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3 years ago

As you increase current, what happens to the strength of an electromagnet?

Physics

1 answer:

Maksim231197 [3]3 years ago

3 0

As you increase current, the strength of an electromagnet intensifies.

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Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

svp [43]

The answer is C, because they moved from a stand still to down the hill

4 0

3 years ago

The largest hailstone every measured fell in Vivian, Nebraska in 2010. The circumference of that hailstone was 19 inches. Using

Crazy boy [7]

Answer:

6.04788 in

$115.82654\ in^3$

78.38779 m/s

0.88159 kg

34.55294 J

Explanation:

Circumference is given by

$c=2\pi r\\\Rightarrow r=\dfrac{c}{2\pi}\\\Rightarrow r=\dfrac{19}{2\pi}\\\Rightarrow r=3.02394\ in$

Diameter is given by

$d=2r\\\Rightarrow d=2\times 3.02394\\\Rightarrow d=6.04788\ in$

The diameter is 6.04788 in

$6.04788\times 2.54=15.3616152\ cm$

Volume of sphere is given by

$v=\dfrac{4}{3}\pi r^3\\\Rightarrow v=\dfrac{4}{3}\pi 3.02394^3\\\Rightarrow v=115.82654\ in^3$

The volume is $115.82654\ in^3$

$115.82654\times \dfrac{1}{1728}=0.06702\ ft^3$

Fall velocity is given by

$V=k\sqrt{d}\\\Rightarrow V=20\sqrt{15.3616152}\\\Rightarrow V=78.38779\ m/s$

The velocity of the fall will be 78.38779 m/s

Mass is given by

$m=\rho v\\\Rightarrow m=29\times 0.06702\\\Rightarrow m=1.94358\ lb$

$1.94358\ lb=1.94358\times \dfrac{1}{2.20462}=0.88159\ kg$

The mass is 0.88159 kg

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}0.88159\times 78.38779\\\Rightarrow K=34.55294\ J$

The kinetic energy is 34.55294 J

4 0

3 years ago

If you pull a wool sweater over your head, you will often end up with a wild hairdo. Your hair is hard to comb and wants to stan

Sergio [31]

Becasue when you rubbed your hair while you were putting on your sweater it caused it to rub against together causing electricity thingy lol. and thats why your hair goes straigh tup.

5 0

3 years ago

Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31

choli [55]

Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:

8 payments of $24,225: $193,800

Par value at maturity: $570,000

Total repaid: $763800 (193,800 + 570,000)

Less amount borrowed: $508050

Total bond interest expense: $255750 (763800 - 508,050)

2)Prepare a straight-line amortization table for the bonds' first two years.

Semiannual Interest Period End; Unamortized Discount; Carrying Value

01/01/2019 61,950 508,050

06/30/2019 54,206 515,794

12/31/2019 46,462 523,538

06/30/2020 38,718 531,282

12/31/2020 30,974 539,026

3) Record the interest payment and amortization on June 30:

June 30 Bond interest expense, dr 31969

Discount on bonds payable, Cr (61950/8) 7743.75

Cash, Cr ( 570000*8.5%/2) 24225

4) Record the interest payment and amortization on December 31:

Dec 31 Bond interest expense, Dr 31969

Discount on bonds payable, Cr 7744

Cash, Cr 24225

6 0

3 years ago

A car travels east at a constant velocity. The net force on the car is:

Scrat [10]

F=ma
As velocity is constant, a=0
So, F=0

Hope this helps!

6 0

3 years ago