This question is incomplete, the complete question is;
Styrofoam has a density of 32kg/m³.
What is the maximum mass that can hang without sinking from a 50-cm diameter Styrofoam sphere in water
Answer: the required maximum mass is 63.3459 kg
Explanation:
Given that;
diameter d = 50 cm = 0.5 m
radius r = 0.5 / 2 = 0.25 m
Styrofoam density = 32kg/m³
Volume of sphere V = 4/3 πr³.
we substitute
V = 4/3 π(0.25)³
V = 0.06544 m³
Mass of sphere M = Volume × Density
we substitute
Mass of sphere M = 0.06544 m³ × 32kg/m³
Mass of Sphere M = 2.09408 kg
Mass of the water it displace will be;
volume × density
we know that density of water = 1000 kg/m³
so
Mass of the water it displace = 0.06544 m³ × 1000kg/m³
= 65.44 kg
Now the difference between mass of water and mass of the Styrofoam will be the amount of mass that the sphere can support
so
65.44 kg - 2.09408 kg = 63.3459 kg
Therefore, the required maximum mass is 63.3459 kg
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:
Explain how scientific knowledge develops through making observations about the natural world. FOR SCIENCE!
Answer number four because it is one of the factors evaluating science
A projectile is launched at an angle of 60° from the horizontal and at a velocity of
12.0 m/s. What is the horizontal velocity of the projectile? *
Answer:
60*12.0= 720 = v/60 * 12.0 squared which is 1,728
Explanation:
Horizontal velocity component: Vx = V * cos(α)
The horizontal velocity of the projectile is 6.0m/s
If a projectile is launched at an angle from the horizontal and at a velocity v, the horizontal velocity of the projectile is expressed as:
[tex]v_x =vcos \theta[/tex]
Given the following parameters
v = 12.0m/s
[tex]\theta=60^0[/tex]
Substitute the given parameters into the formula to have:
[tex]v_x=12.0cos60\\v_x=12.0(0.5)\\v_x=6.0m/s\\[/tex]
Hence the horizontal velocity of the projectile is 6.0m/s
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Sound waves are simultaneously interpreted through air (going through the external,
middle, then inner ear) and through the bone in the inner ear, stimulating the
acoustic nerve.
True
False
Answer:
its true
Explanation:
i took it
When the 5.0 kg cylinder fell 100 m, the final temperature of the water was °C and the change in temperature was °C.
When the 5.0 kg cylinder fell 500 m, the final temperature of the water was °C and the change in temperature was °C.
Answer:
26.17
1.17
30.86
5.86
Explanation:
said they were correct on my assignment
g Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of 0.624c and determines its lifetime to be 159 ns. (a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory
Answer:
[tex]D=29.7648m[/tex]
Explanation:
From the question we are told that
Speed of the particle[tex]V=0.624c[/tex]
Lifetime [tex]t=159ns[/tex]
Generally the equation for distance is mathematically given by
[tex]D=Vt[/tex]
[tex]D=0.624c*159*10^-^9[/tex]
[tex]D=0.624*3*10^8*159*10^-^9[/tex]
[tex]D=29.7648m[/tex]
Therefore the markers are [tex]D=29.7648m[/tex] apart
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 .
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.
A standard 1 kilogram weight is a cylinder 47.0 mm in height and 49.0 mm in diameter. What is the density of the material
Answer:
11,627.91 kg/m^3
Explanation:
The computation of the density of the material is shown below
Given that
Mass , m = 1 kg
Height, H - 47 mm = 0.047 m
Diameter, d = 49 mm = 0.049 m
Now radius, r = D ÷ 2
= 0.049 ÷ 2
= 0.0245
Volume = πr^2h
= 3.14 × (0.0245)^2 × 0.047
= 0.000086m^3
Now the density of the material is
= mass ÷ volume
= 1÷ 0.000086m^3
= 11,627.91 kg/m^3
Answer: 11,282.8 kg/m^3
Explanation:
We know:
Density = mass/volume
1 kg = mass
(π)(r^2)(h) = volume
Density = m/(πr^2h)
Plug in your values:
Density = 1kg/(π)(24.5 mm)^2(47 mm)
^we have to cut 49 (the diameter) in half because the equation calls for radius
Remember to multiply the density by 10^9 mm/ 1 m because the answer asks for kg/m^3
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.
An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the motor radius is 7.165 cm, what is the tangential component of acceleration at the edge of the motor at t = 1.5 seconds? (b) How long will it take, in seconds, to decrease its angular velocity by 75%?
Answer:
a) The tangential component of acceleration at the edge of the motor at [tex]t = 1.5\,s[/tex] is -1.075 meters per square second.
b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.
Explanation:
The angular aceleration of the electric motor ([tex]\alpha[/tex]), measured in radians per square second, as a function of time ([tex]t[/tex]), measured in seconds, is determined by the following formula:
[tex]\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right][/tex] (1)
The function for the angular velocity of the electric motor ([tex]\omega[/tex]), measured in radians per second, is found by integration:
[tex]\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right][/tex] (2)
Where [tex]\omega_{o}[/tex] is the initial angular velocity, measured in radians per second.
a) The tangential component of aceleration ([tex]a_{t}[/tex]), measured in meters per square second, is defined by the following formula:
[tex]a_{t} = R\cdot \alpha[/tex] (3)
Where [tex]R[/tex] is the radius of the electric motor, measured in meters.
If we know that [tex]R = 7.165\times 10^{-2}\,m[/tex], [tex]\alpha = 10\cdot t[/tex] and [tex]t = 1.5\,s[/tex], then the tangential component of the acceleration at the edge of the motor is:
[tex]a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)[/tex]
[tex]a_{t} = -1.075\, \frac{m}{s^{2}}[/tex]
The tangential component of acceleration at the edge of the motor at [tex]t = 1.5\,s[/tex] is -1.075 meters per square second.
b) If we know that [tex]\omega_{o} = 104.720\,\frac{rad}{s}[/tex] and [tex]\omega = 26.180\,\frac{rad}{s}[/tex], then the time needed is:
[tex]26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}[/tex]
[tex]5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}[/tex]
[tex]t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5}[/tex]
[tex]t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5} }[/tex]
[tex]t \approx 3.963\,s[/tex]
The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.
What can we conclude from observing an attractive force between a positively charged rod and some object
Answer: that the object is negatively charged.
Explanation:
We know that the force between objects that have the same type of charge is a repulsive force, while for objects with an opposite charge, the force is attractive.
In this case, we know that we have an attractive force between an object and a positively charged rod.
Then the only conclusion we can take in this situation is that the object is negatively charged.
If an ice cube has a mass of 5g and is left in the sun in a beaker, what will happen to its mass?
Answer:
The mass remains the same
Explanation:
An ice cube of mass 5g left in the sun in a beaker will have no mass difference.
For this system, mass is conserved.
The ice cube begins to melt and the state changes. The state change is from solid to liquid. Since no mass is loss, the mass of the liquid formed and the ice will remain the same. In retrospect, this physical change ensures that mass is conserved.Calculate the momentum of a 48.5 kg passenger seated on a train that is stopped.
Answer:
0kgm/s
Explanation:
Given parameters:
Mass of the passenger = 48.5kg
Unknown:
Momentum = ?
Solution:
The momentum of a body is the quantity of motion a body possesses. A body not in motion has no momentum.
Mathematically;
Momentum = mass x velocity
So;
Since the train is stopped, velocity = 0m/s
Momentum = 48.5 x 0 = 0kgm/s
A piece of putty and a tennis ball with the same mass are thrown against a wall with the same velocity. Which object experiences a greater force from the wall or are the forces equal
Answer:
Explanation:
Firstly, we have to define momentum.
Momentum is define as the product of mass and velocity.
That is P = mass×velocity
Also considering the third law of motion which states that: For every action, there is equal and opposite reaction.
Moreso, considering the 2nd law of motion which states that the rate of change in the momentum of a body is equal to the applied force and takes place in the direction of the applied force.
Now, applying P = mass×velocity
They both have same mass and velocity definitely, they will both experience same momentum.
Also from the question, the both share same velocity hence, the will both hit the wall with same velocity meaning the will both feel the same impact from the wall as well. Hence the third law of motion proves this right.
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!
if a tomato is considered a fruit... then is ketchup considered a smoothie?
Answer:
depends... do you add suger to your ketchup?
Explanation:
How is property belonging to individuals protected under the Fifth Amendment?
By preventing the government from taking property without fair payment.
By preventing a police search of private property without a warrant
By prohibiting the quartering of soldiers without permission
By guaranteeing every citizen the right to apply for a home loan
Answer:
By preventing the government from taking property without fair payment.
Explanation:
Answer 1
Please mark me as brilliant and thank you
Answer:
this is under the wrong section
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
why is red light used in photographic dark rooms
Photographic paper reacts to light to produce the image taken by a camera and stored on film. Darkrooms used red lighting to allow photographers to control light carefully, so that light-sensitive photographic paper would not become overexposed and ruin the pictures during the developing process.
Answer:
Photographic paper reacts to light to produce the image taken by a camera and stored on film.
Explanation:
Darkrooms used red lighting to allow photographers to control light carefully so that light-sensitive photographic paper would not become overexposed and ruin the pictures during the developing process.
Brainliest brainliest help help help mememememememme
Answer:
????????????????????,
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:
A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.3 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
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 !!
Potential energy diagrams for a particle are useful in predicting the motion of that particle.
a. True
b. False
Answer:
True
Explanation:
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
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
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.
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A person runs up 33 stairs of height 0.25 m. The person weighs 599.85 N. How much energy does it take for the person to reach the top of the stairs? I need the answer ASAP
Answer:
149.96J
Explanation:
Given parameters:
Number of stairs = 33
Height = 0.25m
Weight = 599.85N
Unknown:
Energy the person takes to reach the top = ?
Solution:
To solve this problem;
The energy it takes is given as:
Energy = weight x height
Energy = mass x gravity x height
Now,
Insert the given parameters and solve;
Energy = 599.85 x 0.25 = 149.96J
If a man traveled to a different planet
Answer:
He would be in space.
Explanation:
Answer: What would happen if a human traveled to Uranus?
As an ice giant, Uranus doesn't have a true surface. The planet is mostly swirling fluids. While a spacecraft would have nowhere to land on Uranus, it wouldn't be able to fly through its atmosphere unscathed either. The extreme pressures and temperatures would destroy a metal spacecraft.
Explanation:
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.