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3 years ago

Where did the missouri compromise imaginary line run

1 answer:

7 0

In the Missouri Compromise, the slavery line for future US states ran along the southern border of Missouri at 36 degrees north 30 minutes

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It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

$W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m$

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

$W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J$

So, the new work is more than 130 J.

6 0

3 years ago

which of the following are vector quantities? check all that apply. a. force b. acceleration c. displacement d. mass

Ratling [72]

Force, acceleration, and Displacement are all vector quantities.

4 0

4 years ago

A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tal

olganol [36]

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance = 3.6 m

v = Image distance

f = Focal length = 35 mm

$h_u$ = Object height = 1.8 m

a) Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm$

The CCD sensor is 35.34 mm from the lens

b) Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}$

$m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm$

The person appears 17.67 mm tall on the sensor

7 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

4 years ago

Que fuerza será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s²

WARRIOR [948]

Answer:

La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.

Explanation:

La segunda ley de Newton, llamada ley fundamental o principio fundamental de la dinámica, plantea que un cuerpo se acelera si se le aplica una fuerza.

De esta manera, esta ley establece que las aceleraciones que experimenta un cuerpo son proporcionales a las fuerzas que recibe. Dicho de otra forma, la aceleración de un cuerpo es proporcional a la fuerza neta que se le aplica. Cuanto mayor es la fuerza que se le aplica a un objeto con una masa dada, mayor será su aceleración.

La segunda Ley de Newton se expresa matemáticamente como:

F = m*a

Donde:

F es la fuerza neta. Se expresa en Newton (N)
m es la masa del cuerpo. Se expresa en kilogramos (Kg.).
a es la aceleración que adquiere el cuerpo. Se expresa en metros sobre segundo al cuadrado (m/s²).

En este caso:

m= 20 kg
a= 4 m/s²

Reemplazando:

F= 20 kg* 4 m/s²

Resolviendo:

F= 80 N

<u><em>La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.</em></u>

4 0

3 years ago