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MrRissso [65]
4 years ago
12

An Arrow (0.5 kg) travels with velocity 60 m/s to the right when it pierces an apple (1 kg) which is initially at rest. After th

e collision, the arrow and the apple are stuck together. Assume that no external forces are present and therefore the momentum for the system is conserved. What is the final velocity (in m/s) of apple and arrow after the collision
Physics
1 answer:
Rom4ik [11]4 years ago
5 0

Answer:

Velocity after collision will be 20 m/sec

Explanation:

We have given mass of arrow m_1=0.5kg

Mass of arrow v_1=60m/sec

Mass of apple m_2=1kg

Apple is at rest so velocity of apple v_2=0m/sec

According to conservation of momentum

Momentum before collision is equal to momentum after collision

m_1v_1+m_2v_2=(m_1+m_2)v

0.5\times 60+1\times 0=(0.5+1)v

1.5v=30

v = 20 m/sec

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Serjik [45]

Answer:

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Explanation:

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4 0
3 years ago
Hello, I need help answering and explaining the question in the attached image. It includes the figures plus the question and th
Flauer [41]

Answer:

d.

Explanation:

Since the dart's initial speed v at angle has both vertical and horizontal components v₀sinθ and v₀cosθ respectively, the vertical component of the speed continues to decrease until it hits the target. It's displacement ,s is gotten from

s = y - y₀ = (v₀sinθ)t - 1/2gt² where y₀ = 0 m

y - 0 = (v₀sinθ)t - 1/2gt²

y = (v₀sinθ)t - 1/2gt²

which is the parabolic equation for the displacement of the dart.

Note that the horizontal component of the dart's velocity does not change during its motion.

Since the target falls vertically, with initial velocity u = 0 (since it was stationary before the string cut), it's displacement ,s' is gotten from

s' = y - y₀' = ut - 1/2gt² where y₀' = initial height of target above the ground

= (0 m/s)t - 1/2gt²

= 0 - 1/2gt²

y - y₀' = - 1/2gt²

y = y₀' - 1/2gt²

which is the parabolic equation for the displacement of the target.

The equation for both the displacement of the dart and the target can only be gotten if we considered vertical motion. So, the displacement component of both the dart and target are both vertical.

So, the answer is d.

8 0
3 years ago
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat
N76 [4]

Answer:

p_k=\sqrt{p_x^2+p_y^2}}

Explanation:

Apply the momentum in each direction knowing that the impact is at the same time for the pieces so

p_x=m_1*v_1

p_x=200g*2.0m/s=0.4kgm/s

p_y=m_2*v_2

p_y=235g*1.5m/s=0.3525kgm/s

So the momentum in the other piece can be find knowing that

p_x^2+p_y^2=p_k^2

So:

p_k=\sqrt{p_x^2+p_y^2}}

p_k=\sqrt{0.4^2+0.3525^2}}=\sqrt{0.2842 kg^2*m^2/s^2}

p_k=0.5331kg*m/s

To find the velocity knowing the mass

p_k=m_k*v_k

v_k=\frac{p_k}{m_k}=\frac{0.5331 kg*m/s}{0.10kg}

v_k=5.331 m/s

4 0
3 years ago
The sun has sufficient hydrogen to continue fusing into helium for how much longer?
vichka [17]

Answer:

4500 million years

Explanation:

The Sun shines thanks to the thermonuclear conversion of hydrogen to helium inside. It is currently 4,500 million years old and has reservations for a similar period of time. When this fuel is exhausted in the central region, the heart of the Sun, constituted of helium and in an inert state, will contract and put more external fuel reserves within reach of the star, with which this mass of helium will grow over time . When that happens, the Sun will evolve into a giant star that will reach the orbit of Mars and, therefore, destroy the planet Earth.

As the helium heart mass increases so do the central density and temperature. When it reaches 100 million degrees, helium fuses thermonuclearly with itself and becomes a mixture of carbon and oxygen.

When the helium runs out in the center, the previous operation is reproduced approximately. The carbon / oxygen heart contracts and the helium and hydrogen of the surrounding layers are placed within the reach of thermonuclear combustion. The difference is that this double combustion is unstable and the density is so high that electrons can, alone, stabilize the heart of carbon and oxygen. The end result is that the outer layers, which originate a planetary nebula, are expelled, and the old thermonuclear reactor becomes visible, which becomes a white dwarf that slowly cools like the embers of a fire over billions of millions. of years.

8 0
3 years ago
Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal spee
e-lub [12.9K]

Answer:

   r₁ = 20.5 cm

Explanation:

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

        Em₀ = U_g + U_e + K = -G \frac{m_1m_2}{r} +k \frac{q_1q_2}{r} - 2 ( \frac{1}{2}  m v^2)

the two in the kinetic energy is because they are two particles

final point. When it is detained

        Em_f = U_g + U_e = -G \frac{m_1m_2}{r_1} + k \frac{q_1q_2}{r_1}

the energy is conserved

        Em₀ = em_f

the charges and masses of the two particles are equal

         -G \frac{m^2}{r} + k \frac{q^2}{r} + m v^2 = - G \frac{m^2}{r_1} + k \frac{q^2}{r_1}        

         

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

    -5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁  + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

      3.24 10⁻⁵ - 7.2 10⁻⁵ =  8.1 10⁻⁶ / r₁

      -3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

      r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

      r₁ = 2.045 10⁻¹ m

      r₁ = 20.5 cm

4 0
3 years ago
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