Answer:
b. There will be more electrons emitted.
Explanation:
In photoelectric emission, the energy of the emitted electron is dependent on the frequency of the wave incident on the plate; but not the intensity. The rate of electron emission per unit time however depends on the intensity of the incident light. So increasing the intensity of the light at constant frequency will only affect the number of electrons emitted per unit time.
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?
This question is incomplete; here is the complete question:
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?
A. The wave has traveled 32.4 cm in 3 seconds.
B. The wave has traveled 32.4 cm in 9 seconds.
C. The wave has traveled 97.2 cm in 3 seconds.
D. The wave has traveled 97.2 cm in 1 second.
The answer to this question is D. The wave has traveled 97.2 cm in 1 second.
Explanation:
The frequency of a wave, which is in this case 3 hertz, represents the number of waves that go through a point during 1 second. According to this, if the frequency of the wave is 3 hertz this means in 1 second there were 3 waves. Moreover, if you multiply the wavelength (32.4cm) by the frequency (3) you will know the distance the wave traveled in 1 second: 32.4 x 3 = 97.2 cm. This makes option D the correct one as the distance in 1 second was 97.2 cm.
Answer:Is D!
Explanation:TEAT(Sorry) -_-*
A particular celestial body orbits at a particular speed. For every two orbits it makes, another celestial body orbits three times. This orbital resonance would correspond to which musical interval?
Answer:
Explanation:
frequency of first body f₁ = 2 / T where T is time taken by it for making two orbits
frequency of second body f₂ = 3 / T
ration of two frequency
f₁ / f₂ = 2 / 3
This ratio is called perfect fifth in musical interval .
A butterfly is flying around and its velocity(v) as a function of time(t) is given in the graph below where rightwards is the positive velocity direction. What is the butterfly's displacement x from t=2 to 4s? Answer with two significant digits.
Answer: 19 meters.
Explanation:
We want to find the total displacement between t = 2s and t = 4s.
To do it, we can integrate our function, first write our velocity equation.
for t ≤ 3s, we have a linear equation, let's write it:
A linear relationship can be written as:
y = a*t + b
where a is the slope and b is the y-axis intercept.
For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:
a = (y2 - y1)/(x2 - x1).
Now we can see that our line passes through the points (1, 0) and (0, -2)
then the slope is:
a = (0 -(-2)/(1 - 0) = 2/1 = 2
and knowing that when t = 0s, v(0s) = -2m/s, then our equation is:
v(t) = (2m/s^2)*t - 2m/s for t ≤ 3s
now, for t ≥3s the equation is constant, v(t) = 4m/s.
then we have
v(t) = (2m/s^2)*t - 2m/s -------if t ≤ 3s
v(t) = 4m/s ----- if t ≥ 3s
Now we integrate over time to get the position:
for t ≤ 3s we have:
p(t) = (1/2)*(2m/s^2)*t^2 - 2m/s*t + C
where C is a constant of integration, as we are calculating the displacement this constant actually does not matter, so we can use C = 0m
p(t) = (1m/s^2)*t^2 - 2m/s*t for t ≤ 3s
and p(3s) = (1m/s^2)*3s^2 - 2m/s*3s = 9m - 6m = 3m is the initial position of the other part of the function.
for t ≥ 3s we have:
p(t) = 4m/s*t + p(3s) = 4m/s*t + 3m
then the position equation is:
p(t) = (1m/s^2)*t^2 - 2m/s*t ---- t ≤ 3s
p(t) = 4m/s*t + 3m --- if t ≥ 3s
Now the displacement will be:
p(4s) - p(2s) where for each time, you need to use the correct function:
p(4s) = 4m/s*4s + 3m = 16m + 3m = 19m
p(2s) = (1m/s^2)*2s^2 - 2m/s*2s = 4m - 4m = 0m
p(4s) - p(2s) = 19m - 0m = 19m
The butterfly displacement x from t=2 to 4s is 19 meters.
What is displacement?The spacing between two specified points is represented by the one-dimensional quantity of displacement (symbolised as d or s), commonly known as length or distance.
The total displacement between t = 2s and t = 4s.
Integrate our function, the velocity equation.
for t ≤ 3s, we have a linear equation, let's write it:
A linear relationship can be written as:
y = a x t + b
where a is the slope and b is the y-axis intercept.
For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:
a = (y2 - y1)/(x2 - x1).
The line passes through the points (1, 0) and (0, -2)
The slope is:
a = (0 -(-2)/(1 - 0) = 2/1 = 2
When t = 0s, v(0s) = -2m/s, then our equation is:
v(t) = (2m/s²) x t - 2m/s for t ≤ 3s
now, for t ≥3s the equation is constant, v(t) = 4m/s.
v(t) = (2m/s²) x t - 2m/s -------if t ≤ 3s
v(t) = 4m/s ----- if t ≥ 3s
Now we integrate over time to get the position:
for t ≤ 3s we have:
p(t) = (1/2) x (2m/s²) x t^2 - 2m/s x t + C
where C is a constant of integration, to calculate the displacement this constant actually does not matter,
p(t) = (1m/s²)*t^2 - 2m/s x t for t ≤ 3s
and p(3s) = (1m/s^2) x 3s² - 2m/s x 3s = 9m - 6m = 3m is the initial position of the other part of the function.
for t ≥ 3s we have:
p(t) = 4m/s x t + p(3s) = 4m/s x t + 3m
then the position equation is:
p(t) = (1m/s^2) x t² - 2m/s x t ---- t ≤ 3s
p(t) = 4m/s x t + 3m --- if t ≥ 3s
Now the displacement will be:
p(4s) - p(2s) where for each time, you need to use the correct function:
p(4s) = 4m/s x 4s + 3m = 16m + 3m = 19m
p(2s) = (1m/s²) x 2s²- 2m/s x 2s = 4m - 4m = 0m
p(4s) - p(2s) = 19m - 0m = 19m
Thus, the displacement is 19 m.
To learn more about displacement, refer to the link:
https://brainly.com/question/11934397
#SPJ2
Monochromatic light of wavelength 649 nm is incident on a narrow slit. On a screen 2.25 m away, the distance between the second diffraction minimum and the central maximum is 1.99 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.
Answer:
a)0.51°
b)1.47×10^-4m
Explanation:
a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.
bsin(θ)= mλ.........................(*)
Where b= width of the slit
The distance on the screen from Central angle can be expressed as
Sin(θ)= y/d............. (**)
d and y is the horizontal distance between slit and screen
If we input eqn(**) into equation (*) we have
y= mλd/b................(z)
In order to find angle (θ) we have
(θ)= sin-(1.99×10^-2)/2.25
= 0.51°
Therefore, angle of diffraction θ of the second minimum is 0.51°
b)to find the width of the sloth using eqn(z) by substitute the values, we have
b= (2)(649×10^-9)(2.25)/1.99×10^-2
b= 1.47×10^-4m
Therefore, the width of the slit is 1.47×10^-4m
A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net force is applied?(1 point) The acceleration is double its original value. The acceleration is half of its original value. The acceleration is the square of its original value. The acceleration remains the same.
Answer:
The answer is The acceleration is double its original value.
Explanation:
It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.Hope this helps....
Have a nice day!!!!
Answer:
The acceleration is half of its original value
Explanation:
element X has two isotopes: X-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and X-29 has an atomic mass of 29.018 and a relative abundance of 17.67%. calculate the atomic mass of element X. show your work
Answer:
27.34 (no unit)
Explanation:
26.975*82.33%+29.018*17.67%
=27.34
Eli and Andy want to find out which of the two is stronger. Eli pushes a table with a force of 120 newtons while Andy pushes the table from the opposite side with a force of 125 newtons. Ignoring the masses of Eli and Andy, what is the resultant acceleration of the table if its mass is 10.0 kilograms?
Answer:
a = 0.5 m/s²
Explanation:
the type of problem is called a Newtons second law of motion.
and the equation would be the sum of F = m * a where m = mass and a = acceleration
forces are 125N and the opposite direction is 120N
Eli pushes the table with a force of 120N towards Andy
and
Andy pushes the table with a force of 125N towards Eli
mass of table given as 10 kg.
using the equation
120N - 125N = 10kg * a
a = (120-125) / 10
a = -0.5 m/s² so the acceleration is in the direction of Andy's force towards Eli.
therefore a = 0.5 m/s²
Answer:
B.
0.50 meters/second2
Explanation:
On the graph of voltage versus current, which line represents a 2.0 resistor?
Answer:
answer is B
Explanation:
Acc. to Ohm's Law:
V = I.R and therefore R=( V/I )
For line A, R = 6/2 = 3
line B, R = 4/2 = 2 ( for all the points plotted on this line the ratio of V and I is 2)
Line C = 2/2 = 1
Line D < 1
Hence, the correct answer is Line B :-)
in how many ways can five basketball players be placed in three postitions?
Answer:
Well if they playing a game like that
Calculate the current passing through a conductor of resistance 4ohms. If a potential difference of 15V its ends______
Explanation:
current = velocity/resistance
I = V/R
15/4
current = 3.75A
hope this helps...
Express the time spent to daily life activities during holidays as a fraction of whole day 1. Represent the fractions in pictorial form. 2. Write the fractions in simplest form. 3. Arrange them in ascending order.
Answer:
(II). The fractions in simplest form
[tex]\dfrac{31}{35}[/tex], [tex]\dfrac{91}{165}[/tex] and [tex]\dfrac{13}{5}[/tex]
(III). The fractions in ascending order
[tex]\dfrac{91}{165}<\dfrac{31}{35}<\dfrac{13}{5}[/tex]
Explanation:
Given that,
(I). Represent the fractions in pictorial form
(II). Write the fractions in simplest form.
(III). Arrange them in ascending order.
Suppose, The fractions in pictorial form
(a). [tex]\dfrac{3}{5}+\dfrac{2}{7}[/tex]
(b). [tex]\dfrac{9}{11}-\dfrac{4}{15}[/tex]
(c). [tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}[/tex]
We need to write in simplest form
Using given fraction
(a). [tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{3\times7}{5\times7}+\dfrac{2\times5}{7\times5}[/tex]
[tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21}{35}+\dfrac{10}{35}[/tex]
[tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21+10}{35}[/tex]
[tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}[/tex]
(b). [tex]\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{9\times15}{11\times15}-\dfrac{4\times11}{15\times11}[/tex]
[tex]\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{135-44}{165}[/tex]
[tex]\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{91}{165}[/tex]
(c). [tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{7}{10}+\dfrac{2\times2}{5\times2}+\dfrac{3\times5}{5\times2}[/tex]
[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{7}{10}+\dfrac{4+15}{10}[/tex]
[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{7+4+15}{10}[/tex]
[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{26}{10}[/tex]
[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{13}{5}[/tex]
We need to arrange them in ascending order
Using simplest form
The simplest fraction in ascending order
[tex]\dfrac{91}{165}<\dfrac{31}{35}<\dfrac{13}{5}[/tex]
Hence, This is required solution.
Question 14 of 30
A bundle of roofing shingles slides off a roof and is falling to the ground. As it
falls, what kind of energy does it possess?
O A. Kinetic only
O B. Potential only
O C. Radiant
D. Kinetic and potential
Answer:
kinetic and potential
Q1 b name the three tiny particles which make up atoms ? a write the charge alongside each one ?
Answer:
Protons
Neutrons
Electrons.
Protons + charge(positively charged)
Neurons no charge.
Electrons - ( negatively charged).
Explanation:
Atom is the smallest unit of a matter that constitutes chemical elements.
It is composed of three particles which are protons,neutrons and electrons.
Protons is positively charged,neuron has no charge and electrons is negatively charged.
Proton and neuron are found in the nucleus while electron is found outside the neuron.
you walk 6 block east, 2 blocks north, 3 blocks west and then 2 blocks north. the total distance you travel is blocks
Answer:
The answerI travel 13 blocksA student measures the depth of a water well with an adjustable frequency audio oscillator. 2 successive resonant frequencies are heard at 40Hz and 50Hz. What is the depth of the well?
Answer:
17.15 m
Explanation:
Assuming the well is empty (full of air instead of water), the speed of sound is v = 343 m/s.
The water well acts as a pipe closed at one end. Therefore, the depth of the water well must be an odd multiple of a quarter of the resonant wavelength.
L = (2n − 1) λ/4, n = 1, 2, 3, etc.
v = λf, so λ = v/f. Substituting:
L = (2n − 1) v/(4f)
Solving for frequency:
f = (2n − 1) v/(4L)
The difference between two successive resonant frequencies is therefore:
Δf = (2(n+1) − 1) v/(4L) − (2n − 1) v/(4L)
Δf = (2n + 1) v/(4L) − (2n − 1) v/(4L)
Δf = 2 v/(4L)
Δf = v/(2L)
Plugging in values:
50 Hz − 40 Hz = 343 m/s / (2L)
2L = 34.3 m
L = 17.15 m
an object weights 0.250 kgf in air 0.150 in water and 0.125 in an oil.find out the density of the object and the oil
Answer:
1) The density of the object = 2500 kg/m³
2) The density of the oil = 1250 kg/m³
Explanation:
1) The information relating to the question are;
The mass of the object in air = 0.250 kgf
The mass of the object in water = 0.150 kgf
The mass of the object in the oil = 0.125 kgf
By Archimedes's principle, we have;
The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced
The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf
∴ The weight of the water displaced = 0.1 kgf
Given that the object is completely immersed in the water, we have;
The volume of the water displaced = The volume of the object
The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)
The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³
The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³
The density of the object = 2500 kg/m³
2) Whereby the mass of the object in the oil = 0.125 kgf
The upthrust of the oil = The weight of the oil displaced
The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil
The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf
The weight of the oil displaced = The upthrust of the oil
Given that the volume of the oil displaced = The volume of the oil, we have;
The volume of the oil displaced = 0.0001 m³
The mass of the 0.0001 m³ = 0.125 kg
Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.
The density of the oil = 1250 kg/m³.
uestloh 1
Which is taller, a 20ft giraffe or a 240 inch pole?
(Show your conversion factor in fraction form!)
Answer:
Height of pole = Height of giraffe
Explanation:
Given:
Height of giraffe = 20 ft
Height of pole = 240 inch
Find:
Which is taller
Computation:
Height of giraffe = 20 ft
We know that 1 ft = 12 inch
So,
Height of giraffe = 20 × 12 inch
Height of giraffe = 240 inch
and
Height of pole = 240 inch
Height of pole = Height of giraffe
THE LENGTH OF A PENDULUM IS (1.5±0.01)m AND THE ACCELERATION DUE TO GRAVITY IS TAKEN AS (9.8±0.1)ms-² calculate the time period of the pendulum with uncertainty in it
Answer:
2.4583 ± 0.0207 seconds
Explanation:
The time period of a pendulum is approximately given by the formula ...
T = 2π√(L/g)
The maximum period will be achieved when length is longest and gravity is smallest:
Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds
The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:
Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds
If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.
T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2
T ≈ 2.4583 ± 0.0207 . . . seconds
__
We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.
Suppose an experiment is designed to test the durability of batteries in different conditions. All of the batteries tested are double-A (AA) Brand X. All sets of batteries are preconditioned in different environmental conditions for exactly 168 hours (1 week).
Set 1: 0°C (freezing point of water)
Set 2: 24°C (approximately room temperature)
Set 3: 37°C (approximately body temperature)
The batteries are then continuously used to power identical mechanical drummer toys. As long as the toy keeps drumming the battery is considered functional. The drumming time for each toy is measured as an indication of battery durability. In this experiment, which condition is not controlled?
A.) temperature
B.) brand of batteries
C.) test for durability
D.) type of battery (battery size)
Answer:
I assume its c. Since its talking about testing.
Explanation:
Answer:
The answer is test of durability
Explanation:
A copper-nickel alloy of composition 50 wt% Ni-50 wt% Cu is slowly heated from a temperature of 1200°C (2190 °F). (a) At what temperature does the first liquid phase form? (b) What is the composition of this liquid phase? (c) At what temperature does complete melting of the alloy occur? (d) What is the composition of the last solid remaining prior to complete melting?
Answer:
HELLO BELOW IS THE ATTACHED DIAGRAM USED TO ANSWER YOUR QUESTION AS IT WAS MISSING
A) 1270⁰c
B) 65%
C) 1320⁰c
D) 62%
Explanation:
Nickel alloy composition : 50 wt% Ni - 50 wt%
initial temperature = 1200⁰c = 2190⁰F
A) The temperature at which the first liquid phase form
from the attached diagram the temperature at which the first liquid if formed is 1270⁰c ( at point 2 )
B ) The composition of this liquid phase ( THE FIRST LIQUID )
the composition is found at point 3
wt % of Nickel = 35%, wt% of copper = 100 - 35 = 65%
C ) The temperature at which the alloy melts completely
from the attached diagram the temperature = 1320⁰c ( point 4 )
D) The composition of the last solid remaining prior to complete melting
this can be gotten at point 5 and calculated as
wt % of Ni = 62%
wt % of Cu = 100 - 62 = 38%
Hi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.
Answer:
The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.
hooke's law is described mathematically using the formula fsp=-kx. which statement is correct about spring force, fsp? A. It is always a positive force B. It is larger than the applied force C. It is the force doing the push or pull D. It is a vector quantity
Answer:
A. It is always a positive force
Explanation:
Hooke's law describes the relation between an applied force and extension ability of an elastic material. The law states that provided the elastic limit, e, of a material is not exceeded, the force, F, applied is proportional to the extension, x, provided temperature is constant.
i.e F = - kx
where k is the constant of proportionality, and the minus sign implies that the force is a restoring force.
The applied force can either be compressing or stretching force.
Describe the motion of water waves.
Answer:
Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.
A car accelerates at a rate of 3 m/s^2. If its original speed is 8 m/s, how many seconds will it take the car to reach a final speed of 25 m/s?
Answer:
[tex]\Large \boxed{\mathrm{5.67 \ seconds }}[/tex]
Explanation:
[tex]\displaystyle \mathrm{acceleration \ = \ \frac{final \ velocity - initial \ velocity }{elapsed \ time}}[/tex]
[tex]\displaystyle A \ = \ \frac{V_f - V_i }{t}[/tex]
[tex]\displaystyle 3 \ = \ \frac{25 - 8 }{t}[/tex]
[tex]\displaystyle 3 \ = \ \frac{17 }{t}[/tex]
[tex]\displaystyle t \ = \ \frac{17 }{3} \approx 5.67[/tex]
At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counselors measure the two "dry" legs of a right triangle. What is the length in meters of the swimming course in the figure below?
Answer:
47.17 m
Explanation:
From the diagram of the question attached, The length of the legs are 25 m and 40 m . This legs form a right angle triangle with the length of the swimming course (L).
Pythagoras theorem states that for a right angle triangle with hypotenuse a and legs b and c, then:
a² = b² + c²
In the triangle, the length of the swimming course (L) is the hypotenuse and the two legs are 25 m and 40 m. Using Pythagoras:
L² = 25² + 40²
L² = 625 + 1600
L² = 2225
L = √2225
L = 47.17 m
observe the virual skateboarder coming down the hill and over the ramp describe how each of newton’s laws of motion can be observed in this action you can choose the dry wet or muddy conditions or some combination of these
Answer:
first part the skater goes down a constant slope ramp, initially he has Newton's second law
pply Newton's third law, the normal is the reaction to the support of the body on the surface
the ramp shoots off. axis becomes zero and therefore with Newton's first law its speed
Explanation:
It is the description of this movement let's write Newton's laws.
* The first law that a body goes at constant speed or zero if the sum of the external forces is zero
* the second law is F = m a
* The third law states that the forces act in pairs of equal magnitude and opposite direction, one applied to each body.
Let's apply these laws to our case
In the first part the skater goes down a constant slope ramp, initially he has Newton's second law when he accelerates from the initial velocity of zero to a terminal velocity.
The expression for this is
Wₓ - fr = ma
W sin θ - μ W cos θ = m a
W = mg
g (sin θ - μ cos) = a
the value of the coefficient of kinetic friction depends on the condition of the surface, dry, wet or muddy
This is Newton's second law
On the Y axis, which is perpendicular to the ramp we have
N- [tex]W_{y}[/tex] = 0
If we apply Newton's third law, the normal is the reaction to the support of the body on the surface, note that it can be different from the weight.
In the second part when he is on the ramp.
In the ramp the skater enters with a speed v, suppose that the ramp has an incline so that the skater can jump, in this case the angle is positive with respect to the axis x
In this case the analysis is similar to the previous one
Newton's second law gives the acceleration of the skater, who when he reaches the end of the ramp shoots off.
At this point the force in the x (horizontal) axis becomes zero and therefore with Newton's first law its speed this axis remains constant and the force in the y axis is the force of gravity and has an acceleration that changes if velocity according to Newton's second law
Answer:look at explanations
Explanation:
(b) A cylinder of cross-sectional area 0.65m2 and
height 0.32m has a mass of 2. Ikg. If there is a
cavity inside, find the volume of the cavity.
(Density of cylinder = 11.053 kg/m^3)
Answer:
The volume of the cavity is 0.013m^3
Explanation:
To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:
Step one:
Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.
Volume of the cylinder = 2.1 / 11.053 =0.19[tex]m^{3}[/tex]
Step two:
From the volume of the cylinder, we can get the radius of the cylinder.
[tex]radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m[/tex]
Step three:
From the cross-sectional area, we can obtain the radius of the cavity.
Let the radius of the cavity be = r, while the radius of the cylinder be = R
CSA of cavity =
[tex]\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m[/tex]
Step Four:
calculate the volume of the cavity using volume =[tex]\pi r^2 \times h[/tex]
Recall that the cavity has the same height as the original cylinder
[tex]volume = \pi \times 0.115^2\times 0.32= 0.013m^3[/tex]
lab safety test Safety goggles and an apron must be worn when handling chemicals labeled corrosive because they:________.
Answer: React with the skin
Explanation:
lab safety test Safety goggles and an apron must be worn when handling chemicals labeled corrosive because they React with the skin
what are some factors that affect the frequency of sound
Answer:
1. direction of propagation of sound
2.medium through which sound trsnsmitted
An object of mass 25kg is at rest. What is its momentum ?
Answer:
[tex]\boxed{0}[/tex]
Explanation:
Momentum is the measure of mass in motion.
[tex]\sf momentum = mass \times velocity[/tex]
An object at rest has a velocity of 0.
[tex]p=mv[/tex]
[tex]p = 25 \times 0[/tex]
[tex]p=0[/tex]
The momentum of an object at rest is always 0.
