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A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

4 0

3 years ago

the amount of water returning to the earth through precipitation is blank the amount of water leaving the earth through evaporat

nalin [4]

<u>Answer:</u>

<em>The amount of water entering the earth through precipitation is equal to the amount of water leaving earth through transpiration.</em>

<u>Explanation:</u>

Rates of precipitation and evaporation vary widely according to regions and seasons. But in a global scale the rates are equal. Thus the total amount of earth’s water maintains its constancy even though there is a continuous change in forms of water.

Evaporation and transpiration are the forms in which Water leaves the earth and it returns to the earth in various forms of precipitation like rain, snow, dew, fog etc. This water then reaches ocean and land. The water that reaches the land flows as surface run off into rivers and water bodies or seep into the ground replenishing the ground water table.

5 0

3 years ago

g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for

lbvjy [14]

Answer:

2

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

$a\text{sin}\theta=\frac{2N+1}{2}\lambda$

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

$0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}$

Solving the above equation we get

we N= 2.313≅ 2

4 0

3 years ago

A student took a calibrated 250.0 gram mass, weighed it on a laboratory balance, and found it read 266.5 g. What was the student

Kruka [31]

Answer:

B. 6.6%

Explanation:

The percentage error of a measurement can be calculated using the formula;

Percent error = (experimental value - accepted value / accepted value) × 100

In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.

Hence, the percentage error can be calculated thus;

Percent error = (266.5-250.0/250.0) × 100

Percent error = 16.5/250 × 100

Percent error = 0.066 × 100

Percent error = 6.6%

7 0

4 years ago

Match the vocabulary terms to their definitions.

Vilka [71]

See the attached picture:

8 0

3 years ago

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