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storchak [24]
3 years ago
12

a solid passes from point A to B by VA=2m/s. AB=1m;m=2kg;F=2N and f=0.5N.Calculate the speed at point B​

Physics
1 answer:
slava [35]3 years ago
4 0

Answer:

a and E

Explanation:

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When you hear an ambulance siren, it alternates between high and low tones, depending on the frequency of sound waves. this is c
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<span>Frequency of a sound wave is called the pitch. Higher frequencies have a higher pitch and lower frequencies have the opposite. When an ambulance travels by a listener, the frequencies are oscillating rapidly and causing the shrill, loud sounds that emanate from the sirens.</span>
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The process of _____ is involved in the operation of all heat engines.
faltersainse [42]

The process of heat exchange is involved in the operation of all heat engines.

<u>Explanation:</u>

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3 years ago
If an object moves 40 m north, 40 m west, 40 m south, and 40 m east, what's the total displacement?
scoundrel [369]
80,000 Is the answer I believe
6 0
3 years ago
Read 2 more answers
A 3-kg object is attached to a spring and moving in simple harmonic motion. Its angular frequency is 20 rads/sec. When the mass-
Trava [24]

Answer:19.5 J

Explanation:

Given

mass of block=3 kg

angular frequency=20 rad/sec

spring constant k=\omega _n^2m=1200 N/m

we know total energy remain conserved

E_T at x=0.1 m

E_T=E_P+E_K

Where E_K=kinetic energy

E_P=potential Energy

E_P=\frac{1}{2}kx^2

E_P=600\times 0.01=6 J

E_K=\frac{1}{2}mv^2

E_K=\frac{1}{2}\times 3\times 3^2=13.5 J

E_T=13.5+6=19.5 J

When mass reaches amplitude its velocity becomes zero

there is only potential energy which is equal to Total energy

E_T=E_P=19.5 J

7 0
3 years ago
A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past t
marta [7]

Answer:

Explanation:

Given that

The window height is 2m

And the window is 7.5m from the ground

Then the total height of the window from the ground is 7.5+2=9.5m

It takes the ball 0.32sec travelled pass the window.

When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')

Now using the equation of free fall during this window travels

S=ut-½gt² against motion.

S=2, g=9.81, t=0.32sec

Then,

S=u't-½gt²

2=u'×0.32-½×9.81×0.32²

2=0.32u'-0.5023

2+0.5032=0.32u'

Then, 0.32u'=2.5032

u'=2.5032/0.32

u'=7.82m/s

This is the initial velocity as the ball got the the window

Now, let analyse from the window bottom to the ground which is a distance of 7.5m

Using the equation of free fall again

v²=u²-2gH

In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,

While u is the original initial velocity from the throw of the ball

Then,

u'²=u²-2gH

7.82²=u²-2×9.81×7.5

61.146=u²-147.15

61.146+147.15=u²

Then, u²=208.296

So, u=√208.296

u=14.43m/s

The initial velocity of the ball form the throw is 14.43m/s

6 0
3 years ago
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