Answer:
[tex] MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093[/tex]
[tex] MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73[/tex]
Explanation:
Assuming this problem: "Carbon dioxide enters an adiabatic nozzle at 1200 K with a velocity of 50 m/s and leaves at 400 K. Assuming constant specific heats at room temperature, determine the Mach number (a) at the inlet and (b) at the exit of the nozzle. Assess the accuracy of the constant specific heat assumption."
Part a
For this case we can assume at the inlet we have the following properties:
[tex] T_1 = 1200 K, v_1 = 50 m/s [/tex]
We can calculate the Mach number with the following formula:
[tex] MA_1 = \frac{v_1}{c_1} = \frac{v_1}{\sqrt{kRT}}[/tex]
Where k represent the specific ratio given k =1.288 and R would be the universal gas constant for the carbon diaxide given by: [tex] R= 188.9 J/ Kg K[/tex]
And if we replace we got:
[tex] MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093[/tex]
Part b
For this case we can use the same formula:
[tex] MA_2 = \frac{v_2}{c_2} [/tex]
And we can obtain the value of v2 from the total energy of adiabatic flow process, given by this equation:
[tex] c_p T_1 + \frac{v^2_1}{2}=c_p T_2 + \frac{v^2_2}{2}[/tex]
The value of [tex] C_p = 0.8439 K /Kg K = 843.9 /Kg K[/tex] and the value fo T2 = 400 K so we can solve for [tex] v_2[/tex] and we got:
[tex] v_2= \sqrt{2c_p (T_1 -T_2) +v^2_1}=1163.074 m/s[/tex]
And now we can replace on this equation:
[tex] MA_2 = \frac{v_2}{c_2} [/tex]
And we got:
[tex] MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73[/tex]
Final answer:
To determine the Mach number at the exit of the nozzle, we can use the isentropic flow equations.
Explanation:
To determine the Mach number at the exit of the nozzle, we need to use the isentropic flow equations. The Mach number (M) at the exit of the nozzle can be calculated using the equation:
M = sqrt( 2/(k-1) * ( (P/Pref) ^ ((k-1)/k) - 1) )
Where:
k is the specific heat ratio (given as 1.288)P is the pressure at the exit of the nozzlePref is the reference pressure (1 atm)Given the information provided in the question, we can substitute the values into the equation to calculate the Mach number.
A 2-m telescope can collect a given amount of light in 1 hour. Under the same observing conditions, how much time would be required for a 6-m telescope to perform the same task? A 12-m telescope?
Answer
given,
diameter of telescope = 2 m
time to collect area = 1 hour
Diameter of the another telescope = 6 m
The light collected by the telescope is directly proportional to the area of its primary mirror.
Area of the mirror is directly proportional to the square of diameter.
so,
6 m diameter telescope will carry (6/2)² = 9 times more light than 2 m telescope.
Time taken to collect light = 60/9 = 6.67 minutes.
now, For 12 m telescope
12 m diameter telescope will carry (12/2)² = 36 times more time than 2 m telescope.
Time taken by 12 m telescope to collect light= 60/36 = 1.7 minutes.
To find the time required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use a formula and rearrange it to solve for the time taken.
Explanation:To find out how much time would be required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use the formula:
(light collected by telescope 1) × (time taken by telescope 1) = (light collected by telescope 2) × (time taken by telescope 2)
Let's solve for the time taken by the 6-m telescope and the 12-m telescope:
For the 6-m telescope: (light collected by 2-m telescope) × (1 hour) = (light collected by 6-m telescope) × (time taken by 6-m telescope)For the 12-m telescope: (light collected by 2-m telescope) × (1 hour) = (light collected by 12-m telescope) × (time taken by 12-m telescope)By rearranging the equation and solving for the time, we can find the answer.
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The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by various atmospheric processes, including lightning. What is the excess charge on the surface of the earth?
Answer:
The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C
Explanation:
Using the formula for an electric field;
E = kQ/r²
k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²
E = 100N/C
r = radius of the earth = 6400 km = 6400000m
Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)
Q = 455617.4 C = 4.56 × 10⁵ C
Hope this helps!!!
The excess charge on the surface of the earth is 4.55×10⁵C
To calculate the charge on the surface of the earth, we apply the equation of electric field strength, that is:
[tex]E = k\frac{Q}{r^2}[/tex]
here, E is the electric field strength = 100N/C (given)
k = 9×10⁹ Nm²/C²
Q is the charge, and
r = distance = radius of earth = 6.4×10⁶ m
Now,
[tex]Q = \frac{r^2E}{k}\\ \\=\frac{(6.4*10^6)^2*100}{9*10^9}\\\\Q=4.55*10^5 C[/tex] is the charge acquired on the surface of the earth.
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A very large sheet of a conductor carries a uniform charge density of 4.00 pC/mm2 on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
451977.40113 N/C
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
[tex]\sigma[/tex] = Surface charge density = [tex]4\ pc/mm^2[/tex]
Electric field near the surface of a charged conductor is given by
[tex]E=\dfrac{\sigma}{\epsilon_0}\\\Rightarrow E=\dfrac{4\times 10^{-12}\times 10^{6}}{8.85\times 10^{-12}}\\\Rightarrow E=451977.40113\ N/C[/tex]
The electric field is 451977.40113 N/C
A rhinoceros is at the origin of coordinates at time t1 = 0. For the time interval from t1 = 0 to t2 = 12.0 s, the rhino’s average velocity has x-component - 3.8 m/s and y-component 4.9 m/s. At time t2 = 12.0 s, (a) what are the x- and y-coordinates of the rhino? (b) How far is the rhino from the origin?
Answer:
a) [tex](x_2,y_2)=(-45.6,58.8)[/tex]
b) [tex]s=74.4097\ m[/tex]
Explanation:
Given:
x-component of avg. velocity, [tex]v_x=-3.8\ m.s^{-1}[/tex]x-component of avg. velocity, [tex]v_y=4.9\ m.s^{-1}[/tex]initial position of rhino, [tex](x_1,y_1)=(0,0)[/tex]initial time, [tex]t_1=0\ s[/tex]final time, [tex]t_2=12\ s[/tex]a)
As we know that average velocity is total displacement per unit time.
Position of x-coordinate of rhino:
[tex]v_x=\frac{x_2}{t_2}[/tex]
[tex]-3.8=\frac{x_2}{12}[/tex]
[tex]x_2=-45.6\ m[/tex]
Position of y-coordinate of rhino:
[tex]v_y=\frac{y_2}{t_2}[/tex]
[tex]4.9=\frac{y_2}{12}[/tex]
[tex]y_2=58.8\ m[/tex]
b)
Now the distance from the origin:
[tex]s=\sqrt{x_2^2+y_2^2}[/tex]
[tex]s=\sqrt{(-45.6)^2+(58.8)^2}[/tex]
[tex]s=74.4097\ m[/tex]
The question concerns vector kinematics in physics. The rhinoceros's coordinates at 12 seconds are (-45.6,58.8), and its distance from the origin is approximately 74.2 meters.
Explanation:The phenomenon described in your question essentially relates to Vector Kinematics. The rhinoceros is said to have a velocity in the x-direction of -3.8 m/s and in the y-direction of 4.9 m/s. These velocities are constant and last for 12 seconds.
To find the x- and y-coordinates, we can simply multiply the component velocities by time.
For the x-coordinate:
x = velocity_x * time = -3.8 m/s * 12 s = -45.6 m
And for the y-coordinate:
y = velocity_y * time = 4.9 m/s * 12 s = 58.8 m
So, for part (a) of your question, the rhino's coordinates at t2 = 12.0 s are (-45.6, 58.8).
Now, for part (b), to calculate the rhino's distance from the origin, we can use the Pythagorean theorem which is rooted in the geometrical interpretation of vectors. The distance from the origin (r) can be given by:
r = sqrt[x² + y²] = sqrt[(-45.6 m)² + (58.8 m)²] = approximately 74.2 m
Therefore, the rhino is approximately 74.2 m from the origin.
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To how many significant figures should each answer be rounded?
Equation:
A: ( 6.626 x 10^− 34 J * s ) ( 2.9979 x 10^8 m / s ) / 4.630 x 10^− 7 m = 4.290299222462 x 10^− 19 J ( unrounded )
1. After rounding, the answer to equation A should have __________.
Answer: After rounding, the answer to equation A should have [tex]4.290\times 10^{-19}J[/tex]
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s preceding the first integers are never significant.
All zero’s after the decimal point are always significant.
The rule apply for the multiplication and division is :
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
Thus for [tex]\frac{6.626\times 10^{-34}Js}{4.630\times 10^{-7}m}\times 2.9979\times 10^8m/s}=4.290299222462\times 10^{-19}J[/tex]
In this problem, 6.626 has 4 significant figures, 4.630 has 4 significant figures and 2.9979 has 5 significant digits thus product will have the least number of significant figures which is 4. So, the answer will be in 4 significant figures. Thus the answer is [tex]4.290\times 10^{-19}J[/tex]
The answer to equation A should have four significant figures, as the least precise measurement in the equation has four significant figures.
Explanation:When you are calculating with measured values, the rule is that your result cannot be more precise than the least precise measurement. This means when using values in calculations, you need to look at the number of significant figures in the original measurements to determine how many significant figures to use in the final answer.
In the given equation, A: (6.626 x 10−34 J * s) (2.9979 x 108 m / s) / 4.630 x 10−7 m = 4.290299222462 x 10−19 J (unrounded), the least precise measurement is 4.630 x 10−7 m, which has four significant figures. Therefore, after rounding, the answer to equation A should also have four significant figures.
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Calculate the speed of an electron (in m/s) after it accelerates from rest through a potential difference of 160 V.
Answer:
v = 7.5*10⁶ m/s
Explanation:
While accelerating through a potential difference of 160 V, the electron undergoes a change in the electric potential energy, as follows:
ΔUe = q*ΔV = (-e)*ΔV = (-1.6*10⁻¹⁹ C) * 160 V = -2.56*10⁻¹⁷ J (1)
Due to the principle of conservation of energy, in absence of non-conservative forces, this change in potential energy must be equal to the change in kinetic energy, ΔK:
ΔK = Kf -K₀
As the electron accelerates from rest, K₀ =0.
⇒ΔK =Kf = [tex]\frac{1}{2}*me*vf^{2}[/tex] (2)
From (1) and (2):
ΔK = -ΔUe = 2.56*10⁻¹⁷ J = [tex]\frac{1}{2}*me*vf^{2}[/tex]
where me = mass of the electron = 9.1*10⁻³¹ kg.
Solving for vf:
[tex]vf =\sqrt{\frac{2*(2.56e-17J)}{9.1e-31kg} } =7.5e6 m/s[/tex]
⇒ vf = 7.5*10⁶ m/s
Final answer:
The speed of an electron after accelerating from rest through a potential difference of 160 V can be calculated using the conservation of energy principle, resulting in a velocity of approximately 5.93 × 10^6 m/s.
Explanation:
To calculate the speed of an electron after it accelerates from rest through a potential difference, we use the concept of conservation of energy where the electrical potential energy converted into kinetic energy is expressed as qAV = ½mv². Here, q is the charge of the electron (-1.60 × 10^-19 C), V is the potential difference (160 V), m is the mass of the electron (9.11 × 10^-13 kg), and v is the velocity of the electron we need to find. Rearranging the equation to solve for v and substituting the given values gives:
v = sqrt(2 × q × V / m) = √(2 × (-1.60 × 10^-19 C) × 160 V / (9.11 × 10^-13kg))
Performing the calculation yields a velocity of approximately 5.93 × 10^6 m/s.
A police car in a high-speed chase is traveling north on a two-lane highway at 35.0 m/s. In the southbound lane of the same highway, an SUV is moving at 18.0 m/s. Take the positive x-direction to be toward the north. Find the x-velocity of the police car relative to the SUV.
Answer:
53 m/s
Explanation:
Since we take the positive x-direction to be toward the north, an SUV travelling to the south would have a negative velocity, -18m/s, relative to Earth. Velocity of Earth relative to the SUV would be 18m/s. And velocity of police car relative to Earth would be positive, 35 m/s.
Velocity of police relative to SUV would equal to velocity of police car relative to Earth plus velocity of Earth relative to SUV
= 35 + 18 = 53 m/s
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s. A seismograph records the arrival of the transverse waves 56.4 s after that of the longitudinal waves. How far away was the earthquake? Answer in units of km.
Answer:
[tex]d=691.71km[/tex]
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
[tex]t=\frac{d}{v_t}-\frac{d}{v_l}[/tex]
Here d is the distance at which the earthquake take place and [tex]v_t, v_l[/tex] is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
[tex]t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km[/tex]
What key ingredient in the modern condensation theory was missing or unknown in the nebula theory?
Answer:
Interstellar dust
Explanation:
In modern solar system theory of condensation, interstellar dust, which was lacking in nebular theory, would be the essential component.Cosmic dust is dust that exists in outer space or that has fallen to earth, also called extra-terrestrial dust or spatial powder. The majority of the cosmic dust particle sizes range from a few molecules to 0.1 μm.
The period of a carrier wave is T=0.005 seconds. Determine the frequency and wavelength of the carrier wave.
Answer:
f = 200[Hz]; L=1500[km]
Explanation:
We know that frequency is the reciprocal of the period, therefore.
[tex]f=\frac{1}{T} \\f=1/0.005\\f=200[Hz][/tex]
And the wavelength is determined using the following equation.
[tex]L=\frac{c}{f} \\where:\\c = light speed = 3*10^{8}[m/s]\\ f = frecuency [Hz]\\L = wavelength [m]\\L= 3*10^{8}/ 200\\ L = 1500000[m] = 1500[km][/tex]
Final answer:
The frequency of a carrier wave with a period of 0.005 seconds is calculated to be 200 Hz. The wavelength cannot be determined without additional information on the wave's speed.
Explanation:
The question involves calculating the frequency and wavelength of a carrier wave given its period (T=0.005 seconds). The frequency (f) of a wave is the reciprocal of its period, defined by the equation f = 1/T. Therefore, for a period of 0.005 seconds, the frequency would be f = 1/0.005 Hz, which equals 200 Hz. To determine the wavelength (λ), we would need the speed (v) of the wave, which is typically the speed of light (c) for electromagnetic waves, but since the speed and specific type of wave are not provided, we can't calculate the wavelength directly from the given information here.
A point charge is placed at the center of a spherical Gaussian surface. The electricflux ΦEischangedif(a) a second point charge is placed outside the sphere(b) the point charge is moved outside the sphere(c) the point charge is moved off center, but still inside the original sphere(d) the sphere is replaced by a cube of one-tenth the volume (the original charge remains in thecenter)
Answer:
(b) the point charge is moved outside the sphere
Explanation:
Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.
[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]
According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.
If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.
Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.
The electric field near the surface of Earth points downward and has a magnitude of 152 N/C. What is the ratio of the magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron?
Answer:
[tex]2.7\times 10^{12}[/tex]
Explanation:
We are given that
Electric field =[tex]E=152N/C[/tex]
We have to find the ratio of magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron.
We know that
F=qE
Charge on an electron,q=[tex]1.6\times 10^{-19}C[/tex]
Using the formula
Upward electric force=[tex]152\times 1.6\times 10^{-19}[/tex] N
Upward electric force=[tex]F_e=2.43\times 10^{-17}[/tex] N
Mass of electron=[tex]m_e=9.1\times 10^{-31} kg[/tex]
[tex]g=9.8m/s^2[/tex]
Gravitational force=[tex]F=mg[/tex]
Using the formula
Gravitational force=[tex]F_g=9.1\times 10^{-31}\times 9.8=8.92\times 10^{-30} N[/tex]
Ratio of Fe to the Fg=[tex]\frac{2.43\times 10^{-17}}{8.92\times 10^{-30}}[/tex]
[tex]\frac{F_e}{F_g}=2.7\times 10^{12}[/tex]
g If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?
Answer:
Range will become 4 times of initial range
Explanation:
Let the velocity of projection is u
And angle at which projectile is projected is [tex]\Theta[/tex]
And acceleration due to gravity is [tex]g\ m/sec^2[/tex]
So range of projectile is equal to [tex]R=\frac{u^2sin2\Theta }{g}[/tex]........eqn 1
Now in second case it is given that velocity of launching is doubled
So new velocity [tex]u_{new}=2u[/tex]
So new range will be equal to [tex]R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}[/tex] .....eqn 2
Now dividing eqn 2 by eqn 1
[tex]\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }[/tex]
[tex]R_{new}=4R[/tex]
So if we double the initial launch speed then range will become 4 times
Doubling the initial launch speed while keeping the launch angle fixed quadruples the range of a projectile due to the direct proportionality between the range and the square of the initial speed, considering factors like launch angle, gravity, and air resistance.
If you keep the launch angle fixed, but double the initial launch speed, the range of a projectile will increase. When the initial speed is doubled, the range increases four times because the range is directly proportional to the square of the initial speed. This phenomenon is influenced by factors such as launch angle, gravity, and air resistance.
Nuclear reactors use fuel rods to heat water and generate steam. Is this process endothermic or exothermic?
Explanation:
Exothermic reaction is defined as the reaction in which release of heat takes place. This also means that in an exothermic reaction, bond energies of reactants is less than the bond energies of products.
Hence, difference between the energies between the reactants and products releases as heat and therefore, enthalpy of the system will decrease.
Whereas in an endothermic reaction, heat is supplied from outside and absorbed by the reactant molecules. Hence, enthalpy of the system increases.
As water acts as a coolent and when fuel rods in a nuclear reactor are immersed in it then heat created by coolent is absorbed by water and then it changes into steam.
Since, absorption of heat occurs in the nuclear reactor. Therefore, it is an endothermic reaction.
Thus, we can conclude that nuclear reactors use fuel rods to heat water and generate steam. This process is endothermic.
Answer:
The heat produced by the nuclear reactor is an exothermic process while the heat absorbed by the water to convert into steam is an endothermic process.
Explanation:
Nuclear reactor being the heat of a nuclear power plant uses the radioactive uranium fuel to generate the heat by the process of nuclear fission in a controlled manner.
The processing of uranium is carried out into small ceramic pellets which are stacked together into sealed metal tubes known as fuel rods.
Usually more than 200 such rods are bunched together leading to the formation of a fuel assembly.
The core of the reactor is often made up of a couple hundred assemblies, according to its power level.
Inside the reactor vessel, these fuel rods are immersed into water which serve as both a coolant and moderator. The moderator helps slow down the neutrons produced by fission to sustain the chain reaction.
You throw a rock upward. The rock is moving upward. but it is slowing down. If we define the ground as the origin, theposition of the rock is ____ and the velocity of the rock is ____ .A. positive, positiveB. positive, negativeC. negative, positiveD. negative. negative
Answer:
.A. positive, positive.
Explanation:
When we throw a rock upward , it will decelerate due to gravitation . It will have acceleration in downward direction or - ve acceleration in upward direction.
If we define the ground as the origin and upward direction as positive , anything in upward direction will be positive and in downward direction will be negative . For example , in the case described above , acceleration of rock thrown upward is negative because is in downward direction .
Its position is in upward direction , so its position is positive with respect to ground.
It is going in upward direction . So velocity too will be positive.
A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00 km. After dropping off a package, he drives 9.50 km 35.0° north of east to Starbucks. What is the direction relative to the positive x axis?
Answer:
[tex]\theta=7^o[/tex]
Explanation:
Displacement
It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.
The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is
[tex]\vec r_1=<0,2>\ km[/tex]
Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by
[tex]\vec r_2=<7cos60^o,-7sin60^o>\ km=<3.5\ ,-6.06>\ km[/tex]
Finally, he drives 9.5 km 35° north of east.
[tex]\vec r_3=<9.5cos35^o,9.5sin35^o>\ km=<7.78\ ,\ 5.45>\ km[/tex]
The total displacement is
[tex]\vec r_t=<0,2>\ km+<3.5\ ,-6.06>\ km+<7.78\ ,\ 5.45>\ km[/tex]
[tex]\vec r_t=<11.28,1.39>\ km[/tex]
The direction can be calculated with
[tex]\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232[/tex]
[tex]\boxed{\theta=7^o}[/tex]
What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ0 = 4π × 10-7 T • m/A)
Answer:
[tex]L=4.19*10^{-6}H[/tex]
Explanation:
The self-inductance of a solenoid is defined as:
[tex]L=\frac{\mu N^2A}{l}[/tex]
Here [tex]\mu_0[/tex] is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:
[tex]L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H[/tex]
The self-inductance of the given solenoid would be approximately 1.395x10^-5 H. The calculation involved substituting known variables into the formula for self-inductance and solving.
Explanation:The self-inductance of a solenoid can be calculated using the formula: L = µ₀n²A/l where L is the self-inductance, µ₀ is the permeability of free space, n is the number of turns per unit length, A is the cross-sectional area, and l is the length of the solenoid.
In this scenario, the cross-sectional area (A) = 1.00 × 10⁻⁴ m², length of the solenoid (l) = 30.0 cm = 0.3m, and the number of turns (n) = 100, so n (number of turns per unit length) = 100/0.3 = 333.33 turns/m. µ₀, the permeability of free space is a constant which is given as 4π × 10⁻⁷ T m/A. So, substituting these values into the formula will give:
L (self-inductance) = 4π × 10⁻⁷ T m/A * (333.33 /m)² * 1.00 × 10⁻⁴ m² / 0.3 m = 1.395x10^-5 H, (where H indicates henries, the unit for inductance)
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A 5 kg ball approaches a wall at a speed of 4 m/s. It then bounces off the wall in the opposite direction at the same speed. What is the magnitude of the average force exerted on the ball if it is in contact with the wall for 0.1 s?
Answer:
Average force = 2 kg m/s²
Explanation:
Given
mass = 5 kg
initial velocity = 0 m/s
final velocity = 4 m/s
time = 0.1 sec
Find
average force = ?
Formula
Average force = m (final velocity - initial velocity)/t₂- t₁
= (5kg)(0 m/s - 4 m/s)/0 - 0.1
= 5 kg (- 4 m/s)/(- 0.1 sec)
= 2 kg m/s²
Final answer:
The magnitude of the average force exerted on the ball by the wall is 400 N, computed using the impulse-momentum theorem and the ball's change in momentum during the collision.
Explanation:
To calculate the magnitude of the average force exerted on the ball, we need to first determine the change in momentum of the ball and then use the impulse-momentum theorem. The ball changes direction after hitting the wall, so the final velocity is -4 m/s (negative sign indicating the opposite direction) and the initial velocity is 4 m/s. The total change in velocity is thus -4 m/s - 4 m/s = -8 m/s. The change in momentum (impulse) is given by mass × change in velocity, which is 5 kg × -8 m/s = -40 kg·m/s. The impulse experienced by the ball equals the average force exerted on the ball multiplied by the time of contact, so:
Average Force = - Impulse / Time
Therefore, the magnitude of the average force is:
|Average Force| = 40 kg·m/s / 0.1 s = 400 N.
The negative sign indicates that the force is in the opposite direction of the initial motion, but since the question asks for the magnitude, we take the absolute value.
falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.040 s. The average force exerted on him by the ground is 18 000 N, where the upward direction is taken to be the positive direction. From what height did the student fall
Answer:
y = 2,645 10⁴ / m²
m=80 kg, y = 4.13 m
Explanation:
We must solve this problem in two parts, one when it is in free fall and another for the collision with the floor
Let's start by analyzing the crash with the floor,
Initial instant When it arrives but if you start to stop
p₀ = m v
Final moment. When he stopped
[tex]p_{f}[/tex] = 0
The momentum is related to the moment by
I = Δp = p_{f} –p₀
F t = 0 - mv
v = -F t / m
Let's calculate
v = -18000 0.040 / m
v = -720 / m
The sign indicates that the speed goes down
Now we use energy conservation at two points
Lowest point. Just before crashing
Em₀ = K = ½ m v²
Highest point. From where it began to fall
Em_{f} = U = m g y
Energy is conserved in the fall
Em₀ = Em_{f}
½ m v² = m g y
y = ½ v² / g
y = ½ (720 / m)² /9.8
y = 2,645 10⁴ / m²
For an explicit height value, the object's mass must be known, suppose the masses are m = 80 kg
y = 2,645 10⁴/80²
y = 4.13 m
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across a 2.25 mm length of the wire? VV = nothing VV SubmitRequest Answer Part B What would the potential difference in part A be if the wire were silver instead of copper, but all else was the same?
Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
[tex]A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\[/tex]
a) The Potential difference across a 2.00 in length of a 14-gauge copper
wire:
L= 2.00 m
From Table Copper Resistivity [tex]p[/tex]= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
[tex]R=\frac{pL}{A}[/tex]
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
[tex]R=\frac{pL}{A}[/tex]
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4
According to a simplified model of a mammalian heart, at each pulse approximately 20 g of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heat muscle?
Answer:
0.02 N.
Explanation:
Given:
m = 20 g
= 0.02 kg.
vi = 0.35 m/s
vo = 0.25 m/s
t = 0.1 s
Force, F = m * a
Where,
m = mass of the blood
a = acceleration.
Acceleration, a = (vi - vo)/t
Where,
vi = final velocity
vo = initial velocity
t = time.
a = (0.35 - 0.25)/0.1
= 0.1/0.1
= 1 m/s².
F = 0.02 * 1
= 0.02 N.
A 230mg piece of gold is hammered into a sheet measuring 23cm x 17cm. What is the thickness of the sheet in meters? Density of gold is 19.32g/cm3
Answer:
t= 0.0003 mm
Explanation:
Given that
mass ,m = 230 mg
m = 0.23 g
Area ,A= 23 x 17 cm²
A= 391 cm²
Density ,ρ = 19.32 g/cm³
Lets take thinness of the sheet = t cm
We know that
Mass = Density x Volume
m = ρ A t
Now by putting the values in the above equation we get
0.23 = 19.32 x 391 x t
[tex]t=\dfrac{0.23}{19.32\times 391}\ cm[/tex]
t=0.000030 cm
t= 0.0003 mm
That is why the thickness of the sheet will be 0.0003 mm.
My sling shot shoots pellets at 50m/s. Find two angles of elevation that can be used to hit a target 65m away. (Assume air resistance is negligible. Assume the gravitational constant ????=10m/s2 for this problem.)
Answer:
Explanation:
Given
velocity of launch [tex]u=50\ m/s[/tex]
Target is [tex]R=65\m\ away[/tex]
Suppose [tex]\theta [/tex] is the launch angle
We know Range of Projectile is
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]65=\frac{50^2\times \sin 2\theta }{9.8}[/tex]
[tex]\sin 2\theta =0.254[/tex]
because [tex]\sin \theta =\sin (180-\theta )[/tex]
so either [tex]2\theta =14.71[/tex]
[tex]\theta =7.35^{\circ}[/tex]
or [tex]180-2\theta =14.71[/tex]
[tex]\theta =82.64^{\circ}[/tex]
A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.00 mm to the bottom of the incline is 3.80 m/s.What is the speed of the block when it is 3.00 m from the top of the incline?
Answer:
Explanation:
Given
Speed of block at bottom is [tex]v=3.8\ m/s[/tex]
Distance traveled [tex]s=6\ m[/tex]
initial velocity is zero
using equation of motion
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](3.8)^2-0=2\times a\times 6[/tex]
[tex]a=1.203\ m/s^2[/tex]
when it is 3 m from top then
[tex]v^2-u^2=2as[/tex]
[tex]v^2-0=2\times 1.203\times 3[/tex]
[tex]v=2.68\ m/s[/tex]
The problem is a physics question on kinematics, which requires calculating the speed of a block half-way down a frictionless incline using the kinematic equation for constant acceleration.
Explanation:The student's question involves finding the speed of a block at a certain point as it slides down a frictionless incline. The kinematics of motion on an inclined plane is the focus here. We are given the speed of the block after traveling 6.00 mm and we need to calculate its speed after it has traveled 3.00 m down the incline.
To solve this, we can use the kinematic equation for constant acceleration, which is stated as:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the block starts from rest, u = 0. We can find the acceleration by using the given final speed and distance from the first part of the trip (6.00 mm to the bottom). We substitute the acceleration into the kinematic equation again using s as 3.00 m to find the speed at that point.
A sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz. What is the maximum speed of the needle?
Answer:
v = 0.2035 m/s
Explanation:
given,
Amplitude,A = 0.0127 m
Frequency, f = 2.55 Hz
maximum speed of the needle = ?
we know,
x = A cos ω t
velocity of the motion
v = - A ω sin ω t
for maximum speed
sin ω t = 1
v = - A ω
ω = 2 π f
now,
v = A 2 π f
v = 0.0127 x 2 π x 2.55
v = 0.2035 m/s
Hence, the maximum speed of the needle is equal to 0.2035 m/s
A 6 kg bowling ball moves with a speed of 3 m/s. How fast does a 7 kg bowling ball need to move so that it has the same kinetic energy?
Answer: 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy.
Explanation:
Kinetic energy is the energy possessed by a body by virtue of its motion.
[tex]K.E=\frac{1}{2}mv^2[/tex]
m = mass of object
v= velocity of the object
[tex]K.E=\frac{1}{2}\times 6kg\times (3m/s)^2=27Joules[/tex]
b) for a 7 kg bowl to have kinetic energy of 27 Joules:
[tex]27J=\frac{1}{2}\times 7kg\times v^2[/tex]
[tex]v^2=7.7[/tex]
[tex]v=2.8m/s[/tex]
Thus 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy
A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.50 m above. The second student catches the keys 1.10 s later. (a) With what initial velocity were the keys thrown? magnitude m/s direction (b) What was the velocity of the keys just before they were caught? magnitude m/s direction
Answer:
a) 8.58 m/s upward
b) -2.211 m/s downward
Explanation:
Let gravitational acceleration g = -9.81m/s2. This is negative because it deceleration the upward motion of the key.
(a)We have the following equation of motion
[tex]s = v_0t + gt^2/2[/tex]
where [tex]v_0[/tex] is the initial upward velocity of the keys, t = 1.1s is the time it takes for the keys to travel a distance of s = 3.5 m
[tex]3.5 = v_01.1 - 9.81*1.1^2/2[/tex]
[tex]3.5 = 1.1v_0 - 5.94 [/tex]
[tex]1.1v_0 = 3.5 + 5.94 = 9.44[/tex]
[tex]v_0 = 9.44 / 1.1 = 8.58 m/s[/tex]
So the keys were thrown initially upward with a speed of 8.58 m/s
(b) If the initial velocity of the key is 8.58 m/s and it is subjected to a deceleration of 9.81m/s2 for 1.1s then the velocity right at the 1.1s instant is
[tex]v = v_0 + gt = 8.58 - 9.81*1.1 = -2.211 m/s[/tex]
So they keys would have a downward speed of 2.211 m/s
You are on the ground and observing a jet traveling in air at a constant height of 4500 m with a speed of 650 m/s. How long after the spaceship has passed directly overhead will the shock wave reach you
Answer:
[tex]t=11.1s[/tex]
Explanation:
Given data
Jet Speed v=650 m/s=1.89504 mach
height h=4500 m
Speed of the Sound V=343 m/s
To find
time t
Solution
As we know that
[tex]Vs_{Velocity}=d_{distance} /t_{time} \\d=Vs*t[/tex]
where
[tex]Vs=1.895*V[/tex]
Also from trigonometric properties we know that
[tex]tan\alpha =\frac{Perpendicular}{Base}[/tex]
We have use height h as perpendicular and distance d is base
So
[tex]tan\alpha =h/d\\tan\alpha=h*(1/Vs*t)\\t=\frac{h}{tan\alpha } \frac{1}{Vs}\\[/tex]
First we need to angle α
Since
[tex]Sin\alpha =V/Vs[/tex]
[tex]Sin\alpha =\frac{v}{1.895v}\\\alpha =32^{o}[/tex]
Substitute given values and angle to find
[tex]t=\frac{h}{tan\alpha } \frac{1}{Vs}\\t=\frac{4500m}{tan(32) } \frac{1}{1.895*343m/s}\\t=11.1s[/tex]
Final answer:
To calculate the time for a shock wave to reach an observer on the ground, divide the altitude of the jet by the speed of sound. For a jet at 4500 m and a sound speed of 343 m/s, it will take approximately 13.12 seconds for the shock wave to reach the ground.
Explanation:
The question pertains to the time it will take for the shock wave from a jet traveling at a constant speed and altitude to reach an observer on the ground.
To calculate this, one has to consider the speed of sound and the altitude of the jet from the ground. Using Pythagoras' theorem, one can determine the distance the sound has to travel to reach the observer. Then divide the distance by the speed of sound to find the time for the shock wave to reach the ground.
Let's denote the speed of sound as v and the altitude of the jet as h. The time t it takes for the shock wave to reach the observer can be calculated using the formula:
t = h / v
As given, the jet is flying at an altitude of 4500 m (4.5 km) and the speed of sound is generally taken to be 343 m/s. Plugging these values into the formula:
t = 4500 m / 343 m/s
t approx 13.12 seconds
Therefore, it will take approximately 13.12 seconds for the shock wave to reach the observer on the ground.
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
If a seed is launched at an angle of 0° with the maximum initial speed, how far from the plant will it land? Ignore air resistance, and assume that the ground is flat. (a) 20 cm; (b) 93 cm; (c) 2.2 m; (d) 4.6 m.
The seed launched at an angle of 0° will land 0 cm from the plant regardless of the initial speed because in a projectile motion, angle of 0° results in zero horizontal distance traveled.
Explanation:This question can be solved by using the formulas of projectile motion. In such motion, the horizontal distance traveled by an object (range) can be calculated using the formula:
R = (v² sin(2θ))/g
where 'v' is the speed of the object, 'θ' is the angle of projection, and 'g' is the acceleration due to gravity.
Here, the launch angle is 0° and the speed is 4.6 m/s. Since the sin(2*0°) is 0, whatever the speed would be, the seed will drop straight down, having no horizontal distance traveled.
So the seed will not move horizontally and it will land 0 cm away from the plant (option not given in choices).
Learn more about Projectile Motion here:https://brainly.com/question/20627626
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To solve this problem, we can use the kinematic equations of motion. When a projectile is launched at an angle from the horizontal, we can split its initial velocity into horizontal and vertical components. Since the launch angle is 0°, the initial velocity will have only a horizontal component, and the vertical component will be zero.
Given:
Initial speed,
v 0=4.6m/s
Launch angle,
θ=0∘
Initial height,
h=20cm=0.20m
Acceleration due to gravity,
g= 9.8m/s²
First, we find the time
t it takes for the seed to hit the ground:
Since the initial vertical velocity is 0, we can use the equation:
h= 1/2gt²
Substituting the values:
0.20= 1/2×9.8×t²
0.20=4.9t²
Solving for:
t²= 4.9/0.20
t²=0.040816
t²≈0.202s
Now, we can use this time to find the horizontal distance travelled x. Since the launch angle is 0°, the horizontal velocity remains constant:
x=v₀×t
x=4.6×0.202
x≈0.9292m
So, the seed will land approximately 0.9292m away from the plant.
The closest option provided is (b) 93 cm. However, the correct value is 92.92 cm, which would round to 93 cm. Therefore, the correct answer is (b) 93 cm.
A 2-ft3 tank contains a gas at 2 atm(g) and 60 oF. This tank is connected to a second tank containing the same gas at atmospheric pressure and 60 oF. The two tanks are connected and allowed to reach equilibrium. The final conditions are measured to be 1 atm(g) and 60oF. What is the volume of the second tank
Answer : The volume of the second tank is, [tex]4ft^3[/tex]
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex]
or,
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 2 atm
[tex]P_2[/tex] = final pressure of gas = 1 atm
[tex]V_1[/tex] = initial volume of gas = [tex]2ft^3[/tex]
[tex]V_2[/tex] = final volume of gas = ?
Now put all the given values in the above equation, we get:
[tex]2atm\times 2ft^3=1atm\times V_2[/tex]
[tex]V_2=4ft^3[/tex]
Therefore, the volume of the second tank is, [tex]4ft^3[/tex]