Answer:he conducts the investigation to see the effect acidic water has on limestone
Explanation:
A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are crossed electric and magnetic fields. If the electric field magnitude is E=450E=450 V/m in the negative yy ‑direction and the protons move at a constant speed of v=7.9×105v=7.9×105 m/s, what must the direction and magnitude of the magnetic field be in order for the beam of protons to continue undeflected along its straight-line trajectory? Select the direction of the magnetic field BB .
Answer:
The magnitude is [tex]B = \frac{450}{7.9* 10^5}[/tex]
The direction is the positive x axis
Explanation:
From the question we are told that
The electric field is E = 450 V/m in the negative y ‑direction
The speed of the proton is [tex]v= 7.9* 10^5\ m/s[/tex] in the positive z direction
Generally the overall force acting on the proton is mathematical represented as
[tex]F_E = q(\vec E + \vec v * \vec B)[/tex]
Now for the beam of protons to continue un-deflected along its straight-line trajectory then [tex]F_E =0[/tex]
So
[tex] 0 = q( E (-y) + v(z) * \vec B)[/tex]
=> [tex]E\^y = v \^ z * \vec B[/tex]
Generally from unit vector cross product vector multiplication
[tex]\^ z \ * \ \^ x = \^ y[/tex]
So the direction of B (magnetic field must be in the positive x -axis )
So
[tex]E\^y = v \^ z * B\^ x [/tex]
=> [tex]E\^y = vB ( \^ z * \^ x) [/tex]
=> [tex]E\^y = vB ( \^y) [/tex]
=> [tex]E = vB [/tex]
=> [tex]B = \frac{450}{7.9* 10^5}[/tex]
=> [tex]B = 0.0005696 \ T [/tex]
Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answer the following questions. You hold a small metal ball of mass a height above the floor. You let go, and the ball falls to the floor. Choose the origin of the coordinate system to be on the floor where the ball hits, with up as usual. Express all results in terms of , , and . Just after release, what are and
Answer:
y(i) = h
v(y.i) = 0
Explanation:
See attachment for elaboration
The sun produces energy via nuclear fusion at the rate of 4×1026 J/s . Based on the proposed overall fusion equation, how long will the sun shine in years before it exhausts its hydrogen fuel? (Assume that there are 365 days in the average year.)
Answer:
2 x 10^10 years.
Explanation:
Given that the sun produces energy via nuclear fusion at the rate of 4×1026 J/s . Based on the proposed overall fusion equation, how long will the sun shine in years before it exhausts its hydrogen fuel? (Assume that there are 365 days in the average year.)
Let us first calculate energy from hydrogen gas.
4(1.007825) + 2(0.00549) - 4.002603 = 0.029795 amu
Since 4.0313 amu H+ > 0.029795 amu = ratio of 1 to 0.00739
dE = ((2*10^30 kg) x 0.8 x 0.25)(3.00x10^8 m/s)^2 = 3.6x10^46 kg x m^2/s^2
3.6x10^46 kg x m^2/s^2 x 0.00739 = 2.6604 x 60^44 kg x m^2/s^2
4x10^26 J/s x s = 2.6604x10^44 kg x m^2/s^2 solve for s.
s = 6.651x10^17 seconds
6.651x10^17 seconds x 1 min/60 s x 1 hr/ 60 min x 1 day/ 24 hr x 1 yr / 365 days = 2.1x10^10 years
Based on the proposed overall fusion equation, it will therefore take the sun shine 2 x 10^10 years before it exhausts its hydrogen fuel.
1 identify two ways (factors) you can change in order to the charge the objects experience
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attached file:
In point (a), The one factor is the charge on each of the objects. In point (b), at this point, the second factor is used as the distance between the charges.The two ways to charge an object are Friction and Induction.
Charging an object:To charge and object generally we can think of three ways, Friction, Conduction, and Induction. Here I am going to explain charging by friction and induction.
(i) Charging by Friction:
In this case, two objects are rubbed against each other. The friction between the two objects creates enough energy for the transfer of charges ( electrons) from one object's surface to the other. So the imbalance in the electrons produces a charge on the objects.
(ii) Charging by induction:
In this case, an already charged object is brought near a neutral object but there is no contact between the objects. The charged object attracts or repels the electrons of the neutral object, depending upon the type of charge it has. Thus creating a charge imbalance in the neutral object.
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what is a proper way and a safe way to dispose batteries
Answer:
throw them away............
A 66-N ⋅ m torque acts on a wheel with a moment of inertia 175 kg ⋅ m2. If the wheel starts from rest, how long will it take the wheel to make one revolution?
Answer:
t = 5.77 s
Explanation:
This exercise will use Newton's second law for rotational motion
τ = I α
α = τ / I
α = 66/175
α = 0.3771 rad/s²
now we can use the rotational kinematics relations, remember that all angles must be in radians
θ = 1 rev = 2π radians
θ = w₀ t + ½ α t²
as the wheel starts from rest w₀ = 0
t = √ (2θ/α)
let's calculate
t = √ (2 2π / 0.3771)
t = 5.77 s
The volume of a cube is found by multiplying length times
width times height. If an object has a volume of 1.44 m?,
what is the volume in cubic centimeters? Remember to mul-
tiply each side by the conversion factor.
Answer:
The volume in cubic centimeters is 1,440,000
Explanation:
You know that an object has a volume of 1.44 m³ and you want to calculate the volume in cubic centimeters cm³.
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other), the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
So, [tex]x=\frac{c*b}{a}[/tex]
Since 1 m³ is equal to 1,000,000 cm³ ( this is the conversion factor) , then you can apply the following rule of three: if 1 m³ is equal to 1,000,000 cm³, 1.44 m³ is equal to how many cm³?
[tex]cm^{3} =\frac{1.44m^{3}*1,000,000 cm^{3} }{1m^{3} }[/tex]
cm³=1,440,000
The volume in cubic centimeters is 1,440,000
If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p
Answer:
8.93*10^13 N.
Explanation:
Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:[tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]
where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:[tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]
Fg = 8.93*10^13 N.m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27 degC and the temperature of the water is 50 degC . The electrical resistor transfers 100 kilojoule to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature, in ∘C.
Answer:
T₂ = 49.3°C
Explanation:
Applying law of conservation of energy to the system we get the following equation:
Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water
E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)
where,
E = Energy Supplied by Resistor = 100 KJ = 100000 J
m = mass of copper tank = 13 kg
C = Specific Heat of Copper = 385 J/kg.°C
T₂ = Final Temperature of Copper Tank
T₁ = Initial Temperature of Copper Tank = 27°C
T'₂ = Final Temperature of Water
T'₁ = Initial Temperature of Water = 50°C
m' = Mass of Water = 4 kg
C' = Specific Heat of Water = 4179.6 K/kg.°C
Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂
Therefore,
(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)
100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J
100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂
(1071055 J)/(21723.4 J/°C) = T₂
T₂ = 49.3°C
PLEASE HELP 25 POINTS WHAT IS THE ANSWER
The answer is letter C vro
thanks me.later
Objects want to keep doing the same thing is a way of stating ....
Answer:
Objects want to keep doing the same thing is a way of stating Newtons First Law.
The correct option is (a) inertia. "Objects want to keep doing the same thing" it implies a connection to the principle of inertia in physics. Inertia is the property of matter that resists changes in its state of motion or rest.
According to Newton's first law of motion, an object will remain at rest or continue moving in a straight line at a constant velocity unless acted upon by an external force. This property is commonly known as "the law of inertia." In other words, objects tend to maintain their current state of motion (whether at rest or in motion) unless influenced by an external force.
When we say "Objects want to keep doing the same thing," we're drawing an analogy between this scientific principle and human behavior. It suggests that objects, like humans, have a natural inclination to resist change and continue their current course of action. This analogy helps convey the idea that objects exhibit a tendency to persist in their existing state.
"Objects want to keep doing the same thing" to inertia emphasizes the notion that objects, like humans, tend to maintain their current state of motion or rest unless influenced by an external force.
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The correct question is:
Objects want to keep doing the same thing is a way of stating ......
a) inertia
b) force
c) power
d) moment
Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years
Answer:
T = 1.4696 10⁴ years
Explanation:
For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law
F = ma
G m M / r² = m a_c = m v² / r
G M / r = v²
the speed of the circular orbit is
v = 2π r / T
we substitute
G M / r = 4π² r² / T²
T² = (4π² / G M) r³
Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse
T² = (4π² /GM) a³
the constant is worth
(4π² / GM) = 2.97 10⁻¹⁹ s² / m³
let's reduce the distance to SI units
AU is the distance from the Earth to the Sun
a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)
a = 8.976 10¹³ m
T² = 2.97 10⁻¹⁹ (8.976 10¹³)³
T² = 21.4786 10²²
T = 4.63 10¹¹ s
Let's reduce to years
T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)
T = 1.4696 10⁴ years
Consider the four scenarios involving visible light. In scenario A, visible light has a wavelength of 715.3 nm. Determine its frequency, energy per photon, and color.
Answer:
1. frequency (f) = 4.194 × 10^14Hz
2. Energy of photon (E) = 2.779 × 10^-19 J
3. Color is RED
Explanation:
1. Based on the information in this question, a visible light has a wavelength of 715.3 nm.
λ = v/f
Where; λ = wavelength (nm)
v = speed of light (3 × 10^8m/s)
f = frequency (Hz)
f = v/λ
f = 3 × 10^8/715.3 × 10^-9
f = 0.004194 × 10^(8+9)
f = 0.004194 × 10^17
f = 4.194 × 10^14Hz
2. Energy per photon is calculated thus;
E = hf
Where; E = energy of photon (J)
h = Planck's constant (6.626 × 10^-34 J/s)
f = frequency (Hz)
E = 6.626 × 10^-34 × 4.194 × 10^14
E = 27.789 × 10^(-34+14)
E = 27.789 × 10^-20
E = 2.779 × 10^-19 J
3. Based on the wavelength range of the visible spectrum, wavelength of 715.3nm falls between 625-740nm, which is RED color of light. Hence, the color of the visible light is RED.
The frequency is "[tex]419.5\times 10^{12} \ s^{-1}[/tex]", energy per photon is "[tex]2.780\times 10^{-19} \ J[/tex]" and the color is "red".
Given:
Wavelength,
[tex]\lambda = 715.1 \ nm[/tex]or,
[tex]= 715.1 \ nm\times (10^-9 \ m/nm)[/tex]
As we know the formula,
→ [tex]Frequency = \frac{Speed \ of \ light}{wavelength}[/tex]
By substituting the values, we get
[tex]= \frac{(3\times 10^8) }{715.1\times 10^{-9} }[/tex]
[tex]= 419.5\times 10^{12} \ s^{-1}[/tex]
Now,
The energy per photon will be:
→ [tex]E = Planck's \ constant\times Frequency[/tex]
[tex]= 6.626\times 10^{-34}\times 419.5\times 10^{12}[/tex]
[tex]= 2.780\times 10^{-19} \ J[/tex]
Thus the above answer is correct.
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until a train is a safe distance from the station, it must travel at 5 m/s. once the train is on open track, it can speed up to 45m/s. if it takes 8 seconds to reach 4 m/s, what is the acceleration of the train?
Answer:
5 meters per second squared
Explanation:
We calculate the acceleration using the formula:
a = (vf - vi) / t
where "vf" is the final velocity, "vi" the initial velocity, and "t" the time it took to change from the initial velocity to the final one.
In our case:
a = (45 - 5) / 8 = 40 / 8 = 5 m/s^2
If a car is traveling at an average speed of 70 kilometers per hour how long does it take the car to travel 14 kilometers
Answer:
Explanation:
O.20 hour A
If a car is traveling at an average speed of 70 kilometers per hour, 0.2 hours it takes the car to travel 14 kilometers.
What is average speed?By multiplying the distance that an item travels in one unit by the amount of time it takes to go that distance, one may determine the speed of the object. The speed of the item on this voyage, denoted by the letter "s," is equal to s = D/T if "D" is indeed the distance traveled in certain time "T."
Understanding average speed will help you better comprehend the pace of a travel. On a travel, the pace could occasionally change. Knowing the average speed then becomes crucial to getting an idea of how quickly the route will be finished.
Distance covered = average speed × Time travelled
14=70× Time travelled
Time travelled = 0.2 hours
Therefore, 0.2 hours it takes the car to travel 14 kilometers.
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A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?
Answer:
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) E=0
Explanation:
Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.
With this analysis we can answer the specific questions
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside
d) If it is very reasonable and this system configuration is called a Faraday Cage
e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.
Basalt is a rock that cooled quickly after lava erupted through a volcano.
What is the best description of its texture?
when the lava erupted from the volcano basalt was the rock that cooled quickly or basalt was the only one rock that cooled quickly after the lava erupted from the volcano
Explanation:
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In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.9 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 16.0 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?
Answer:
The velocity is [tex]v_b = 20.17 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 0.245 \ kg[/tex]
The radius is [tex]r = 59.8 \ cm = 0.598 \ m[/tex]
The force is [tex]F = 30.9 \ N[/tex]
The speed of the ball is [tex]v = 16.0 \ m/s.[/tex]
Generally the kinetic energy at the top of the circle is mathematically represented as
[tex]K_t = \frac{1}{2} * m * v^2[/tex]
=> [tex]K_t = \frac{1}{2} * 0.245 * 16.0 ^2[/tex]
=> [tex]K_t = 31.36 \ J[/tex]
Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as
[tex]W = F * d[/tex]
Here d is the length of a semi - circular arc which is mathematically represented as
[tex]d = \pi * r[/tex]
So
[tex]W = 30.9 * 0.598 [/tex]
[tex]W = 18.48 \ J [/tex]
Generally the kinetic energy at the bottom is mathematically represented as
[tex]K_b = \frac{1}{2} * m * v_b^2[/tex]
=> [tex]K_b = \frac{1}{2} * 0.245 * v_b^2[/tex]
=> [tex]K_b = 0.1225 * v_b^2[/tex]
From the law of energy conservation
[tex]K_t + W =K_b[/tex]
=> [tex]31.36+ 18.48 = 0.1225 * v_b^2[/tex]
=> [tex]v_b = 20.17 \ m/s[/tex]
If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by the ordinary matter (Mord = 1.90 ✕ 1041 kg), find the speed of the star. Assume a circular orbit and assume all the Galaxy's matter is concentrated at the center.
Answer:
The value is [tex]v = 1.309*10^{5}\ m/s[/tex]
Explanation:
The radius is [tex]r = 7.40 *10^{20} \ m[/tex]
The mass of the ordinary matter is [tex]M_{rod} = 1.90 *10^{41}\ kg[/tex]
Generally the speed of the star is mathematically represented as
[tex]v = \sqrt{\frac{G * M}{r} }[/tex]
Here G is the gravitational constant with a value
[tex]G = 6.67384 * 10^{-11}[/tex]
So
[tex]v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }[/tex]
=> [tex]v = 1.309*10^{5}\ m/s[/tex]
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the projectile will be ("E4" means "*10^4") *
1 point
4E4 J
1E4 J
1E3 J
None of the above
Answer:
[tex]K=4\times 10^4\ J[/tex]
Explanation:
Given that,
Mass of cannon is,M = 2000 kg
Mass of projectile, m = 2 kg
Velocity of the projectile, v = 200 m/s
We need to find the kinetic energy of the projectile. We know that the formula for the kinetic energy of an object is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\\text{Putting all the values, we get}\\\\K=\dfrac{1}{2}\times 2\times (200)^2\\\\K=40000\ J[/tex]
or
[tex]K=4\times 10^4\ J[/tex]
So, the kinetic energy of the projectile will be [tex]4\times 10^4\ J[/tex].
why the bodies of water important for recreation
Explanation:
Recreational water activities can have substantial benefits to health and well-being. Swimming pools, beaches, lakes, rivers and spas provide environments for rest and relaxation, physical activity, exercise, pleasure and fun. Yet they also present risks to health.
pls make it the brainliest of it has helped you !!!!
A 100 kg man climbs the stairs at a football stadium 50 m tall in 45 s. What was the man's required power output?
.Answer:
1111.11 W
Explanation:
Power is the rate of doing work with respect to time, its S.I unit is in watts but it can also be expressed in J/s. Power is calculated using the formula:
[tex]Power=\frac{energy}{time}[/tex]
Power is also the rate at which energy is used per second.
Given that mass (m) = 100 kg, height (h) = 50 m, acceleration due to gravity (g) = 10 m/s and time = 45 s.
The man climb the stairs and experiences a potential energy, hence:
potential energy = mgh = 100 * 10 *50 = 50000 J,
[tex]Power=\frac{energy}{time}=\frac{50000}{45}=1111.11\ W[/tex]
1.
Which of these is a relative adverb?
where
lovely
who
Answer:
Where
Explanation:
An electromotive force E(t) = 200, 0 ≤ t ≤ 50 0, t > 50 is applied to an LR-series circuit in which the inductance is 50 henries and the resistance is 5 ohms. Find the current i(t) if i(0) = 0.
Answer:
The answer is below
Explanation:
An LR series circuit has a differential equation in the form of:
[tex]L\frac{di}{dt}+iR=E(t)\\ \\Given\ that\ 50H,R=5\ ohms,E(t)=200,for\ 0 \leq t \leq\ 50. Hence:\\\\50\frac{di}{dt}+5i =200\\\\\frac{di}{dt}+0.1i =4\\\\Solving\ the\ differential\ equation:\\\\The\ integrating\ factor(I)=e^{\int\limits {0.1} \, dt }=e^{0.1t}. The\ DE\ becomes:\\\\e^{0.1t}\frac{di}{dt}+e^{0.1t}(0.1i) =4e^{0.1t}\\\\e^{0.1t}i=\int\limits {4e^{0.1t}} \, dt \\\\e^{0.1t}i=40e^{0.1t}+A\\\\i(t)=40+Ae^{-0.1t}\\\\but\ i(0)=0\\\\0=40+Ae^{-0.1(0)}\\\\A=-40\\\\i(t)=40-40e^{-0.1t}\\\\[/tex]
At 50 seconds:
[tex]i(50)=40-40e^{-0.1*50}\\\\i(50)=40-40e^{-5}[/tex]
[tex]L\frac{di}{dt}+iR=E(t)\\ \\Given\ that\ 50H,R=5\ ohms,E(t)=0,for\ t> 50. Hence:\\\\50\frac{di}{dt}+5i =0\\\\\frac{di}{dt}+0.1i =0\\\\\frac{di}{dt}=-0.1i\\\\\frac{di}{i}=-0.1dt\\\\\int\limits {\frac{di}{i}} =\int\limits {-0.1} \, dt\\ \\ln(i)=-0.1t+A\\\\taking\ exponential:\\\\i=e^{-0.1t+A}\\\\i=e^{-0.1t}e^A\\\\i(t)=Ce^{-0.1t}\\\\i(50)=40-40e^{-5}=Ce^{-5}\\\\C=40(e^5-1)\\\\i(t)=40(e^5-1)e^{-0.1t}\\\\[/tex]
[tex]i(t)=\left \{ {{40-40e^{-0.1t}\ \ \ \ 0 \leq t \leq 50 } \atop {40(e^5-1)e^{-0.1t}\ \ \ \ t>50}} \right.[/tex]
Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days. If it were to undergo gravitational collapse to a neutron star of radius 13 km , losing three-quarters of its mass in the process, what would its rotation speed be
Answer:
Its rotation will be 3.89x10⁴ rad/s.
Explanation:
We can find the rotation speed by conservation of the angular momentum:
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}\omega_{i} = I_{f}\omega_{f} [/tex] (1)
The initial angular speed is:
[tex] \omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d} [/tex]
The moment of inertia (I) of a sphere is:
[tex] I = \frac{2}{5}mr^{2} [/tex] (2)
Where m is 9 times the sun's mass and r is the sun's radius
By entering equation (2) into (1) we have:
[tex] \frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f} [/tex]
[tex]9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}[/tex]
[tex]\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s[/tex]
Hence, its rotation will be 3.89x10⁴ rad/s.
I hope it helps you!
If the speed of a car is increased by 75%, by what factor will its minimum braking distance be increased, assuming all else is the same
Answer:
breaking distance will increase by a factor 3.0625
Explanation:
From Newton's equation of motion, we can say that;
v² = u² + 2as
Where initial velocity is zero, we have;
v² = 2as
s = v²/2a
Where s is the distance and v is the final speed.
We will assume the deceleration(negative acceleration) is the same value,
Now, we see that the distance is directly proportional to the square of the velocity.
Now, 75% increase in speed means it has increased by a factor of 1.75
Thus means;
1.75²v² is directly proportional to the 1.75²d
Distance =
1.75²v² = 3.0625d
Thus, breaking distance will increase by a factor 3.0625
What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75
Answer:
1800 m/s
Explanation:
The equation is v = fλ
λ= 0.75
f = 2400 Hz
V = 2400 × 0.75
V = 1800 m/s
[ you did not give units for wavelength, I assumed it would be m/s]
If a projectile hits a stationary target, and the projectile continues to travel in the same direction, the mass of the projectile is less than the mass of the target. the mass of the projectile is equal to the mass of the target. the mass of the projectile is greater than the mass of the target. nothing can be said about the masses of the projectile and target without further information. this is an unphysical situation and will not actually happen.
The correct arrangement of the question is;
If a projectile hits a stationary target, and the projectile continues to travel in the same direction,
A) the mass of the projectile is less than the mass of the target.
B) the mass of the projectile is equal to the mass of the target.
C) the mass of the projectile is greater than the mass of the target.
D) nothing can be said about the masses of the projectile and target without further information.
E) this is an unphysical situation and will not actually happen.
Answer:
Option C: The mass of the projectile is greater than the mass of the target.
Explanation:
We want to find what will happen when a projectile continues in motion after it hits a target.
Now, for the projectile to keep moving in that direction after it hits the target, it means it had a force bigger than the force of the target to overpower it and force it to move with it.
Now, from law of inertia, Force = ma.
But in this case acceleration is 0 because the speed of the projectile is constant.
Thus, the force depends on the mass. So for a higher force, the mass of the projectile has to be more than that of the stationary object.
Thus, option C is correct
What determines the paths of electrons around a nucleus?
a. the fundamental quantum of energy
b. interaction with other atoms
c magnetism
d. the number of neutrons in the nucleus
Answer:
* positive repulsion, which makes the electrons move as far away as possible
* Attraction with the core atomizes
he analysis is the foundations of quantum theorizing
Explanation:
Electrons are deposited in an atom following two principles.
* positive repulsion, which makes the electrons move as far away as possible
* Attraction with the core atomizes
The mathematical model used for the analysis is the foundations of quantum theorizing
What is the final velocity (in m/s) of a hoop that rolls without slipping down a 3.50-m-high hill, starting from rest
Answer:
8.29m/s
Explanation:
Given
Height of hill H = 3.50m
Initial velocity u = 0m/s
Required
Final velocity v
Using the equation of motion
v² = u²+2gH
v² = 0²+2(9.81)(3.5)
v² = 0+7(9.81)
v² = 68.67
v = √68.67
v = 8.29m/s
Hence the final velocity of the hoop is 8.29m/s