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Alex73 [517]
3 years ago
9

Teenagers

Physics
1 answer:
Sophie [7]3 years ago
5 0
A would be the best
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A 0.150 kg mass is attached to a spring with k = 18.9 N/m. At the equilibrium position, it moves 2.39 m/s. How much mechanical e
Nastasia [14]

Answer:0.428

Explanation: got it right on Acellus

5 0
3 years ago
After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample
alina1380 [7]

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

5 0
3 years ago
A tennis ball of mass 0.060 kg travels horizontally at a speed of 25m/s. The ball hits a tennis
stepladder [879]
It would be A because a is perfect
7 0
3 years ago
Read 2 more answers
Help wit these questions someone.
romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, \frac{1}{Req}  = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}

R₂ + 2 = 12

R₂ = 10Ω

<h3>Q8.</h3>

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}

Req = 1.7Ω

7 0
3 years ago
You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract
stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

#SPJ4

3 0
2 years ago
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