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Keith_Richards [23]
3 years ago
7

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

r the values of t that are separated by the step Δt = 2s.

Physics
1 answer:
timama [110]3 years ago
3 0

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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Explanation:

From Ohm's law,

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where \rho is the resistivity of the material, l  and A are the length and cross-sectional area, respectively.

A = \rho\dfrac{l}{R}

A = \rho\dfrac{l\timesI}{V}

Since the wire is assumed to have a circular cross-section, its area is given by

A = \pi\dfrac{d^2}{4} where d is the diameter.

\pi\dfrac{d^2}{4} = \rho\dfrac{l\timesI}{V}

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Resistivity of copper = 1.68\times10^{-8}. With these and other given values,

d = \sqrt{\dfrac{4\times1.68\times10^{-8}\times1.5\times290}{3.14\times 0.55}}

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12. A boy of mass 50 kg running 5m/s jumps on to a 20kg trolley
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Answer:

4 m/s

Explanation:

Momentum is conserved.

m₁ v₁ + m₂ v₂ = (m₁ + m₂) v

(50)(5) + (20)(1.5) = (50 + 20) v

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The final velocity is 4 m/s.

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A man inside the water seea a fish in water. will the fish appear to man in it's actual position? will the man appear to fish in
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kompoz [17]

Answer:

0.5 m/s².

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

a = (v – u) / t

a = (10 – 0) / 20

a = 10/20

a = 0.5 m/s²

Therefore, the acceleration of the car is 0.5 m/s².

6 0
2 years ago
An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte
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Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2

a. The angular acceleration is Torque divided by moments of inertia:

\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2

b. 5 revolution would be equals to 10\pi rad, or 31.4 rad. Since the engine just got started

\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5

\omega = \sqrt{1893.5} = 43.5 rad/s

c. Work done during the first 5 revolution would be torque times angular displacement:

W = T*\theta = 1930 * 31.4 = 60633 J

d. The time it takes to spin the first 5 revolutions is

t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s

The average power output is work per unit time

P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

P_i = T*\omega = 1930*43.5=83983 W or 84 kW

7 0
2 years ago
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