The displacement of a certain object is described by y(t) = 23 sin 5t, where t is measured in seconds. Compute its period and its oscillation frequency in rad/sec and in hertz

Answer:

true

Explanation:

Answer:

1. TURN ON YOUR HAZARD/EMERGENCY LIGHTS

Turn on your hazard lights to warn other drivers as soon as you sense something's wrong. Keep them on until help arrives, recommends the National Motorists Association (NMA).

2. SLOW DOWN AND PULL OFF THE ROAD

Aim for the right shoulder of the road. Consumer reports recommends that you pull over to a safe, flat location that is as far away from moving traffic as possible.

3. TURN YOUR WHEELS AWAY FROM THE ROAD AND PUT ON THE EMERGENCY BRAKE

The California Department of Motor Vehicles (DMV) recommends pulling your emergency brake, sometimes called the parking brake. If you have to park on a hill or slope, turn the car's wheels away from the road to help prevent the care from rolling into traffic, says the California DMV.

4. STAY IN YOUR VEHICLE

If you're on a highway or crowded road, the Insurance Information Institute (III) recommends that you avoid getting out of your vehicle to look at the damage or fix a mechanical problem. If you need to get out of the car, get your vehicle to a safe place and make sure the road around you is completely clear. If you're stopped on the right-hand side of the road, get out through the passenger-side door.

5. BE VISIBLE

Once you're safely out of the vehicle, prop up your hood to let other drivers know they should proceed with caution. This will alert other drivers that you're broken down, according to the NMA.

6. SET UP FLARES OR TRIANGLES

Place flares or triangles with reflectors behind your car to alert other drivers to the location where you've stopped, says the III.

7. CALL FOR HELP

Call or use an app to get a tow truck, mechanic or roadside assistance to come help. your insurance company or other provider who may be able to help. If you're in an emergency situation or are not sure who to contact, call 911 or the local police for help.

Hope this helps :)

Answer:

Install drywall

insert wood panels

unscrew nails

Explanation:

Answer:

Hence, the steady state temperature distribution between concentric spheres with radii 1 and 4 is:

u = [tex]\frac{C1}{r }[/tex] + C2

u = [tex]\frac{\frac{-320}{3} }{r}[/tex] + [tex]\frac{320}{3}[/tex]

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve this question. But I have found a similar question and will try to solve it.

This question has this following part missing:

"If the temperature of the outer sphere is maintained at 80 degrees and the inner sphere at 0 degrees.

Now, this question is complete.

Let's find out the steady state temperature distribution.

As, we know the spherical symmetric heat equation:

[tex]\frac{du}{dt}[/tex] = [tex]\frac{k}{r^{2} }[/tex] [tex]\frac{D}{Dr}[/tex] ([tex]r^{2}[/tex] [tex]\frac{Du}{Dr}[/tex])

Where, small (d) represents the partial differentiation and Capital D represents simple derivative.

And the ordinary differential equation (ODE) for steady state temperature distribution is:

[tex]\frac{k}{r^{2} }[/tex] [tex]\frac{D}{Dr}[/tex]([tex]r^{2}[/tex] [tex]\frac{Du}{Dr}[/tex]) = 0

So it can be said that:

[tex]r^{2}[/tex] [tex]\frac{Du}{Dr}[/tex]

Consequently,

[tex]\frac{Du}{Dr}[/tex] = [tex]\frac{C1}{r^{2} }[/tex]

Taking derivative of the above equation we get:

u = -[tex]\frac{C1}{r }[/tex] + C2

Solution of the ordinary differential equation

u = [tex]\frac{C1}{r }[/tex] + C2 (Consuming the negative sign into C1 constant)

As, it is given in our question that our boundary conditions are: 1 and 4

So,

Putting the boundary conditions into the variable (r) to find the constants we get:

u1 = [tex]\frac{C1}{1} }[/tex] + C2 = 0 (degrees)

u1 = C1 + C2 = 0 (degrees)  equation (1)

Similarly,

u4 = [tex]\frac{C1}{4}[/tex] + C2 = 80 (degrees)

u4 = [tex]\frac{C1}{4}[/tex] + C2 = 80 (degrees)  equation (2)

Solving for C1, we get:

Equation 1 - Equation 2

(C1 - C2 = 0) - ([tex]\frac{C1}{4}[/tex] + C2 = 80)

[tex]\frac{-3}{4}[/tex]C1 = 80

Solving for C1

C1 = -[tex]\frac{320}{3}[/tex]

With the help of value of C1, we get value of C2

Put the value of C1 in equation (1) to get value of C2

C1 + C2 = 0

-[tex]\frac{320}{3}[/tex] + C2 = 0

Solving for C2

C2 = [tex]\frac{320}{3}[/tex]

Hence, the steady state temperature distribution between concentric spheres with radii 1 and 4 is:

u = [tex]\frac{C1}{r }[/tex] + C2

Plugging in the values of C1 and C2

u = [tex]\frac{\frac{-320}{3} }{r}[/tex] + [tex]\frac{320}{3}[/tex]

Answer:

Explanation:

Given that:

[tex][\sigma] = \left[\begin{array}{cc}800&300\\ \\300&-400\end{array}\right] (MPa)[/tex]

[tex][\sigma] = \left[\begin{array}{cc}\sigma_{xx}&\sigma_{xy}\\ \\\sigma_{yx}&\sigma_{yy}\end{array}\right][/tex]

Using Determinant method

The principal stress is the maximum or minimum normal stress acting on any plane. For the 2D stress system, the 2-principle plane always carries zero shear stress.

For principal stress [tex]( \sigma_1, \sigma_2)[/tex]

[tex]\sigma_{1,2} = \dfrac{\sigma_x+ \sigma_y}{2} \pm \sqrt{( \dfrac{\sigma_x - \sigma_y}{2} )^2 + \sigma^2_{xy}}[/tex]

[tex]\sigma_{1} = \dfrac{800+(-400)}{2} \pm \sqrt{( \dfrac{800 -(-400)}{2} )^2 + (300)^2}[/tex]

[tex]\sigma_{1} = \dfrac{400}{2} \pm \sqrt{ (600)^2 + (300)^2}[/tex]

[tex]\sigma_{1} = 200+ 670.82 \\ \\ \sigma_{1} = 870.82 \ MPa[/tex]

[tex]\sigma_{2} = \dfrac{800+(-400)}{2} \pm \sqrt{( \dfrac{800 -(-400)}{2} )^2 + (300)^2}[/tex]

[tex]\sigma_{2} = 200- 670.82 \\ \\ \sigma_{1} = - 470.82 \ MPa[/tex]

According to Mohr's circle;

Mohr's circle is the locus provided that the position of the normal stress and the shear stress is acting on any plane.

Center = (a,0)

[tex]a = \dfrac{\sigma_{x}+\sigma_{y}}{2}[/tex]

[tex]a = \dfrac{800+(-400)}{2}[/tex]

a = 200 MPa

radius (r) = [tex]\sqrt{ (\dfrac{\sigma_{x}-\sigma_{y}}{2})^2 + \sigma^2 _{xy}}[/tex]

[tex]=\sqrt{ (\dfrac{800-(-400)}{2})^2 + (300)^2}[/tex]

[tex]=\sqrt{ (600)^2 + (300)^2}[/tex]

r = 670.82 MPa

[tex]\sigma_1 = a +r \\ \\ \sigma_1 = 200 + 670.82 \\ \\ \sigma_1 = 870.82 \ MPa[/tex]

[tex]\sigma_2 =-(r-a)[/tex]      (it is negative because of the negative x-axis)

[tex]\sigma_2 =670.82 - 200 \\ \\ \sigma_1 = 470.82 \ MPa[/tex]    

[tex]\tau_{max} = radius \ of \ Mohr's \ circle[/tex]

[tex]\tau_{max} = 670.82 \ MPa[/tex]

Answer:

100 V

Explanation:

V = IR

100 x 1 = 100

100 volts

The voltage across a 100-ohm circuit element that draws a current of 1 A is 100 V.

Electric current is a term used to describe how much electricity flows across a circuit and how it flows in an electronic circuit. Amperes are used to measure it (A).

The more electricity flowing across the circuit, the higher the ampere value. The difference in electric potential between two places is known as voltage, electric potential difference, electric pressure, or electric tension.

Ampere is the SI unit of electric current and ohm is the SI unit of electric resistance.

The circuit element is 100 ohms

The current is 1 A.

Voltage = current (I) x resistance (r)

V = IR

100 x 1 = 100

100 volts

Therefore, the voltage across a 100-ohm circuit element is 100 volts.

To learn more about electric current, refer to the below link:

https://brainly.com/question/28019870

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