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2 years ago

A 11,000-watt radio station transmits at 880 kHz. Determine the number of joules transmitted per second.

Mathematics

2 answers:

storchak [24]2 years ago

5 0

1 watt = 1 joules per second

The number would be the same:

11,000 watts = 11,000 joules per second

melamori03 [73]2 years ago

4 0

Answer:

The number of J/s is 11000.

A is correct option.

Step-by-step explanation:

Given that,

Power P = 11000 watt

We know that,

Joule :

Joule is the unit of SI system. Joule is the unit of energy.

Energy is the product of power and time.

1 J = 1 N-m

One watt is equal to the one joule per second.

So, the number would be same

11000 watt = 11000 J/s

Hence, The number of J/s is 11000.

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If x+y=9, and 8(x+3y)= 120, what is the value of x-y

Triss [41]

Answer:

Explanation:

Apply Multiplicative Distribution Law:

{x + y =9

{8x + 24y = 120

Reduce the greatest common factor on both sides of the equation.

{x + y = 9

{x + 3y = 15

Subtract the two equations: x + y - (x +3y) = 9 - 15

Remove paranthesis: x + y - x - 3y = 9 - 15

Cancel the unknown variables: y - 3y = 9 - 15

Combine like terms: -2y=9 - 15

Calculate the sum or difference: -2y = -6

Reduce the greatest common factor on both sides of the equation: y =3

Substitute one unknown quantity into the elimination: x + 3 = 9

Rearrange unknown terms to the left side of the equation: x = 9 - 3

Calculate the sum or difference: x = 6

Write the solution set of equations: {x = 6

{y = 3

Substitute: 6 - 3

Calculate the sum or difference: 3

Answer: 3

6 0

1 year ago

If a certain cannon is fired from a height of 8.8 meters above the ground, at a certain angle, the height of the cannonball ab

Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

$h(t)=-4.9t^2+30.5t+8.8$

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

$-4.9t^2+30.5t+8.8=0$

Multiply both sides by 10.

$-49t^2+305t+88=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a = -49, b = 305 and c=88

$t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}$

Ignore the negative value of t as time can't be a negative number.

Thus,

$t=\frac{305+\sqrt{110273}}{98}\approx6.50$

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

6 0

2 years ago

What is the slope-intercept equation for the line below

rusak2 [61]

Answer:

the answer is A

Step-by-step explanation:

7 0

2 years ago

-cep Leu<br>2. 3x -1 = -10

nekit [7.7K]

Answer:

-3

Step-by-step explanation:

3x-1=-10

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x=-3

7 0

2 years ago

Missing numbrr of unit rate for 32/8= ?/1

galina1969 [7]

Answer:

4/1

Step-by-step explanation:

8/8=1, so 32/8=4

3 0

2 years ago

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