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Alex
3 years ago
7

In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom

of his arc, he picks up his 55.0-kg costar in an inelastic collision. What maximum height do they reach after their upward swing?
Physics
1 answer:
Arada [10]3 years ago
3 0

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V  = Velocity of actor

h_i = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s

As the linear momentum is conserved

MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s

Applying conservation of energy again

\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m

The maximum height they reach is 1.57772 m

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Scilla [17]

Answer:

Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.

Explanation:

When an object is placed 25cm from Gbenga's eye, the glasses lens must produce an image 61cm away (Gbenga's eye near point).

An image 61cm from the eye will be (61cm - 1.6cm) from the glasses.

i.e. d_{i}=61cm-1.6cm=59.4cm

and d_{o} = 25cm - 1.6cm = 23.4cm

note d_{i} will be negative because the image is formed on the same side as the object.

finally, d_{i}=-59.4cm\\d_{o}=23.4cm

the formula for finding the focal length f is given as

f=\frac{d_{i}d_{o}  }{d_{o}+d_{i}  }

f=\frac{-59.4*23.4}{23.4-59.4} \\

f=\frac{-1389.96}{-36}

f=38.61cm

The focal length is positive which indicates converging lens

power p=\frac{1}{f}

but f\\ must be in metres

Therefore, f=38.61cm=0.3861m

p=\frac{1}{0.3861}

p=2.59 Diopter

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4 years ago
Which force controls the size of an atomic nucleus?
IrinaVladis [17]

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2 years ago
A) In the figure below, a cylinder is compressed by means of a wedge against an elastic constant spring = 12 /. If = 500 , deter
Radda [10]

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline.  Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left.  Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

6 0
3 years ago
What do radio waves carry
iren [92.7K]

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4 0
3 years ago
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IceJOKER [234]

Answer:

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Explanation:

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