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padilas [110]
3 years ago
9

1. Which of the following is an accurate statement? A. Step-down voltage transformers have a different number of turns in the pr

imary than in the secondary winding, so they change the incoming voltage to a higher voltage. B. DC Generators produce current with a voltage that changes from positive to negative. C. Step-up voltage transformers have a lower number of turns in the primary than in the secondary winding. D. AC Generators produce current that is pulsating, but always remains positive.
Physics
1 answer:
Snowcat [4.5K]3 years ago
4 0
'C' is the only true statement on the list.

Step-up voltage transformers have a lower number of turns
in the primary than in the secondary winding.

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What is 3.75 x 10^-7?
pogonyaev

Answer:

Explanation:

3.75 * 10^-7

=3.75 * 1/10^7

=3.75/10000000

=3/800000000

any base which has it's power negative do it's reciprocal then the power will be positive.

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3 years ago
The moon has a weaker gravitational force than earth. sofia weighs 50 lbs. on earth. how much will she weigh on the moon?
kupik [55]
If Sofia weighs 50 lbs on Earth, then she would weigh 8.3 lbs, because you are 6 times lighter on the moon than you are on Earth!
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3 years ago
Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi
Mamont248 [21]

Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

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4 0
3 years ago
Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral
Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit  such that

P=VI=V^2/R

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

Current2=\frac{V}{2R}

Power2=Voltage * Current2

Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2

TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}

6 0
3 years ago
The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?
Likurg_2 [28]
\frac{7.85g }{1.59 \frac{g}{ cm^{3} } } = 4.937 cm^{3}
4 0
3 years ago
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