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pogonyaev
3 years ago
14

All circuits include

Physics
1 answer:
valina [46]3 years ago
4 0

Answer:

a battery, wires, and a switch.

Explanation:

All circuits include?

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An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start
galina1969 [7]

Answer:

it moves 25 inches.

Explanation:

the east west bit isn't important, ignore it. if an ant starts at 6 then moves to 19 then we need to subtract 19 from 6, that's 13. then it moves to 7. the difference between 19 and 7 is 12. add that to 13 and you get 25. it's important to remember that there is no such thing as negative distance. if it moved, then it counts.

3 0
3 years ago
Just need to know net force, unbalanced or balanced and accelerated or not accelerated :)
jonny [76]

Answer:

9. 45 N North East

Unbalanced

Accelerate

10. 45 N right and 6 N down

Unbalanced

Accelerate

Explanation:

Hope this helps!

5 0
3 years ago
The sampling rate of an ADC is 8.1 kHz. What will be an appropriate cut-off frequency (break frequency) for the anti-aliasing fi
KiRa [710]

Answer:

4000 Hz

Explanation:

An anti-alias filter is usually added in front of the ADC to limit a certain range of input frequencies in order to avoid aliasing. This filter is usually a low pass filter that passes low frequencies but attenuates the high frequencies.

The Nyquist sampling criteria states that the sampling rate should be at least twice the maximum frequency component of the desired signal.

Sampling rate = 2(max input frequency)

From the relation we can find out the cut-off frequency for the anti-aliasing filter.

max input frequency = sampling rate/2

max input frequency = 8100/2 = 4050 Hz

Therefore, 4000 Hz would be an appropriate cut-off frequency for the anti-aliasing filter.

3 0
3 years ago
s A steel ball with mass m=5.21 g is moving horizontally with speed ????=412 m/s when it strikes a block of hardened steel with
denis23 [38]

Answer:

Vf₂ = 0.29 m/s :Speed of the block immediately after the collision.

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction and direction as velocity) and its magnitude is calculated like this:

P=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀=Pf  Formula (1)

P₀ :Initial  linear momentum quantity

Pf : Initial  linear momentum quantity

Nomenclature and data

m₁: ball mass= 5.21 g= 5.21*10⁻³kg

V₀₁: initial ball speed,  =412 m/s

Vf₁: final ball speed

m₂: block mass =  14.8 kg  

V₀₂: initial block speed, = 0

Vf₂: final block speed

Problem development

For this problem the collision is perfectly elastic ,then, In addition to the linear moment, the kinetic energy is also conserved.

We assume that the ball moves to the right before the collision y continues moving  to the right after the collision .(+)

We assume that the block moves to the right before the collision.(+)

We apply furmula (1)

P₀=Pf

m₁*V₀₁+m₂*V₀₂=m₁*Vf₁+m₂*Vf₂

5.21*10⁻³*412+14.8*0= 5.21*10⁻³*Vf₁+14.8*Vf₂

2.15= 5.21*10⁻³*Vf₁+14.8*Vf₂ Equation (1)

For perfectly elastic collision the coefficient of elastic restitution (e) is equal to 1, and e is defined like this:

e=\frac{v_{f2}- v_{f1} }{v_{o1} -v_{o2} }

1*(V₀₁-V₀₂) =Vf₂-Vf₁  , V₀₂=0, V₀₁ =412 m/s

412=Vf₂-Vf₁

Vf₁=Vf₂-412 Equation (2)

We replace Equation (2) in Equation (1)

2.15= 5.21*10⁻³(Vf₂-412)+14.8*Vf₂

2.15= 5.21*10⁻³*Vf₂-2.15+14.8*Vf₂

4.3=14.805Vf₂

Vf₂ =4.3/14.805= 0.29 m/s : (+) ,with equal direction  of the movement of the ball before the collision.

8 0
3 years ago
2. Another car weighs 2000kg, you can push it .05 m/s?, how much force are you
Vesna [10]

Answer:

\boxed {\boxed {\sf 100 \ Newtons}}

Explanation:

We are asked to calculate the force you are applying to a car. According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, we can use the following formula to calculate force.

F= m \times a

The mass of the car is 2000 kilograms and the acceleration is 0.5 meters per second squared.

  • m= 2000 kg
  • a= 0.05 m/s²

Substitute the values into the formula.

F= 2000 \ kg \times 0.05 \ m/s^2

Multiply.

F= 100 \ kg *m/s^2

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 100 kilogram meters per second square is equal to 100 Newtons.

F= 100 \ N

You apply <u>100 Newtons</u> of force to the car.

7 0
3 years ago
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