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Sedbober [7]
3 years ago
13

A truck decelerates at a rate of 2 m/s2. If it starts with a velocity of 15 m/s, how long will it take to stop?

Physics
1 answer:
Nikitich [7]3 years ago
7 0
Use the formula Vf = Vo + at

Stop => Vf = 0

=> 0 = Vo + at => Vo = -at => t =-  Vo / a

Vo = 15 m/s

a = - 2 m/s^2

=> t = - 15 m/s / ( - 2 m/s^2) = 7.5 s

And you can find the distance travled by using the equation: d = Vo*t + at^2 / 2

=> d = 15 m/s * 7.5 s - 2m/s^2 * (7.5s)^2 / 2 = 56.25 m

Answer: 7.5s (and 56.25m)
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Answer:

False

Explanation:

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3 years ago
Which of the following statements is true about magnetic fields.
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Answer:  

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8 0
3 years ago
When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4
daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

V_{0} = A_{1} V_{1} + A_{2} V_{2} \\

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6  and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of  A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |

CMRR = |\frac{48}{0.4} |\\CMRR = 120

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3 years ago
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Marizza181 [45]

Answer:

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8 0
3 years ago
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