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harina [27]
3 years ago
12

An object is moving in a horizontal direction only. The graph shows its velocity vs. time. (See Graph Below)

Physics
1 answer:
zysi [14]3 years ago
8 0
48 M to get the answer add the area of the triangle and rectangle under the line
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At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th
Korvikt [17]

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

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Consider a constant density gas flowing steadily over an airfoil. Far upstream the velocity is V0. Halfway along the top surface
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Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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3 years ago
A machine has an efficiency of 70%. How much work dose the machine do when 20,000 J of work is done on it ?
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Answer: The machine does 14.000 J

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Anything that moves up and down or back and forth in a rhythmic way is vibrating true false
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Read 2 more answers
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
balandron [24]

Answer with Explanation:

We are given that

Radius of  solid core wire=r=2.28 mm=2.28\times 10^{-3} m

1mm=10^{-3} m

Radius of each strand  of thin wire=r'=0.456 mm=0.456\times 10^{-3} m

Current density of each wire=J=3750 A/m^2

a.Area =\pi r^2

Where \pi=3.14

Using the formula

Cross section area of copper wire has solid core =3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2

Current density =J=\frac{I}{A}

Using the formula

3750=\frac{I}{16.3\times 10^{-6}}

I=3750\times 16.3\times 10^{-6}=0.061 A

Total number of strands=19

Area of strand wire=A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2

J'=\frac{I'}{A'}

3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}

I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2

I'=0.047 A

b.Resistivity of copper wire=\rho=1.69\times 10^{-8}\Omega-m

Length of each wire =6.25 m

Resistance, R=\frac{\rho l}{A}

Using the formula

Resistance of solid core wire=R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega

Resistance of strand wire=R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega

7 0
3 years ago
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