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Degger [83]
3 years ago
6

A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3

.00 m and D = 13.9 m, find H. Submit Answer Tries 0/12 Find her total kinetic energy as she reaches the ground.
Physics
1 answer:
marissa [1.9K]3 years ago
4 0

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

mg(H-h)=\frac{1}{2}mv^2

g(H-h)=\frac{1}{2}v^2

\Rightarrow H=\frac{v^2}{2g}+h

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

=v\frac{2g}{h}

\Rightarrow v^2=\frac{gD^2}{2h}

thus, height H in terms of D  is given by

H=\frac{D^2}{2h}+h

H=\frac{13.9^2}{2\times3}+3

H=35.20 m

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A small metal sphere weighs 0.34 N in air and has a volume of 13 cm3 . What is the acceleration of the sphere as it falls throug
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To solve this problem we will apply the concepts related to the balance of forces. Said balance will be given between buoyancy force and weight, both described as derived from Newton's second law, are given as

Buoyancy force

F_B = V\rho g

Here,

V = Volume

\rho=Density of air

g = Acceleration due to gravity

Weight

F_W = mg

m = mass

g = Gravity

Our values are given as,

\text{Weight of the sphere} = W = 0.34 N

\text{Volume} = V = 13 cm^3 = 13*10^{-6}m^3

\text{density of air} =\rho =1.29kg/m^3

\text{gravity}= g = 9.8 m/s^2

Then,

F = V\rho g

Replacing,

F = (13*10^{-6}m^3 )(1.29Kg/m^3)( 9.8 m/s^2) = 1.6434* 10^{-4} N

Now net force is ,

F_{net} = mg - F

Mass of the sphere is

m = \frac{W}{g} = \frac{0.34N}{9.8m/s^2} = 0.03469 kg

Now acceleration of the sphere is

a = \frac{F_{net}}{m}

a = \frac{( 0.34 N)- (1.6434* 10^{-4} N)}{0.03469 kg}

a = 9.822m/s^2

Therefore the acceleration of the sphere as it falls through water is 9.822m/s^2

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When placed in direct sunlight, which an object will absorb the most visible light energy
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What are two reasons why density is a useful physical property for identifying substances?
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The correct choice is ~ A

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P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

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3 years ago
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