What does a lighting bolt and getting shocked by a doorknob have in common?

To determine a waves frequency you must know the??

lutik1710 [3]

Answer: I think, the number of oscillations in a given period of time.

Explanation: Well I guess because in a period time is known as the rate of occurrence of the wave. Hope this helps!

7 0

3 years ago

Read 2 more answers

Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the m

Phantasy [73]

Answer:46.05 mm

Explanation:

Given

$Power=260\ hp\approx 260\times 746=193.96\ KW$

speed $N=550\ rev/min$

allowable shear stress $(\tau )_{max}=15\ kpsi\approx 103.421\ MPa$

Power is given by

$P=\frac{2\pi NT}{60}$

$193.96=\frac{2\pi 550\times T}{60}$

$T=3367.6\ N-m$

From Torsion Formula

$\frac{T}{J}=\frac{\tau }{r}-----1$

where J=Polar section modulus

T=Torque

$\tau$ =shear stress

For square cross section

$r=\frac{a}{2}$

where a=side of square

$J=\frac{a^4}{6}$

Substituting the values in equation 1

$\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}$

$a=0.04605\ m$

$a=46.05\ mm$

7 0

3 years ago

The formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with ra

Archy [21]

Answer:

562 miles per hour.

Explanation:

As given in the question, the formula for the maximum speed on a curved road is

$v=\sqrt{2.3}$ $r$

Given value of $r=370$ feet

So the maximum safe speed will be

$v=\sqrt{2.3} \times 370 = 1.52\times 370 = 562.4$ miles per hour.

Rounding off to the nearest whole number we get the maximum safe speed at the curved road is 562 miles per hour.

6 0

4 years ago

If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will

Anastasy [175]

Assuming that the voltage is constant, and the resistance was doubled, since I=V/R, the current through the circuit will be halved. As P=IV, with the same voltage and halved current, the power dissipated by the circuit will be halved

3 0

3 years ago

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on t

diamong [38]

Answer:

1. 18.5m/s

2. 17.5 m

3. 0 at its highest point

4. Direction is downwards

Explanation:

1. This egg is thrown vertically from a height

Yo = 0. This egg then falls to the point y = -30.0 at t = 5seconds

Y-Yo = V0t - 1/2gt²

-30-0 = V0(5)-1/2(9.8)(5²)

-30 = 5v0 - 4.9x25

-30 = 5V0 - 122.5

-30+122.4 = 5v0

V0 = 92.5/5

= 18.5m/s

this is the initial speed of the egg

2. When the egg is at a maximum height it would have a velocity equal to 0

V² = V0² - 2*g*y

V = 0, V0 = 18.5, g = 9.8

0 = 18.5²-2x9.8*y

342.25-19.6y = 0

342.25 = 19.6y

Divide through by 19.6

Y = 342.25/19.6

Y = 17.5m

this value is how high it rises above starting point

3.

The magnitude of velocity is = 0 at its highest point

4.

This egg falls under gravity. Therefore the acceleration due to gravity has a constant magnitude and direction. Magnitude = 9.8m/s and it's direction is downwards.

5. Please check attachment for graph

6 0

3 years ago