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Brut [27]
3 years ago
11

A tangent line drawn on a velocity-time graph has a rise of 19 m/s and a run of 4.0 m/s. How large is the acceleration? What typ

e of acceleration Is this?​
Physics
1 answer:
klemol [59]3 years ago
8 0

Answer:

Acceleration = 4.8 m/s²

Explanation:

Given:

Change in velocity = 19 m/s

Change in time = 4 s

Find:

Acceleration

Computation:

Acceleration = Change in velocity / Change in time

Acceleration = 19/4

Acceleration = 4.8 m/s²

Positive acceleration

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HELP DUE IN 5 How does a battery supply energy to a circuit?
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You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body
stellarik [79]

A) The change in internal chemical energy is 1.15\cdot 10^7 J

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

P=Fv

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

P=(80)(8.0)=640 W

The energy output is related to the power by the equation

P=\frac{E}{t}

where:

P = 640 W is the power output

E is the energy output

t = 30 min \cdot 60 = 1800 s is the time elapsed

Solving for E,

E=Pt=(640)(1800)=1.15\cdot 10^6 J

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

E=1.15\cdot 10^7 J

Here we want to find the time t' needed to convert an amount of chemical energy of

E'=3.8\cdot 10^5 J

So we can setup the following proportion:

\frac{t}{E}=\frac{t'}{E'}

And solving for t',

t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0
3 years ago
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did
dezoksy [38]

Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

  • \boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}

It divides:

  • \boxed{t=5\ s}

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

8 0
3 years ago
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