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Blizzard [7]
3 years ago
14

What is the average acceleration of the particle within the time interval 1 second to 3 seconds? A. -1 meter/second2 B. 0 meters

/second2 C. 1 meter/second2 D. 1.5 meters/second2 E. 2 meters/second2
Physics
1 answer:
yuradex [85]3 years ago
6 0

The answer is Option B: 0 meters/seconds^2

A= velocity/time

Points: ( 1,3) (3,3)

Velocity: 3-3=0

Time: 3-1=2

0/2= 0 meters/seconds^2

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What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0
ANEK [815]

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

Initial velocity = 0

Time = 3 seconds.  

Using the Formula,  

Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,  

Force = mass × acceleration

= 20 × 1.6

=

Explanation: I REALLY  HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]

8 0
2 years ago
If you have 3 equal plates hanging 2 spaces from the center of a balance, at what distance would you need to place 2 equal plate
Reptile [31]
You answere would be C. 3 spaces
3 0
3 years ago
The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find
lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance  between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is  approximately  v = 343 m/s

so

lowest frequency will be = \frac{v}{2L}   ..............1

put here value in equation 1

lowest frequency will be = \frac{343}{2(0.32)}

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength =  \frac{v}{f}   ..............2

put here value

wavelength =  \frac{343}{4000}  

wavelength = 0.08575 m

so distance =  \frac{wavelength}{2}   ..............3

distance =  \frac{0.08575}{2}  

distance = 0.0425 m

so distance  between adjacent anti nodes is 4.25 cm

3 0
3 years ago
Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri
Artemon [7]
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

Ek=(1/2)*m*v²=12.5 J.

Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

Ep=m*g*h=0.7*9.8*7=48.02 J

The order of examples starting with the lowest energy:

1. book, 2. ball, 3. stone, 4. brick 


4 0
3 years ago
30 POINTS! (Question is located in picture under graph)
harkovskaia [24]
Hi i think its E because the the numbers on both sides are to spread out and are not even one side is 4 and the other is 3
8 0
3 years ago
Read 2 more answers
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