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Strike441 [17]
2 years ago
14

An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of

the other what is the ratio of their masses
Physics
1 answer:
satela [25.4K]2 years ago
8 0
<span>First, we use the kinetic energy equation to create a formula: Ka = 2Kb 1/2(ma*Va^2) = 2(1/2(mb*Vb^2)) The 1/2 of the right gets cancelled by the 2 left of the bracket so: 1/2(ma*Va^2) = mb*Vb^2 (1) By the definiton of momentum we can say: ma*Va = mb*Vb And with some algebra: Vb = (ma*Va)/mb (2) Substituting (2) into (1), we have: 1/2(ma*Va^2) = mb*((ma*Va)/mb)^2 Then: 1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2 We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1: 1/2(ma) = (ma^2)/mb Cancel the ma of the left, leaving the right one with exponent 1: 1/2 = ma/mb And finally we have that: mb/2 = ma mb = 2ma</span>
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A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng
frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

Q_{1} =\frac{W}{n}

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

Q_{1} =\frac{750}{0.35} =2142.86kJ

Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ

c) The efficiency of Carnot engine is:

n=1-\frac{300K}{550K} =0.455

The heat transfer to the heat engine is:

Q_{1c} =\frac{750}{0.455} =1648.35kJ

The heat transfer from the heat engine is:

Q_{2c} =1648.35-750=898.35kJ

4 0
2 years ago
A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational
a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: F = 49.06N

Pebble: F = 29.44N

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: a =9.8m/s^{2}

Pebble:  a =9.8m/s^{2}

Explanation:

The universal law of gravitation is defined as:

F = G\frac{m1m2}{r^{2}}  (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em>m = 5.0 Kg<em>:</em>

m1 will be equal to the mass of the Earth m1 = 5.972×10^{24} Kg and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth r = 6371000m.

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}

F = 49.06N

Newton's second law can be used to know the acceleration.

F = ma

a =\frac{F}{m} (2)

a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}

a =9.8m/s^{2}

<em>Case for the pebble </em>m = 3.0 Kg<em>:</em>

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}

F = 29.44N

a =\frac{F}{m}

a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}

a =9.8m/s^{2}

3 0
2 years ago
Read 2 more answers
Odległość między kolejnymi grzbietami fal na morzu wynosi 20 m. Łódź opada z grzbietu fali, unosi się i osiąga ponownie najwyższ
Veronika [31]

Answer:

Explanation:

The distance between successive wave crests at sea is 20 m. The boat descends from the crest of the wave, rises and reaches the highest position again within 5 s. Calculate the wave propagation speed.

Given that,

The distance between two successive crest is 20m

Wavelength is the distance between two successive crest or trough

Then, it's wavelength is λ = 20m

The time to reached the maximum height is 5seconds, then it will take (5×4) to complete one period

Then,

Period T = 20seconds

From wave equation

v = fλ

Where

v is speed

f is frequency and

λ is wavelength

The frequency is related to the period

f =  1 / T

Then,

v = λ / T

So, v = 20 / 20

v = 1 m/s

The speed of propagation of the wave is 1m/s

To Polish

Jeśli się uwzględni,

Odległość między dwoma kolejnymi grzebieniami wynosi 20 m

Długość fali to odległość między dwoma kolejnymi grzebieniami lub dolinami

Zatem jego długość fali wynosi λ = 20 m

Czas do osiągnięcia maksymalnej wysokości wynosi 5 sekund, a następnie ukończenie jednego okresu zajmie (5 × 4)

Następnie,

Okres T = 20 sekund

Z równania falowego

v = fλ

Gdzie

v to prędkość

f oznacza częstotliwość, a

λ jest długością fali

Częstotliwość jest związana z okresem

f = 1 / T

Następnie,

v = λ / T

Zatem v = 20/20

v = 1 m / s

Prędkość propagacji fali wynosi 1m/s

6 0
2 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
2 years ago
The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu
larisa [96]

Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

5 0
2 years ago
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