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jeyben [28]
3 years ago
13

What is the formula for acceleration? ( PLEASE HELP! ALSO IF YOU GET B IT'S WRONG BECAUSE I ALREADY PUT THAT AND IT WAS WRONG so

rry caps)
A: Final speed-initial speed ________________________ time

B: Net force = Mass x acceleration

C: Mass x Velocity

D: Force a 5N Force B 10N Net force of forces A and B 15N
Physics
1 answer:
skad [1K]3 years ago
3 0

Answer:

option A is correct

Explanation:

acceleration the time rate of change of velocity or speed so

a=Δv/t

Δv=vf-vi  

Δv= final speed-initial speed

now a=final speed-initial speed/time

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deff fn [24]
The 2nd Law says F=ma, where F is the force in Newtons, m is mass and a is acceleration.  Earth's gravity is an acceleration, 9.8m/s^2.  So you can solve the equation for mass, m=F/a, or m=F/9.8 where you've measured the force (weight) in Newtons.
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3 years ago
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Tridil is infusing at 15 ml/hr on an infusion pump. The drug is mixed 50 mg in 500 ml D5W. How many MCG/minute is the patient re
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Answer:

patient receiving drug 25 MCG/minute

Explanation:

given data

infusing = 15 ml/hr

drug = 50 mg

D5W = 500 ml

to find out

How many MCG/minute

solution

we know infusing rate is 15 ml/hr = 0.25 ml/min

so 0.25 ml drug content = 50 /500 × 0.25

0.25 ml drug content = 0.025 mg

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rate of drug will be 0.025 mg

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rate of drug = 25 MCG/minute

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8 0
2 years ago
A car that brakes suddenly comes to a screeching halt. Is the sound energy produced in this conversion a useful form of energy?
Nezavi [6.7K]

Answer:

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7 0
2 years ago
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

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3 years ago
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Neptune's atmosphere is made up mostly of hydrogen and helium with just a little bit of methane.

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