Answer:
16.86 ft/s
Step-by-step explanation:
Given that
Speed of the carousel, s = 14 rpm
Distance from the carousel, d = 23 ft
To start with, we convert the given speed to revolution per second, so that
s = 14/60 rev/sec
To find the needed speed, we use the formula
v =2πnR, remember that Diameter = 2R, so we substitute D for 2R and get πnD
v = πnD, and we use this to find our needed velocity
v = 3.142 * 14/60 * 23
v = 72.266 * 14/60
v = 16.86 ft/s
Therefore the boy must run at a speed of 16.86 ft/s in order to be able to match the carousel and jump on it
Answer: The voltage across an identical resistor = 60 Volts
Step-by-step explanation:
Step 1
Given that voltage across the resistor is directly proportional to the current running through the resitor.
Mathematically, it would be represented as
Voltage ∝ Current
Introducing the constant of proportionality, k which represents the Resistor as it is constant.
we have that
Voltage = K x Current
When current = 12 amps and voltage = 480v, the constant of proportionality K
480 = k x 12
k = 480/12
k= 40
Step 2 ,
when current = 1.5 amps
Voltage = ?
Using our equation that
Voltage = K x Current
Voltage = 40 x 1.5
Voltage = 60 Volts
Answer:
x= 5/4 + −1/4√17 or x = 5/4 + 1/4√17
Step-by-step explanation:
Numbers expressed by exponents are powers
Answer:
<em>There is no significant difference in the amount of rain produced when seeding the clouds.</em>
Step-by-step explanation:
Assuming that the amount of rain delivered by thunderheads follows a distribution close to a normal one, we can formulate a hypothesis z-test:
<u>Null Hypothesis
</u>
: Average of the amount of rain delivered by thunderheads without seeding the clouds = 300 acrefeet.
<u>Alternative Hypothesis
</u>
: Average of the amount of rain delivered by thunderheads by seeding the clouds > 300 acrefeet.
This is a right-tailed test.
Our z-statistic is
We now compare this value with the z-critical for a 0.05 significance level. This is a value 
such that the area under the Normal curve to the left of is less than or equal to 0.05
We can find this value with tables, calculators or spreadsheets.
<em>In Excel or OpenOffice Calc use the function
</em>
<em>NORMSINV(0.95)
</em>
an we obtain a value of
= 1.645
Since 1.2845 is not greater than 1.645 we cannot reject the null, so the conclusion that can be drawn when the significance level is 0.05 is that there is no significant difference in the amount of rain produced when seeding the clouds.
