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Delvig [45]
3 years ago
12

An object of mass 4 kg is initially at rest on a frictionless ice rink. It suddenly explodes into three pieces (I have no idea w

hy it happened!). One chunk of mass 1 kg slides across the ice with a velocity 1.2 ms^-1i and another chunk of mass 2kg slides across the ice with a velocity 0.8 m.s^-1j. Determine the velocity of the third chunk in magnitude angle form.
Physics
1 answer:
Rina8888 [55]3 years ago
6 0

We have two events that occurred on different axes, the most convenient is to perform the operations on the two axes, and then look for the resulting force.

Our values are,

m_1=1kg\\m_2 = 2kg\\v_1' = 1.2m/s\\v_1''= 0.8m/s

PAR A) For the X axis, we apply momentum conservation, which is given by,

Total momentum before = Total momentum After

We start from rest, so in X the initial speeds are 0,

m_1v_1+m_2v_2=m_1v_1'+m_2v_2'

0+0 = (1*1.2)+(1)v_2'

V_2' = -1.2m/s

Now we apply for the conservation of the moment, it is part of the rest, so,

m_1v_1+m_2v_2=m_1v_1''+m_2v_2''

0+0=2*0.8+1*v_2''

v_2'' = -1.6

To find the total speed, we simply apply pitagoras,

V = \sqrt{v_2'^2+v_2^2''}

V = \sqrt{1.2^2+1.6^2}

V = 2m/s

PART B) The address is given by,

tan\theta = \frac{V_2''}{V_2'}

\theta = tan^{-1} \frac{V_2''}{V_2'}

\theta = tan^{-1} \frac{1.6}{1.2}

\theta = 53.31\° (Below -x axis)

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When gravitational field lines get farther apart, the gravitational field _________.
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<span>b. becomes weaker            
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3 0
3 years ago
During an episode of turbulence in an airplane you feel 210 n heavier than usual.if your mass is 72 kg, what are the magnitude a
lana66690 [7]
According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.

Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.
8 0
3 years ago
Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temp
coldgirl [10]

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

3 0
3 years ago
how long would it take for a resultant upward force of 100N to increase the speed of 50kg object from 100m/s to 150m/s​
ValentinkaMS [17]

Answer:

25 sec

Explanation:

F = ma

100 = 50 a

a = 2 m/s^2

vf = vo + at

150 = 100 +  2 t

50 = 2 t

t = 25 sec

7 0
2 years ago
Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i
Naddik [55]

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

    G m M / r² = m a

The centripetal acceleration is given by

    a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

    d = 2 π r

Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

    G m M / r² = m a

    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

     T = 1.03 10⁸ s

Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

      r = ∛(0.0543087 10 36)

      r = 0.3787 10¹² m

      r = 3.787 10¹¹ m

7 0
3 years ago
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