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aleksklad [387]

2 years ago

10

Dr. Evil stands on a 150 m high cliff and drops the kittens straight down. Fill out the following table that shows how the veloc

ity and acceleration of the kittens change with time during their fall.

1 answer:

riadik2000 [5.3K]2 years ago

3 0

I would say d if I remember right I think that’s what I put and got right sorry if it’s wrong

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Sound waves cannot carry energy through. A water B air C a mirror D a vacuum

Ber [7]

I looked up the question and got D- a vacuum

3 0

2 years ago

What type of charge is acquired when a rubber is rubbed with a fur

Rus_ich [418]

Answer:

The triboelectic charging process (a.k.a., charging by friction) results in a transfer of electrons between the two objects that are rubbed together. Rubber has a much greater attraction for electrons than animal fur.

Explanation:

HOPE IT HELP^_^

THANKS^_^

4 0

1 year ago

A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change

IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball $v_1=25m/sec$

And velocity after rebound $v_2=-19m/sec$

We have to find the change in momentum

So change in momentum is equal to $=m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec$ ( here negative sign shows only direction )

So option (A) will be correct answer

5 0

2 years ago

Why might a scientist repeat an experiment if she did not make a mistake in the first one?

iren2701 [21]

Answer:

B

Explanation:

4 0

2 years ago

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a

finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$ after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y$

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

= 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

5 0

2 years ago