

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3 years ago

What does g mean in physics

o-na [289]

Answer:

Grams, I believe..! (Meter, liter, gram)

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3 years ago

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A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the tables edge in the

GREYUIT [131]

A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the edge of the table in the horizontal direction,

t= 0.45 seconds.
V=2.22m/s
VT=4.95 m/s

This is further explained below.

<h3>What is its speed when it hits the floor...?</h3>

Generally, the equation for motion is mathematically given as

S= ut + 0.5at²

Therefore

y = Voy t + 0.5gt^2

1 = 0.5x 98 x 6²

1=4.9t^2

$t=\sqrt{0.2041 }$

t= 0.45 seconds.

b) Horizontal motions are uniform.

V=Horizontal displacement/time

V=1/0.45

V=2.22m/s

C)

Vx: 2.22 m/s At bottom,

Vy² = Voy² + 2as

Vy² = 2x95x1

Vy² = 19.6

Total velocity

$VT=\sqrt{( 2.22 m/)^2+19.6}$

VT=4.95 m/s

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4 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

A tangential force of 1500 N exerted upon the upper surface of a cube of 20 cm edge. Calulate the shear modulus of the cube mata

I am Lyosha [343]

<u>Answer:</u>

<em>The shear modulus of the cube material is $7.5 \times 10^6 N/m^2$ . </em>

<u>Explanation:</u>

<em>Given that shearing force applied F = 1500 N </em>

<em>Displacement produced x = 0.1 cm=0.001 m </em>

<em>side of the cube =20 cm = 0.2 m </em>

Since the object is a cube the upper surface is a square and it is on this surface the shearing

force is applied

<em>area of the upper surface $A=a \times a=(20 \times 10^(^-^2^))^2=400 \times 10^(^-^4^) m$ </em>

<em>shear strain = tan⁡ θ = $\frac {x}{h} = \frac {0.001}{0.2} =0.005$ </em>

<em>shearing stress = $\frac {F}{A} = \frac{1500}{0.04} = 37500 N$ </em>

<em>modulus of rigidity η $= \frac{(shearing \ stress)}{(shearing \ strain)}$ </em>

<em> $= \frac{37500}{0.005}=7.5 \times 10^6 N/m^2$ </em>

8 0

3 years ago