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PolarNik [594]
4 years ago
15

The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re

st. its speed on striking the ground is exactly 60 m/s. for what time interval was the ball falling?
Physics
1 answer:
gogolik [260]4 years ago
7 0
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is 

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s. 
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

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How electricity is generated in fossil fuel power stations
Leona [35]
Fossil fuel power plants burn carbon fuels such coal, oil or gas to generate steam that drives large turbines that produce electricity.
6 0
4 years ago
Calculate the wavelength of the electromagentic waves with the given frequencies, and determine the type of electromagnetic radi
ololo11 [35]

To develop this problem we require the concepts related to wavelength and its expression to calculate it.

The wavelength is given by

\lambda =\frac{c}{f}

Where,

c=3*10^8 m/s light velocity

f = frequency.

Our values are given by,

f_1=4.38*10^{14}Hz

f_2= 4.14*10^{20}Hz

f_3 = 3.24*10^{12}Hz

Then,

\lambda_1=\frac{3*10^8}{4.38*10^{14}}= 6.8493*10^{-7}m Visible

\lambda_2=\frac{3*10^8}{4.14*10^{20}}= 7.24*10^{-13}m Gamma Ray

\lambda_3=\frac{3*10^8}{3.24*10^{12}}= 9.259*10^{-5}m Infrared

<em>*Note the designation on the type of rays that are, can be found in consulted via On-line or in the optical books referring to the electromagnetic spectrum table with their respective ranges.</em>

5 0
3 years ago
Which property of sound waves decreases as the square of the distance from the source increases?
Vitek1552 [10]

Answer:

Intensity

Explanation:

The intensity of a sound wave is equal to the ratio between to the power emitted by the source divided by the area of the spherical surface through which the wave propagates:

I=\frac{P}{4\pi r^2}

where

P is the power

4\pi r^2 is the area of the spherical surface

r is the distance from the source

As we see from the formula, the intensity is inversely proportional to the square of the distance from the source:

I\propto \frac{1}{r^2}

so, intensity is the correct answer.

3 0
3 years ago
A spinning disc rotating at 130 rev/min slows and stops 31 s later. how many revolutions did the disc make during this time?
gayaneshka [121]
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
3 0
3 years ago
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
3 years ago
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