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2 years ago

7

A box is at rest at the top of a frictionless inclined plane. As the box slides down the ramp where along the ramp would the box

have half of its maximum kinetic energy?

1 answer:

Yuliya22 [10]2 years ago

3 0

Answer:

I would say near the end of the slide.

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Question 5 (Multiple Choice Worth 3 points) (02.07 MC) Rachel needs to eat fewer carbohydrates to improve her health. Which of t

patriot [66]

Milk, apples, and beans don't have much carbohydrate. So if you
cut down on those, you don't really cut down much on carbohydrates.

If Rachel needs to reduce her intake of carbohydrates, she should
cut down on bread. (Also cake, sugar, corn, pasta, and potatoes.)

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3 years ago

What is the wavelength of a 10 Hz wave that travels with a speed of 5 m/s?

garik1379 [7]

B

V= f x lambda
V= 5m/s
F = 10hz
Lambda = ?
5 = 10 x lamba
5 /10 = lambda
Wavelength =0.5

8 0

2 years ago

Organisms that live on or burrow into the ocean floor are

creativ13 [48]

Benthos

Option b is the answer

8 0

3 years ago

Read 2 more answers

A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco

Sunny_sXe [5.5K]

Answer:

(a)

$x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km$

(b) $r=41.32km$

$\alpha =24.23^o$

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies $45^o$ south of east. The y-component will be negative and the x-component will be positive.

$x_1=25cos45^o=17.68km$

$y_1=-25sin45^o=-17.68km$

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

$x_2=40cos60^o=20km$

$y_2=40sin60^o=34.64km$

Therefore, total displacements:

$x=x_1+x_2=(17.68+20)km=37.68km$

$y=y_1+y_2=(-17.68+34.64)km=16.96km$

Magnitude of displacements,

$r=\sqrt{x^2+y^2}=41.32km$

Direction,

$\alpha =tan^{-1}(\frac{y}{x})=24.23^o$

6 0

4 years ago

the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t

givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

$a_{t} = v = 0.8 m/s^{2}$

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

= 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

= 0.8

at x= 5 Km

dy/dx = 0.8(5)

= 4

$p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }$

now insert the values,

$p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km$

→ Now the normal component of acceleration is given by

$a_{n} = \frac{v^{2} }{p}$

= (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

$a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}$

$a = [(0.8)^{2} + (0.457)^{2}]^{0.5}$

a = 0.921 m/s²

4 0

3 years ago