3 years ago

I need help with number 8

Physics

1 answer:

Mashutka [201]3 years ago

5 0

Answer:

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a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after

STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

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A truck moves 30 km West, and then 40 km North, and then travels in a straight path back to its starting

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Distance = (30+40+50) = 120 km

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As you lift an 88 N box straight upward you produce a power of 72 W. What is the speed of the box?

jeka57 [31]

F = force applied to lift the box = weight of the box = 88 N

P = power produced while lifting the box upward = 72 Watt

v = speed of the box = ?

Speed of the box is given as

v = $\frac{P}{F}$

inserting the values

v = $\frac{72}{88}$

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A magnesium oxide component must not fail when a tensile stress of 10.5 MPa is applied. Determine the maximum allowable surface

Aloiza [94]

Answer:

Maximum permitted surface crack length is 1.29 mm

Explanation:

As per the question:

Tensile stress, $\sigma = 10.5\ MPa = 10.5\times 10^{6}\ Pa$

Surface energy of magnesium oxide, SE = $1.0\ J/m^{2}$

Modulus of elasticity of the material, E = 225 GPa = $225\times 10^{9}\ Pa$

Now,

To calculate the maximum allowable surface crack length:

$L = \frac{2E\times SE}{\sigma^{2}\pi }$

$L = \frac{2\times 225\times 10^{9}\times 1.0}{10.5\times 10^{6}\times \pi } = 1.29\times 10^{- 3}\ m = 1.29\ mm$

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kompoz [17]

The Answer should be A: 0.0

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