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Tanya [424]
3 years ago
14

A car traveling at 3.6 m/s accelerates at the rate of 0.71 m/s2 for an interval of 4.4 s. Find vf. Incorrect: Your answer is inc

orrect. m/s
Physics
1 answer:
alexandr1967 [171]3 years ago
8 0

Answer:

vf = 6.724 m/s

Explanation:

Velocity: This an be defined as the rate of change of displacement. Velocity is a vector quantity, because it can be represented both in magnitude and direction. The S.I unit of velocity is m/s.

From newton's equation of motion,

vf = vi + at ......................................... Equation 1

Where vf = final velocity, vi = initial velocity, a = acceleration, t = time.

Given: vi = 3.6 m/s, a = 0.71 m/s²,  t = 4.4 s.

Substitute into equation 1

vf = 3.6+0.71(4.4)

vf = 3.6+3.124

vf = 6.724 m/s

Hence, vf = 6.724 m/s

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The question is poorly worded, and makes it harder for the student to learn the basics of the subject. "Potential Difference" is the difference in electrical potential between two places in an electric field ... not in electrical potential energy.
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3 years ago
A 0.30-kg object is traveling to the right (in the positive direction) with a speed of 3.0 m/s. After a 0.20 s collision, the ob
andre [41]

Answer:

I = -2.1 kg.m/s

Explanation:

Given,

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initial speed, v_i = 3 m/s

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final speed, v_f = -4 m/s

Impulse = change in momentum

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How many complete revolutions are needed to draw the angle 725°?
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A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br
serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

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klemol [59]

Answer:

true

Explanation:

that's the answear

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