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Lapatulllka [165]
3 years ago
7

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ????????,2=17.9vi,2=17.9

m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle theta=61.9θ=61.9 ° below the horizontal and a speed of ????????,1=35.9vi,1=35.9 m/s. What is the speed of the eagle immediately after it catches its prey?
Physics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

V=27.24 m/s

Explanation:

We need to apply the linear momentum conservation theorem:

m_1*v_{o1}+m_2*v_{o2}=m_t*v_{f}\\

The velocity of the eagle its defined by its two components:

v_x=V*cos(\theta)\\v_y=V*sin(\theta)\\v_x=35.9*cos(61.9^o)=16.9m/s\\v_y=35.9*sin(61.9^o)=31.7m/s

2*(16.9m/s(i)+31.66m/s(j))+17.9m/s(i)=3*v_f\\v_f=17.23m/s(i)+21.10m/s(j)

because speed is a scalar value:

V=\sqrt{(21.10m/s)^2+(17.23)^2}\\V=27.24m/s

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Explanation:

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A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnit
nordsb [41]

Answer:

D

Explanation:

Ball A is a non positively charged non metal while ball B is metal ball.

Given: The ball B positive charge of small magnitude

To prove: Balls will attract each other

IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

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8 0
3 years ago
Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

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4 0
2 years ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
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