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umka21 [38]
3 years ago
6

You use a 1200-watt hair dryer for 10 minutes each day. a. How many minutes do you use the hair dryer in a month? (Assume there

are 30 days in the month.) b. How many hours do you use the hair dryer in a month? c. What is the power of the hair dryer in kilowatts? d. How many kilowatt-hours of electricity does the hair dryer use in a month? e. If your town charges $0.15/kWh, what is the cost to use the hair dryer for a month?
Physics
1 answer:
sertanlavr [38]3 years ago
3 0
300 minuts
5 hours
1.2 killowatt
6 killowatt hour
6*0.15=0.9
$0.9

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v₀ = initial speed of the object = 8 meter/second

v = final speed of the object = 16 meter/second

t = time taken to increase the speed = 10 seconds

d = distance traveled by the object in the given time duration = ?

using the kinematics equation

d = (v + v₀) t/2

inserting the above values in the above equation

d = (16 + 8) (10)/2

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The diameter of Saturn is nearly ten times that of Earth. However, the density of Saturn is much less than that of Earth. What i
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Saturn is a gaseous planet

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Sugar dissolved in water is an example of?
lbvjy [14]

Answer:

D. Solution

Explanation:

Sugar dissolved in water is an example of solution.

A solution is a homogenous mixture of solutes and solvents.

In a solution the solute particles ae distributed uniformly in the solvents. The solute is the substance and it is the sugar here that is dissolved to make a solution.

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Answer: D

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jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
3 years ago
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