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3 years ago

If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is

Physics

1 answer:

expeople1 [14]3 years ago

8 0

Answer:

We can use 2 g H = v2^2 - v1^2 or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

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A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

amm1812

-- Equations #2 and #6 are both the same equation,
and are both correct.

-- If you divide each side by 'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by 'frequency', you get Equation #3,
which is also correct.
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

7 0

3 years ago

Read 2 more answers

A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the magnitude of the displacement (on a

sattari [20]

Answer:

16 km

Explanation:

Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.

The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.

11^2 + 11^2 = c^2

121 + 121 = c^2

242 = c^2

c = 15.56

Rounding the answer makes it 16 km for the hiker's magnitude of displacement.

5 0

3 years ago

If a girl is standing in front of a smooth surface from which a sound is reflected the girl may hear

Harlamova29_29 [7]

If the girl is also near the source of the sound, two alike sets of sounds will be heard.

7 0

3 years ago

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You are traveling at 14 m/s for 20 seconds. What is your displacement?

enyata [817]

Velocity = 14 m/s

Time = 20 s

Displacement = Velocity×Time = (14×20) m = 280 m

The displacement is 280 m towards the direction of motion.

5 0

3 years ago

Read 2 more answers