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elena-s [515]
4 years ago
10

Please help quickly!!! 20 points!!

Physics
2 answers:
kifflom [539]4 years ago
4 0

i believe the second not had a higher frequency than the first note

11111nata11111 [884]4 years ago
4 0

The answer is D/4 The second note vibrated slower than the first note.

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A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler
Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

W=K_f - K_i

Since the cart was initially at rest, K_i = 0, so

W=K_f = \frac{1}{2}mv^2 (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, F_g = 99.5 N:

m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg

So solving eq.(1) for v, we find the final speed of the cart:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s

2) 2.51\cdot 10^7 J

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

F=4.28 \cdot 10^5 N

d=586 m

So the work done is

W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J

3)  2.51\cdot 10^7 J

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

W=\Delta K = K_f - K_i

where

W is the work done

\Delta K is the change in kinetic energy

Therefore, the change in kinetic energy is

\Delta K = W = 2.51\cdot 10^7 J

4) 37.2 m/s

According to the work-energy theorem,

W=\Delta K = K_f - K_i

where

K_f is the final kinetic energy of the train

K_i = 0 is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

W=K_f = \frac{1}{2}mv^2

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s

7 0
4 years ago
what would someones mass be if their weight was 158.4? This is for science homework, they gave me a random weight # but no mass.
bekas [8.4K]

ON EARTH, every kilogram of mass weighs about 2.205 pounds.

So if you know the weight (in pounds), and you know that it was
weighed on Earth, you can simply divide the pounds by 2.205 ,
and you'll get the mass (in kilograms).

I can see that on this science homework, they used 2.2 for the weight
of 1 kilogram expressed in pounds.  So just divide the weight by 2.2
to get the mass in kilograms.

(The weight number they gave you is not so random.  :-)  )


8 0
3 years ago
What term describes an atom’s tendency to hold onto electrons?
avanturin [10]
It's called Electrophily, or electron affinity.
3 0
4 years ago
Read 2 more answers
A projectile is launched horizontally from a cliff top at 12 m/s. Determine the x-y positions at 1-second intervals. The launch
Natalka [10]

The x position will be "12 m" and the y position will be "-4.9 m".

According to the question,

Along x direction, Initial velocity:

  • 12 m/s

Along x direction, Acceleration is:

  • 0

Along y direction, Acceleration is:

  • -10 m/s²

Time,

  • t = 1 second

→ The X coordinate will be:

= 12\times t

= 12\times 1

= 12 \ m

→ The Y coordinate will be:

= -(u_y t+\frac{1}{2}a_y t^2 )

= 0+4.9 t^2

= 4.9 \ t^2

= -4.9\times 1^2

= -4.9 \ m

Thus the above answer is correct.

Learn more:

brainly.com/question/17108357

5 0
3 years ago
If you are looking straight ahead an object that is behind you is in your cone of vision.
Dimas [21]
False
You cannot see behind you, therefore it is not in your cone of vision.
5 0
3 years ago
Read 2 more answers
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