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Art [367]
2 years ago
7

A 480 g peregrine falcon reaches a speed of 69 m/s in a vertical dive called a stoop. If we assume that the falcon speeds up und

er the influence of gravity only, what is the minimum height of the dive needed to achieve this speed?
Physics
1 answer:
Ivanshal [37]2 years ago
3 0

Answer:

elevation = 243[m]

Explanation:

This problem can be solved using the principle of energy conservation, where potential energy becomes kinetic energy. At the point where the speed is equal to 69 [m / s] will be taken as the reference point for potential energy, at this point all potential energy will have been transformed into kinetic energy.

Ek= 0.5*m*v^{2} \\where:\\m = mass = 480[g] = 0.480[kg]\\v = velocity = 69 [m/s]\\Ek=0.5*.480*(69)^{2} \\Ek=1142.64[J]\\

So this kinetic energy will be equal to the potential energy at the beginning.

Ek=Ep\\Ep=m*g*h\\1142.64=0.480*9.81*h\\h=242.66[m]

The falcon was originally at 243 [m] above the point (reference point) where he reached the speed given.

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Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an index of ref
blsea [12.9K]

Answer:

the new diameter of the third ring = 0.607 mm

Explanation:

Consider the radius of \\m ^{th}\\ bright ring when air is in between the lens and the plate ;

r = \sqrt {\frac{(2m+1)\lambda R}{2}}

Using the expression: r (n)= \sqrt {\frac{(2m+1)\lambda R}{2n}} for the radius of the  \\m ^{th}\\ bright ring since the liquid (water ) of the refractive index 'n' is now in between the lens and the plate;

where;

\\m ^{th}\\ = number of fringe

λ = wavelength

R = radius

n = refractive index of water

Now ;

r (n)=\frac {r}{n}}

the radius r of the third bright ring when the air is in between lens and plate = r= \frac{0.700 \ mm}{2} \\\\

r= 0.35 \ mm

The new radius of the third bright fringe is r(3) = \frac{0.35}{\sqrt 1.33}

r(3) = 0.3035 \ mm

Calculating the new diameter ; we have:

d(3) = 2(r(3))

d(3) = 2(0.3035)

d = 0.607 mm

Thus, the new diameter of the third ring = 0.607 mm

7 0
2 years ago
A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".
Alex787 [66]

Answer:

Part a)

U = 13 J

Part b)

v = 2.28 m/s

Part c)

v = 177.66 m/s

Part d)

W = 1012.7 J

Part e)

v = 2.1 m/s

Part f)

E = 1037.2 J

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

\theta = 38^o

so the maximum height reached by the pendulum is given as

h = L(1 - cos\theta)

so we will have

h = L(1 - cos38)

h = 1.25(1 - cos38)

h = 0.265 m

now gravitational potential energy of the pendulum is given as

U = mgh

U = 5(9.81)(0.265)

U = 13 J

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

v = \sqrt{2(9.81)(0.265)}

v = 2.28 m/s

Part c)

now by momentum conservation we can say

mv = (M + m) v_f

0.065 v = (5 + 0.065)2.28

v = 177.66 m/s

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2

W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2

W = 1012.7 J

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

0 = mv_1 + Mv_2

0 = 0.065(177.6) + 5.50 v

v = 2.1 m/s

Part f)

Total energy released in the process of shooting of gun

E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2

E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)

E = 1037.2 J

6 0
2 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
1 year ago
A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t
klasskru [66]

Answer:

a.a. \ \frac{7}{25}

b.\ \ \ P(D_1D_2)=\frac{6}{25}

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let A_i=i^{th} photo is accepted and the probability D_i=i^{th} is defected.

Therefore:

P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0
2 years ago
A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.
myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0
2 years ago
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