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3 years ago

HELP ASAP PLEEEEES

Physics

1 answer:

matrenka [14]3 years ago

6 0

Explanation:

33. The 1.5kg owl is now soaring at 20m/s. What is the owl’s KE?

a. Step 1: Formula ½mv²

b. Step 2: Data m = 1.5 kg, v = 20 m/s

c. Step 3: Solve

KE = (1/2)(1.5)(20)² = 300 J

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At what height does a 1000-kg mass have potential energy of 1J relative to the ground?

Dvinal [7]

Answer:

0.0001 m

Explanation:

Given

M=1000 KG

PE=1 J

G=9.8 m/s

H=?

Formula

PE = MGH

H= PE/MG

H=1/1000x9.8

H= 1/9800

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3 years ago

A 66-foot-tall woman walks at 55 ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th

I am Lyosha [343]

Answer:

a. $\frac{dx}{dt}=20ft/s$

b. $\frac{d(x+L)}{dt}==25ft/s$

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

$tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}$

Solve to find the rate relation

$x=\frac{24}{6}*L$

$x=4*L$

Now the rate of the change rate

$\frac{dx}{dt}=4*\frac{dL}{dt}$

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Finally the rate of her shadow moving

$\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}$

$\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s$

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4 years ago