All categories

3 years ago

Many battery-powered devices come with cords that allow them to be

Physics

1 answer:

lara [203]3 years ago

4 0

Answer:

The change that has to take place inside the power cord in order for the device to function properly include;

The changing of the alternating current (AC) from the electrical outlet to the direct current having a specific voltage and current value with which the device can be powered

Explanation:

A battery powered device makes use of direct current (DC) electric power from a battery, while the power normally given out at an electrical outlet comes as an alternating current (AC) electric power source.

The power cord for battery-powered device, also known as an AC/DC adapter, that allows them to be plugged into electrical outlets converts the AC electric current it obtains from the electrical outlets to DC electrical current of the appropriate voltage and amperage that the device can make use of for electric power to function and for charging the battery which is the power source for the device

Therefore, the change that has to take place in the power cord is the conversion of the electric outlet alternating current (AC) voltage and amperage values, into direct current (DC) of the required voltage and current for the device

You might be interested in

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

In terms of matter and resources, Earth is essentially a(n) ________ system ; in terms of energy, Earth is a(n) ________ system.

Savatey [412]

Answer:

Explanation:

6 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago

a care starting from rest has an acceleration 0.3 m/s square, calculate the velocity and distance travelled by this car after 2

Anna [14]

Answer:

Final velocity (v) = 36 m/s

Distance traveled (s) = 2,160 m

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 0.3 m/s

Time travel (t) = 2 minutes = 120 seconds

Find:

Final velocity (v) = ?

Distance traveled (s) = ?

Computation:

v = u + at

v = 0 + 0.3(120)

v = 0.3(120)

v = 36 m/s

Final velocity (v) = 36 m/s

Distance traveled (s) = ut + (1/2)at²

Distance traveled (s) = (0.5)(0.3 × 120 × 120)

Distance traveled (s) = 2,160 m

3 0

3 years ago

A closed system consisting of 2 lb of a gas undergoes a process during which the relation between pressure and volume is pVn = c

vazorg [7]

Answer:

V1=2.5ft3

V2=1ft3

n=1.51

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=2lb*1.25ft3/lb=2.5ft3

state 2

V2=m.v2

V2=2lb*0.5ft3/lb= 1ft3

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/60)/ ln (1/2.5)

n=1.51

8 0

3 years ago