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Verdich [7]
4 years ago
10

You are facing North. Turn 90 degrees left. Turn 180 degrees right. Reverse direction. Turn 45 degrees left. Reverse direction.

In which direction are you now facing?
Physics
1 answer:
ZanzabumX [31]4 years ago
3 0

You are facing North. Turn 90 degrees left. W.

Turn 180 degrees right. E.

Reverse direction. W.

Turn 45 degrees left. S.W.

Reverse direction. N.E.

In which direction are you now facing? North East

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I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s.
AfilCa [17]

1. 5.5 m/s

We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

v2 = 25 m/s is the final velocity of the bottle

Substituting and re-arranging the equation, we can find the final velocity of the ball:

v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}=\frac{(0.4 kg)(18m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s


2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

m_1 u_1 + m_2 u_2 = (m_1 +m_2) v

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

\frac{m_1 u_1}{m_1 +m_2}=\frac{(0.04 kg)(300 m/s)}{0.04 kg+0.5 kg}=22.2 m/s


3. 6560 N

The impulse exerted on the ball is equal to its change in momentum:

I=\Delta p (1)

The impulse can be rewritten as product between force and time of collision:

I=F \Delta t

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

\Delta p = m\Delta v = m(v_f -v_i)

So, eq.(1) becomes

F \Delta t = m(v_f -v_i)

where:

F = ? is the unknown force

\Delta t = 0.002 s is the duration of the impact

m = 0.16 kg is the mass of the ball

v_f = 44 m/s is the final velocity of the ball

v_i = -38 m/s is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

F=\frac{m (v_f -v_i)}{\Delta t}=\frac{(0.16 kg)(44 m/s-(-38 m/s))}{0.002 s}=6560 N

7 0
4 years ago
A fly is call inside a car that is moving in the positive direction with constant velocity. The fly flies with a constant veloci
balu736 [363]
A. The fly's velocity to the ground is less than the car's velocity to the ground.
4 0
4 years ago
The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a
Setler [38]

Answer:

 x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

         x_{cm} = 1 / M_{total}  ∑ x_{i}  m_{i}

Where M_{total} is the total masses of the system, x_{i} is the distance between the particles and m_{i} is the masses of each body

Let's apply this equation to our problem

        M = Me + m

        M = 5.98 10²⁴ + 7.36 10²²

        M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

       x_{cm} = 1 / 605.36 10²²   [Me 0 + 7.36 10²² 3.82 10⁸]

       x_{cm} = 4.644 10⁶ m

4 0
3 years ago
Which of the following provides the best evidence in support of the big bang
Svetllana [295]

C. The Uniformity of Cosmic Background Radiation

5 0
3 years ago
A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied
faltersainse [42]

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

3 0
3 years ago
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