What are the three symbols used in Ohm's law. Explain what each symbol represents and give the units for each of the variables.

32.What is the area under the curve between 0 seconds and 2 seconds?

Mademuasel [1]

Explanation:

32. The area under the curve from t = 0 and t = 2 is the area of the triangle. The base of the triangle is 2 seconds, and the height of the triangle is 2 m/s. The area is therefore:

A = ½ bh

A = ½ (2 s) (2 m/s)

A = 2 m

33. The car is not moving when the velocity is 0 m/s. This occurs at t = 0, t = 6, and t = 10.

3 0

3 years ago

No file link please

musickatia [10]

Rock 3, because it has the most mass while still moving at the same speed, it is kinetic because there is movement and energy being used

5 0

3 years ago

The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of th

Yuri [45]

Answer:

$R=64.32\ lb\\\\\theta=84.3\°$

Explanation:

Given:

Ratio of lift force to drag force is, $\frac{L}{D}=10$

Lift force on a short section is, $L=64\ lb$

Magnitude of resultant, $R= ?$

The angle of 'R' with the horizontal is, $\theta=?$

We know that, lift force and drag are at right angles to each other. So, the resultant can be computed using Pythagoras theorem.

For calculating 'R', we first compute drag force 'D'.

As per question:

$\frac{L}{D}=10\\\\D=\frac{L}{10}=\frac{64\ lb}{10}=6.4\ lb$

Now, the magnitude of resultant 'R' is given as:

$R=\sqrt{L^2+D^2}$

Plug in the given values and solve for 'R'. This gives,

$R=\sqrt{64^2+6.4^2}\\\\R=\sqrt{4096+40.96}\\\\R=\sqrt{4136.96}=64.32\ lb$

Therefore, the magnitude of the resultant force 'R' is 64.32 lb.

Now, the angle $\theta$ is given as the arctan of the ratio of the lift and drag force.

Therefore,

$\theta=\tan^{-1}(L/D)\\\\\theta=\tan^{-1}(10)\\\\\theta=84.3\°$

Therefore, the angle made with the horizontal is 84.3°.

6 0

3 years ago

ANSWER FAST!!!!

Amiraneli [1.4K]

Power = (voltage) x (current)

Power = (240 volts) x (4 Amp)

Power = 960 watts

4 0

3 years ago

If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine

pickupchik [31]

Answer:

τ = 132.773 lb/in² = 132.773 psi

Explanation:

b = 12 in

F = 60 lb

D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in

d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in

We can see the pic shown in order to understand the question.

Then we get

Mt = b*F*Sin 30°

⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in

Now we find ωt as follows

ωt = π*(R⁴ - r⁴)/(2R)

⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)

⇒ ωt = 2.7114 in³

then the principal stresses in the pipe at point A is

τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)

⇒ τ = 132.773 lb/in² = 132.773 psi

7 0

3 years ago