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deff fn [24]
3 years ago
7

n explosion breaks an object initially at rest into two pieces, one of which has 2.0 times the mass of the other. If 8400 J of k

inetic energy were released in the explosion, how much kinetic energy did the heavier piece acquire
Physics
2 answers:
hichkok12 [17]3 years ago
8 0

Answer:

The heavier piece acquired 2800J of energy.

Explanation:

Let the velocity of the first piece be v1 and that of the second piece be v1 and also let the masses be m1 and m2 respectively.

m2 = 2m1

From the principles of conservation of momentum

m1v1 = m2v2

m1v1 = 2m1v2

Dividing through by m1

That is v1 = 2v2

v2 = 1/2v1

Total kinetic energy = 8400J

1/2m1v1² + 1/2m2v2² = 8400

1/2m1v1² + 1/2(2m1) × (1/2v1)² = 8400

1/2m1v1² + 1/4m1v1² = 8400

Let 1/2m1v1 = K1

K1 + 1/2K1 = 8400

3/2K1 = 8400

K1 = 8400× 2/3 = 5600J

K2 = 8400 – 5600 = 2800J

ANTONII [103]3 years ago
3 0

Answer:

The heavier piece acquired 2800 J  kinetic energy

Explanation:

From the principle of conservation of linear momentum:

0 = M₁v₁ - M₂v₂

M₁v₁ = M₂v₂

let the second piece be the heavier mass, then

M₁v₁ = (2M₁)v₂

v₁  = 2v₂ and v₂ = ¹/₂ v₁

From the principle of conservation of kinetic energy:

¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J

¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400

¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400

K.E₁ + ¹/₂K.E₁ = 8400

Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁

1.5 K.E₁ = 8400

K.E₁ = 8400/1.5

K.E₁ = 5600 J

K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J

Therefore, the heavier piece acquired 2800 J  kinetic energy

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Alex73 [517]

Answer:

1752.14 tonnes per year.

Explanation:

To solve this exercise it is necessary to apply the concepts related to power consumption and power production.

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\dot{P} = m \dot{E}\eta

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The problem gives us the aforementioned values under a production efficiency of 45%, that is,

\dot{P} = \bar{P}

m \dot{E}\eta = \bar{P}

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Solving for m,

m = \frac{ 2*10^{12}}{(8*10^13)(0.45)}

m = 0.0556 \frac{kg}{s}

We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

m = 0.556\frac{kg}{s}*(\frac{3.1536*10^7s}{1year})(\frac{1ton}{1000kg})

m = 1752.14tonnes per year.

8 0
3 years ago
A(n) 77.5 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 37.3 m away from the shuttle
NeX [460]

Answer:

Explanation:

The speed of the astronaut can be found with the help of law of conservation of momentum .

mv = MV , M is mass of astronaut , m is mass of object thrown , v is velocity of object thrown and V is velocity of  astronaut.

Putting the values

77.5 x V = .94 x 12

V = .14554 m /s

This will be the uniform velocity of astronaut.

Distance to be covered = 37.3 m

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= 37.3 / .14554

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4 0
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mario62 [17]

A transform fault boundary is the correct answer.

5 0
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Find the mass of the car.
SpyIntel [72]

\huge\boxed{770.8\overline{3}\ \text{kg}}

We know that force equals mass times acceleration, so substitute in the known values and solve.

\begin{aligned}F&=ma\\1850&=m*2.4\\1850\div2.4&=m*2.4\div2.4\\\boxed{770.8\overline{3}}&=m\end{aligned}

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3 years ago
A 0.3 g mosquito is flying toward a girl with a speed of 4.5 mph. Just before landing on the girl, the fly is swatted straight b
luda_lava [24]

Answer:

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Therefore the force that was exerted is equal to 1.1x10^-2

5 0
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