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3 years ago

5

A car initially traveling at 15 m/s speeds up at a constant rate of 2.0 m/s2 for 3 seconds. The velocity of the car at the end o

f the 3 second interval is _________ m/s.

1 answer:

timofeeve [1]3 years ago

8 0

Answer:

Vf = 21 m/s

Explanation:

Data:

Initial Velocity (Vo) = 15 m/s
Acceleration (a) = 2.0 m/s²
Time (t) = 3 s
Final Velocity (Vf) = ?

Use formula:

Vf = Vo + a * t

Replace:

Vf = 15 m/s + 2.0 m/s² * 3s

Multiply the acceleration with time:

Vf = 15 m/s + 6 m/s

Solve the sum:

Vf = 21 m/s

The velocity of the car at the end of the 3 second interval is <u>21 meters per second.</u>

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harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation: In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ) for I(t)/io=0.0257=Exp(-t/τ) then

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3 years ago

Explain the change of frequency of the wave if the tension of the string is increased

jek_recluse [69]

Answer:

Increasing the tension on a string increases the speed of a wave, which increases the frequency (for a given length). Pressing the finger at different places changes the length of string, which changes the wavelength of standing wave, affecting the frequency.

Explanation:

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3 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

lukranit [14]

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3 years ago

A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.

natima [27]

Answer:

(a) 24.025 m/s. downward.

(b) 31 m

Explanation:

From Newton's equation of motion,

(a)

v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

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Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

v = 24.025 m/s. downward.

(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

s₁ = -0.5(2.5) + 1/2(9.81)(2.5)²

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also,

s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

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Note: Distance between the pelican and fish = s₁ + s₂

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Answer:

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Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

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v and u is final and initial velocity respectively, measured in $ms^{-1}$

t is time measured in seconds.

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4 0

3 years ago