Answer:
The average force exerted by the water on the ground is 17.53 N.
Explanation:
Given;
mass flow rate of the water, m' = 135 kg/min
height of fall of the water, h = 3.1 m
the time taken for the water to fall to the ground;
[tex]h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s[/tex]
mass of the water;
[tex]m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg[/tex]
the average force exerted by the water on the ground;
F = mg
F = 1.789 x 9.8
F = 17.53 N
Therefore, the average force exerted by the water on the ground is 17.53 N.
An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off in the form of light and heatenergy, calculate the voltage drop across the bulb.
Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:
[tex]V=\frac{\Delta W}{\Delta q}[/tex]
Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq
[tex]I=\frac{\Delta q}{\Delta t}[/tex]
Where:
I is the current (2 A)
Δt is the time (20 s)
[tex]2=\frac{\Delta q}{20}[/tex]
[tex]q=40 C[/tex]
Then, we can put this value of charge in the voltage equation.
[tex]V=\frac{4600}{40}=115 V[/tex]
Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
Scientists create models to better understand Earth. Which evidence has led scientists to conclude that there are different layers within Earth's interior?
A.analysis of seismic wave data
B.measurement of Earth's diameter
C.temperatures taken within each layer
D.rock samples taken from Earth's core
Answer:
it is A or D
Explanation:
Answer:
ANswer:A
Explanation:
If you travel from Tucson to Argentina, you will see some different constellations in the night sky. true or false
Answer:
its true!!
Explanation: have a nice day !!
Which is greater, the energy of one photon of orange light or the energy of one quantum ofradiation having a wavelength of 3.36 * 10^-9
The question is incomplete, here is the complete question:
Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]
Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
Explanation:
To calculate the energy of one photon, we use the Planck's equation:
[tex]E=\frac{N_Ahc}{\lambda}[/tex]
where,
E = energy of radiation
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\lambda}[/tex] = wavelength of radiation
For orange light:For 1 photon, the term [tex]N_A[/tex] does not appear
[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex] (Conversion factor: [tex]1nm=10^{-9}m[/tex] )
Putting values in above equation, we get:
[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]
For one quantum of radiation:[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]
Putting values in above equation, we get:
[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]
Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
A 22.0 kg child is riding a playground merry-go- round that is rotating at 40.0 rev/min. What centripetal force must
Answer:
F = 482.51 N
Explanation:
Given that,
Mass of a child, m = 22 kg
Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min[/tex]
Let the radius of the path, r = 1.25 m
We need to find the centripetal force acting on the child. The formula for the centripetal force is given by :
[tex]F=m\omega^2r\\\\=22\times (4.18879)^2\times 1.25\\\\=482.51\ N[/tex]
So, the required centripetal force is 482.51 N.
Brainliest brainliest help help help mememememememme
Answer:
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Explanation:
I need points sorry
Answer:
honestly this was so long ago can i get brainliest i need 2 more until i am at expert level
Explanation:
The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to travel 2 m from your toes to your brain.
Answer:
t = 0.196 s
Explanation:
The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement
v = x / t
t = x / v
calculate
t = 2/102
t = 0.196 s
A hammer strikes a nail with a 10 N force for 0.01 seconds. Calculate the impulse of the hammer.
Answer:
0.1Ns
Explanation:
Impulse is the product of Force and time
Impulse = Force * Time
Given
Force = 10N
Time = 0.01s
Substitute into the formula
Impulse = 10 * 0.01
Impulse = 10 * 1/100
Impulse = 10/100
Impulse = 0.1Ns
hence the impulse of the hammer is 0.1Ns
An inductor is connected to a 120-V, 60-Hz supply. The current in the circuit is 2.4 A. What is the inductive reactance
Answer:
Inductive reactance is 50.00 ohms
Explanation:
Given the following data;
Voltage = 120v
Frequency = 60Hz
Current = 2.4 A
To find the inductive reactance;
Inductive reactance, XL = V/I
Where;
XL represents the inductive reactance. V represents the voltage. I represents the current.Substituting into the equation, we have;
XL = 120/2.4
XL = 50.00 ohms
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?
Answer:
Average speed = 10,000 m/s
Explanation:
Given the following data;
Distance = 2m
Time = 0.0002secs
To find the average speed;
Average speed = distance/time
Average speed = 2/0.0002
Average speed = 10,000 m/s
Therefore, the average speed of the
electron is 10,000 meters per seconds.
2. A 2500 kg car is slowed down uniformly from an initial velocity of 20.0 m/s to
the north by a 6250 N braking force acting opposite the car's motion. Use the
impulse-momentum theorem to answer the following questions:
a. What is the car's velocity after 2.50 s?
b. How far does the car move during 2.50 s?
c. How long does it take the car to come to a complete stop?
Answer:
13.75m/s; 42.2m; 8s
Explanation:
(a) the car's velocity after 2.50 s is 13.75 m/s
(b) The distance traveled by the car is 42.18 m
(c) the time taken for the car to come to complete stop is 8 s.
The given parameters;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
breaking applied on the car, f = 6250 N
The acceleration of the car is calculated as follows;
[tex]F = ma \\\\a = \frac{F}{m} = \frac{6250}{2500} = 2.5 \ m/s^2[/tex]
(a) Using impulse-momentum theorem, the car's velocity after 2.5 s is calculated as follows;
[tex]F = \frac{m(u-v)}{t} \\\\m(u-v) = Ft\\\\u-v = \frac{Ft}{m} \\\\v = u - \frac{Ft}{m} \\\\v = 20 - \frac{6250 \times 2.5}{2500} \\\\v = 13.75 \ m/s[/tex]
(b) The distance traveled by the car during the 2.5 s;
[tex]v^2 = u^2 - 2as\\\\2as = u^2 - v^2\\\\s = \frac{u^2 - v^2}{2a} \\\\s = \frac{20^2 - 13.75^2}{2\times 2.5} \\\\s = 42.18 \ m[/tex]
(c) The time taken for the car to come to a complete stop;
when the car stop's the final velocity, v = 0
v = u - at
0 = 20 - 2.5t
2.5t = 20
[tex]t = \frac{20}{2.5} \\\\t = 8 \ s[/tex]
Thus, the time taken for the car to come to complete stop is 8 s.
Learn more here: https://brainly.com/question/14559060
Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone of frequency f, at a speed of 10 m/s directly towards Jason, but his aim is a bit off. As a result, Jason runs forward towards the speaker at a speed of 6 m/s before catching it. Then, the frequency that Jason hears while running can be written as (m/n)f Hz, where m and n are relatively prime positive integers. Compute m n.
Answer:
Explanation:
We shall apply Doppler's effect of sound .
speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .
apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]
V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .
Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f
apparent frequency = [tex]f\times \frac{340+6}{340-10}[/tex]
= [tex]f\times \frac{346}{330}[/tex]
So m = 346 , n = 330 .