Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The distance which the car skid is [tex]l = \frac{v_i^2 }{2 * \mu_k * g }[/tex]
Explanation:
From the question we are told that
The initial velocity of the car is [tex]v_i[/tex]
The coefficient of kinetic friction is [tex]\mu_k[/tex]
According to the law of energy conservation
The initial Mechanical Energy = The final Mechanical Energy
[tex]M_i = M_f[/tex]
The initial mechanical energy is mathematically represented as
[tex]M_i = KE _o + PE_e[/tex]
where KE is the initial kinetic energy which is mathematically represented as
[tex]KE = \frac{1}{2} m v_i^2[/tex]
And PE is the initial potential energy which is zero given that the car is on the ground
Now
[tex]M_f = W_{\mu}[/tex]
Where [tex]W_{\mu}[/tex] is the work which friction exerted on the car which is mathematically represented as
[tex]W_{\mu} = m* \mu_k * g * l[/tex]
Where [tex]l[/tex] is the distance covered by the car before it slowed down
[tex]\frac{1}{2} m v_i^2 = m* \mu_k * g * l[/tex]
=> [tex]l = \frac{v_i^2 }{2 * \mu_k * g }[/tex]
Related Questions
A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not constant across the cross section of the wire, but rather varies as J=αrJ=αr, where αα is a constant.
(a) By the requirement that J integrated over the cross section of the wire gives the total current I, calculate the constant αα in terms of I and R.
(b) Use Ampere’s law to calculate the magnetic field B(r) for (i) r≤Rr≤R and (ii) r≥Rr≥R. Express your answers in terms of I.
Answers
Answer: (a) α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]
(b) For r≤R: B(r) = μ_0.[tex](\frac{I.r^{2}}{2.\pi.R^{3}})[/tex]
For r≥R: B(r) = μ_0.[tex](\frac{I}{2.\pi.r})[/tex]
Explanation:
(a) The current I enclosed in a straight wire with current density not constant is calculated by:
[tex]I_{c} = \int {J} \, dA[/tex]
where:
dA is the cross section.
In this case, a circular cross section of radius R, so it translates as:
[tex]I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr[/tex]
[tex]I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr[/tex]
[tex]I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}[/tex]
[tex]\alpha = \frac{3I}{2.\pi.R^{3}}[/tex]
For these circunstances, α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]
(b) Ampere's Law to calculate magnetic field B is given by:
[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]
(i) First, first find [tex]I_{c}[/tex] for r ≤ R:
[tex]I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr[/tex]
[tex]I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr[/tex]
[tex]I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr[/tex]
[tex]I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}[/tex]
[tex]I_{c} = \frac{I.r^{3}}{R^{3}}[/tex]
Calculating B(r), using Ampere's Law:
[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]
[tex]B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )[/tex].μ_0
B(r) = [tex](\frac{Ir^{3}}{R^{3}2.\pi.r})[/tex].μ_0
B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0
For r ≤ R, magnetic field is B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0
(ii) For r ≥ R:
[tex]I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr[/tex]
So, as calculated before:
[tex]I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}[/tex]
[tex]I_{c} =[/tex] I
Using Ampere:
B.2.π.r = μ_0.I
B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0
For r ≥ R, magnetic field is; B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0.
A number, which represents a property, amount, or relationship that does not change under certain situations is constant and further calculations as follows:
constant calculation:The Radius of the cross-section of the wire R
Current passing through the wire I
Current Density [tex]J = \alpha r[/tex]
Constant [tex]\alpha[/tex]
Distance of the point from the center [tex]r[/tex]
For part a)
Consider a circular strip between two concentric circles of radii r and r+dr.
Current passing through the strip [tex]dI =\overrightarrow J \times \overrightarrow{dA}[/tex]
[tex]\to\alpha r (2\pi r dr) cos 0^{\circ}[/tex]
Integration
[tex]\to I =2\pi \alpha \int^R_0 r^2\ dr =2\pi \alpha [r^3]^R_0=2\pi \alpha \frac{r^3}{3}\\\\\to \alpha = \frac{3I}{2\pi R^3}\\\\[/tex]
For part b)
The magnetic field at a point distance [tex]'r'^{(r \ \pounds \ R)}[/tex] from the center is B.
We have the value of the line integral of the magnetic field over a circle of radius ‘r’ given as
[tex]\oint \overrightarrow B \times \overrightarrow{dl} = \mu_0 I\\\\[/tex]
where ‘I’ is the threading current through the circle of radius ‘r’
[tex]\oint B \ dl \cos 0^{\circ} = \mu_0 [2\pi \alpha \frac{r^3}{3}]\\\\ B \int dl = \mu_0 [2\pi \frac{3I}{2\pi R^3} \frac{r^3}{3}]\\\\ B \cdot 2\pi r = \mu_0 I [\frac{r}{R}]^3\\\\ B = \frac{\mu_0}{2\pi} I [\frac{Ir^2}{R^3}]\\\\[/tex]
(ii) Similarly, we can calculate the magnetic field at the point at A distance ‘r’ where
[tex]\to r^3 R\\\\\to \int \overrightarrow{B} \overrightarrow{dl} = \mu_0\ I[/tex] [The threading current is the same]
[tex]\to \beta - 2\pi r = \mu_0 I[/tex] As (I)
[tex]\to \beta =\frac{\mu_o \ I}{2\pi \ r}[/tex]
Find out more about the density here:
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A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magnitude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x
Answers
Answer:
The change in potential energy is [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]
Explanation:
From the question we are told that
The magnitude of the uniform electric field is [tex]E = 950 \ N/C[/tex]
The distance traveled by the electron is [tex]x = 2.50 \ m[/tex]
Generally the force on this electron is mathematically represented as
[tex]F = qE[/tex]
Where F is the force and q is the charge on the electron which is a constant value of [tex]q = 1.60*10^{-19} \ C[/tex]
Thus
[tex]F = 950 * 1.60 **10^{-19}[/tex]
[tex]F = 1.52 *10^{-16} \ N[/tex]
Generally the work energy theorem can be mathematically represented as
[tex]W = \Delta KE[/tex]
Where W is the workdone on the electron by the Electric field and [tex]\Delta KE[/tex] is the change in kinetic energy
Also workdone on the electron can also be represented as
[tex]W = F* x *cos( \theta )[/tex]
Where [tex]\theta = 0 ^o[/tex] considering that the movement of the electron is along the x-axis
So
[tex]\Delta KE = F * x cos (0)[/tex]
substituting values
[tex]\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)[/tex]
[tex]\Delta KE = 3.8*10^{-16} J[/tex]
Now From the law of energy conservation
[tex]\Delta PE = - \Delta KE[/tex]
Where [tex]\Delta PE[/tex] is the change in potential energy
Thus
[tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]
how far do you think you would go in a car while sneezing for 2.5 seconds
Answers
Answer: If you are traveling at a speed of 60mph, you will go 220 feet.
Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.
(c) If you want the beam to undergo total internal reflection at the second interface (the interface between sheet 1 and the air), what is the minimum angle the incoming beam (the beam traveling in the sheet 2) must make with the x-axis
Answers
Answer:
33.749
Explanation:
According to the given situation, the solution of the minimum angle is shown below:-
We will apply the law to no and n1 medium which is
[tex]1.8\times sin(\theta_2)=1\times sin90[/tex]
[tex]\theta_2 = sin^{-1} \frac{1}{1.8}[/tex]
After solving the above equation we will get
= 33.749
Therefore for computing the minimum angle we simply applied the above formula.
Hence, the correct answer is 33.749
A sample of iodine is easiest to ship asa powder because it is
Answers
what is thermodynamics?
Answers
Answer:
The branch of physical science that deals with the relations between heat and other forms of energy (such as mechanical, electrical, or chemical energy), and, by extension, of the relationships between all forms of energy.
Answer:
The branch of physical science that deals with the relationship between heat and energy.
A sound wave travels through a column of carbon monoxide at STP. Assuming a density of rho = 1.25 kg/m3 and a bulk modulus of β = 1.42 ✕ 105 Pa, what is the approximate speed (in m/s) of the sound wave?
Answers
Answer:
v = 337.04m/s
Explanation:
In order to calculate the speed of sound for the case of a carbon monoxide medium, you use the following formula:
[tex]v=\sqrt{\frac{\beta}{\rho}}[/tex] (1)
v: speed of sound
β: bulk modulus = 1.412*10^5 Pa
ρ: density of the column of carbon monoxide = 1.25 kg/m^3
You replace the values of the parameters in the equation (1):
[tex]v=\sqrt{\frac{1.42*10^5Pa}{1.25kg/m^3}}=337.04\frac{m}{s}[/tex]
The speed of the sound in the column of carbon monoxide is 337.04 m/s
An electron (mass=9.11 X 10^-31kg) leaves a TV picture tube with zero initial speed and reaches the screen 1.90cm away at 3.00 X 10^6 m/s. Ignore gravity and find the net force. (1): 2.28 X 10^-19N (2): 2.28 X 10^-16N (3): None of the above
Answers
Answer:
2.16 × 10^-16N
Explanation:
The computation of the net force is shown below:
Data provided in the question
Electron mass = 9.11 × 10^-31kg
V_o = 0
V_f = 3.00 × 10^6 m/s
s = 1.90 cm i.e 1.9 × 10^-2
Based on the above information, the force is
As we know that
[tex]Force\ f = ma = \frac{mv^2}{2s}\\\\ = \frac{(9.11\times 10^{-31})(3\times 10^{6})^2}{2(1.9\times 10^{-2})}[/tex]
= 2.16 × 10^-16N
Hence, the last option is correct
Basically we applied the above formula to determine the net force
a point charge q is located at the center of a cube with edge length d. whatis the value of the flux over one face of the cube
Answers
Answer:
q/6Eo
Explanation:
See attached file pls
Two long, straight, parallel wires 7.2 cm apart carry currents of equal magnitude I. They repel each other with a force per unit length of 4.2 nN/m. Find the current I.
Answers
Answer:
current I = 38 mA
Explanation:
given data
distance r = 7.2 cm
repel each other force per unit length \frac{F}{l} = 4.2 nN/m
solution
we know 2 wire is parallel and when current flow through these wire they exert force each other due to magnetic field
and current I(1) = I(2) ................1
so
[tex]\frac{F}{l} = \frac{\mu _o}{2\pi } \times \frac{I(1) \times I(2)}{r}[/tex] ..................2
put here value
4.2 × [tex]10^{-9}[/tex] = [tex]\frac{4\pi \times 10^{-7}}{2\pi } \times \frac{I^2}{7.2\times 10^{-2}}[/tex]
solve it we get
I = 0.038884 A
current I = 38 mA
The current flow in the wires will be:
"38 mA".
Force, Current and DistanceAccording to the question,
Distance, r = 7.2 cm
Force per unit length, [tex]\frac{F}{l}[/tex] = 4.2 nN/m
Current passes, when wire is parallel:
→ I₁ = i₂
We know the relation,
→ [tex]\frac{F}{l}[/tex] = [tex]\frac{\mu_0}{2 \pi}\times \frac{I_1\times I_2}{r}[/tex]
By substituting the values, we get
4.2 × 10⁻⁹ = [tex]\frac{4 \pi\times 10^{-7}}{2 \pi}\times \frac{I^2}{7.2\times 10^{-2}}[/tex]
hence,
The current will be:
I = 0.038884 A or,
= 38 mA
Thus the above answer is appropriate.
Find out more information about current here:
https://brainly.com/question/25922783
25. Which of the following is true for an electromotor? A. It transforms thermal energy to electrical energy. B. It transforms mechanical energy into electrical energy. C. It transforms electrical energy into mechanical energy. D. It transforms electrical energy into potential energy.
Answers
Answer:
C
Explanation:
it transforms electrical energy into mechanical energy.
Find the magnitude of the resultant of forces 6N and 8N acting at 240° to each other
Answers
Answer:
magnitude of the resultant of forces is 11.45 N
Explanation:
given data
force F1 = 6N
force F2 = 8N
angle = 240°
solution
we get here resultant force that is express as
F(r) = [tex]\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}[/tex] ..............1
put here value and we get
F(r) = [tex]\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}[/tex]
F(r) = 11.45 N
so magnitude of the resultant of forces is 11.45 N
What if a solid cylinder of mass M = 2.50 kg, radius R = 2.18 cm, and length L = 2.7 cm, is rolling down from rest instead? With h = 79.60 m and x = 4.64 m, what is the center of mass velocity when the cylinder reaches the bottom?
Answers
Answer:
The center of mass velocity is [tex]v = 32.25 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the cylinder is [tex]m = 2.50 \ kg[/tex]
The radius is [tex]r = 2.18 \ cm = 0.0218 \ m[/tex]
The length is [tex]l = 2.7 \ cm = 0.027 \ m[/tex]
The height of the plane is h = 79.60 m
and the distance covered is [tex]d = 4.64 \ m[/tex]
The center of mass velocity o the cylinder when it reaches the bottom is mathematically represented as
[tex]v = \sqrt{\frac{4gh}{3} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{4 * 9.8 * 79.60}{3} }[/tex]
[tex]v = 32.25 \ m/s[/tex]
A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back toward your friend. What should you do in order to maximize your speed right after your interaction with the ball?
A. You should catch the ball.
B. You should let the ball go past you without touching it.
C. You should deflect the ball back toward your friend.
D. More information is required to determine how to maximize your speed.
E. It doesn't matter. Your speed is the same regardless of what you do.
Answers
Answer:
C You should deflect the ball back toward your friend.
Explanation:
This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,
with m= mass, V=velocity, i=initial, f=final:
mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)
So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision
Answer:
A. You should catch the ball.
Explanation:
Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.
what is pressure? A)Shown in picture B)Shown in picture C)Shown in picture D)Shown in picture
Answers
Answer:
The force exerted on a given area.
Option B is the right option.
Explanation:
About pressurePressure is defined as force acting per unit area. It's SI unit is Pascal or N/m^2.
Formula to find pressure is:
[tex] \frac{thrust(f)}{area(a)} [/tex]
The force acting perpendicular on a surface is called the thrust. It is a vector quantity.
One Pascal pressure
One Pascal pressure is defined as the pressure exerted in a surface of area m^2 by the thrust of 1 N
Hope this helps ..
Good luck on your assignment.....
An electron initially 3.00 m from a nonconducting infinite sheet of uniformly distributed charge is fired toward the sheet. The electron has an initial speed of 390 m/s and travels along a line perpendicular to the sheet. When the electron has traveled 2.00 m , its velocity is instantaneously zero, and it then reverses its direction.
Required:
a. What is the surface charge density on the sheet?
b. Given the same initial velocity, from what distance should the electron be fired if it is to just reach the sheet?
Answers
Answer:
a. σ = 3.82*10^-18C/m^2
b. d = 2.00m
Explanation:
a. In order to calculate the surface charge density of the sheet, you first calculate the acceleration of the electron on its motion.
You use the following formula:
[tex]v^2=v_o^2-2ad[/tex] (1)
v: final speed of the electron = 0m/s
vo: initial speed of the electron = 390m/s
a: acceleration of the electron = ?
d: distance traveled by the electron = 2.00m
You solve the equation (1) for a, and replace the values of all parameters:
[tex]a=\frac{v_o^2-v^2}{2d}\\\\a=\frac{(390m/s)^2}{2(2.00m)}=3.8*10^4\frac{m}{s^2}[/tex]
Next, you calculate the electric field that exerts the electric force on the electron, by using the second Newton law, as follow:
[tex]F_e=qE=ma[/tex] (2)
q: charge of the electron = 1.6*10^-19C
E: electric field of the sheet = ?
m: mass of the electron = 9.1*10^-31kg
You solve the equation (2) for E:
[tex]E=\frac{ma}{q}=\frac{(9.1*10^{-31}kg)(3.8*10^{4}m/s^2)}{1.6*10^{-19}C}\\\\E=2.16*10^{-7}\frac{N}{C}[/tex]
Next, you use the following formula to calculate the surface charge density, by using the value of its electric field:
[tex]E=\frac{\sigma}{2\epsilon_o}[/tex] (3)
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
σ: surface charge density of the sheet
You solve for σ:
[tex]\sigma=2\epsilon_o E=2(8.85*10^{-12}C^2/Nm^2)(2.16*10^{-7}N/C)\\\\\sigma=3.82*10^{-18}\frac{C}{m^2}[/tex]
The surface charge density of the sheet id 3.82*10^-18C/m^2
b. To calculate the required distance for the electron reaches the sheet, you take into account that the electron acceleration is the same in all places near the sheet, then by the result of the previous point, you can conclude that the electron must be fired from a distance of 2.00m.
which of the following is an example of potential energy?
A. a basketball sitting on a shelf
B. a dog running across a field
C. a bowling ball rolling down a lane
D. a teenager riding their bike???
Answers
Answer:
A
Explanation:
potential energy is stored energy so the ball has potential energy to bounce or roll which would then have been converted to kinetic but the rest are in motion meaning energy is no longer stored but used in motion and therefore cannot be potential energy
An electron has a kinetic energy that is twice its rest energy. Determine its speed. Group of answer choices
Answers
Answer:
The speed of the electron will be 6x10^8m/s
Explanation:
See attached file
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c
Answers
Answer:
0.85c
Explanation:
Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²
Rest mass of proton [tex]M_{0P}[/tex] = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV
for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV
Recall that the rest energy, and the total energy are related by..
[tex]E[/tex] = γ[tex]E_{0}[/tex]
which can be written in this case as
[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1
where [tex]E[/tex] = total energy of the kaon, and
[tex]E_{0}[/tex] = rest energy of the kaon
γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]
where [tex]\beta = \frac{v}{c}[/tex]
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2
where [tex]E_{K}[/tex] is the total energy of the kaon, and
[tex]E_{0P}[/tex] is the rest energy of the proton.
From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89
1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1
squaring both sides, we get
3.57( 1 - [tex]\beta^{2}[/tex]) = 1
3.57 - 3.57[tex]\beta^{2}[/tex] = 1
2.57 = 3.57[tex]\beta^{2}[/tex]
[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72
[tex]\beta = \sqrt{0.72}[/tex] = 0.85
but, [tex]\beta = \frac{v}{c}[/tex]
v/c = 0.85
v = 0.85c
some snakes have special sense organs that allow them to see the heat emitted from warm blooded animals what kind of an electromagnetic waves does the sense organs detect?
A. Visible light waves
B. Ultraviolet light waves
C. Infrared Waves
D. Microwaves
Answers
The heat emitted from anything is carried in the form of infrared waves. (C)
An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?
Answers
Answer:
Towards the west.
Explanation:
The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.
Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.
According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward
Two eggs of equal mass are thrown at a blanket with equal velocity. Egg B hits the blanket but egg A hits the wall instead. Compare the work done on the eggs in reducing their velocities to zero.
1. More work was done on A than on B.
2. It is meaningless to compare the amount of work because the forces were so different.
3. Work was done on B, but no work was done on A because the wall did not move.
4. More work was done on B than on A.
5. The amount of work is the same for both.
Answers
Answer:
5. The amount of work is the same for both.
Explanation:
Work done is a measure of change in kinetic energy of each egg
For both egg , the initial speed and mass are same , so they have equal initial Kinetic energy
For both egg , the final speed is 0 and mass are same , so they have equal final Kinetic energy which is 0.
So work done is same for both eggs since they have same change in kinetic energy.
As you drive down the road at 13 m/s , you press on the gas pedal and speed up with a uniform acceleration of 1.02 m/s2 for 0.70 s. If the tires on your car have a radius of 33 cm, what is their angular displacement during this period of acceleration?
Answers
Answer:
The angular displacement is [tex]\theta = 28.33 \ rad[/tex]
Explanation:
From the question we are told that
The speed of the driver is [tex]v =13 \ m/ s[/tex]
The acceleration of the driver is [tex]a = 1.02 \ m/s^2[/tex]
The time taken is [tex]t = 0.70 \ s[/tex]
The radius of the tire is [tex]r = 33 cm = 0.33 \ m[/tex]
The distance covered by the car during this acceleration can be calculated using the equation of motion as follows
[tex]s = v*t +\frac{1}{2} * a * t^2[/tex]
Now substituting values
[tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]
[tex]s = 9.35 \ m[/tex]
Now the angular displacement of the car with respect to the tire movement can be represented mathematically as
[tex]\theta = \frac{s}{r}[/tex]
substituting values
[tex]\theta = \frac{9.35}{0.33}[/tex]
[tex]\theta = 28.33 \ rad[/tex]
the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the surface by the block
Answers
Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
A crane lifts a 425 kg steel beam vertically a distance of 64 m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s2
Answers
Answer:
work done= 48.96 kJExplanation:
Given data
mass of load m= 425 kg
height/distance h=64 m
acceleration a= 1.8 m/s^2
The work done can be calculated using the expression
work done= force* distance
but force= mass *acceleration
hence work done= 425*1.8*64= 48,960 J
work done= 48.96 kJ
Suppose that the separation between two speakers A and B is 4.30 m and the speakers are vibrating in-phase. They are playing identical 103-Hz tones and the speed of sound is 343 m/s. An observer is seated at a position directly facing speaker B in such a way that his line of sight extending to B is perpendicular to the imaginary line between A and B. What is the largest possible distance between speaker B and the observer, such that he observes destructive interference
Answers
Answer:
The largest possible distance is [tex]x = 4.720 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 4.30 \ m[/tex]
The frequency of the tone played by both speakers is [tex]f = 103 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s[/tex]
The wavelength of the tone played by the speaker is mathematically evaluated as
[tex]\lambda = \frac{v}{f}[/tex]
substituting values
[tex]\lambda = \frac{343}{103}[/tex]
[tex]\lambda = 3.33 \ m[/tex]
Let the the position of the observer be O
Given that the line of sight between observer and speaker B is perpendicular to the distance between A and B then
The distance between A and the observer is mathematically evaluated using Pythagoras theorem as follows
[tex]L = \sqrt{d^2 + x^2}[/tex]
Where x is the distance between the observer and B
For the observer to observe destructive interference
[tex]L - x = \frac{\lambda}{2}[/tex]
So
[tex]\sqrt{d^2 + x^2} - x = \frac{\lambda}{2}[/tex]
[tex]\sqrt{d^2 + x^2} = \frac{\lambda}{2} +x[/tex]
[tex]d^2 + x^2 = [\frac{\lambda}{2} +x]^2[/tex]
[tex]d^2 + x^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} + x^2][/tex]
[tex]d^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} ][/tex]
substituting values
[tex]4.30^2 = [\frac{3.33^2}{4} +2 * x * \frac{3.33}{2} ][/tex]
[tex]x = 4.720 \ m[/tex]
A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule with an energy of about 0.8 × 10-19 J. If the velocity of the kinesin is 800 nm/s, can you determine the force the kinesin is exerting, if you assume that all the ATP energy is used (100% efficiency)? If you can, find it and give your answer in newtons. If not, answer with 0.
Answers
Answer:
The force is [tex]F = 8*10^{-12} \ N[/tex]
Explanation:
From the question we are told that
The rate at which ATP molecules are used is [tex]R = 80 ATP/ s[/tex]
The energy provided by a single ATP is [tex]E_{ATP} = 0.8 * 10^{-19} J[/tex]
The velocity of the kinesin is [tex]v = 800 nm/s = 800*10^{-9} m/s[/tex]
The power provided by the ATP in one second is mathematically represented as
[tex]P = E_{ATP} * R[/tex]
substituting values
[tex]P = 80 * 0.8*10^{-19 }[/tex]
[tex]P = 6.4 *10^{-18}J/s[/tex]
Now this power is mathematically represented as
[tex]P = F * v[/tex]
Where F is the force the kinesin is exerting
Thus
[tex]F = \frac{P}{v}[/tex]
substituting values
[tex]F = \frac{6.4*0^{-18}}{800 *10^{-9}}[/tex]
[tex]F = 8*10^{-12} \ N[/tex]
The force exerted by the kinesin is 8 × 10-12 N.
Let us recall that power is defined as the rate of doing work. Hence, power = Energy/Time.
Since;
Energy = 0.8 × 10-19 J/molecule
Number ATP molecules transported per second = 80 ATP molecules/s
Power = 0.8 × 10-19 J/molecule × 80 ATP molecules/s
Power = 6.4 × 10-18 J
Again, we know that;
Power = Force × Velocity
Velocity of the ATP molecules = 800 nm/s or 8 × 10-7 m/s
Force = Power/velocity
Force = 6.4 × 10-18 J/ 8 × 10-7 m/s
Force = 8 × 10-12 N
Learn more: https://brainly.com/question/11897796
Light of wavelength λ=0.01nm, is scattered at 1350 from a stationary electron. What is the kinetic energy of the recoiling electron?
Answers
Answer:
198.9 x 10^-16
Explanation:
E = hc/ wavelength
E =(6.63 x 10^-34 x 3 x 10^8)/(0.01 x 10^-9)
E = 198.9 x 10^-16
Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?
Answers
Answer:
94.248 g/sec
Explanation:
For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:
[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]
And, the velocity of blood pumping is 30 cm^2
Now apply the following formula to solve the total current
[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]
Q = 94.248 g/sec
Basically we applied the above formula So, that the total current could come
Two protons are maintained at a separation of 973 nm. Calculate the electric potential due to the two particles at the midpoint between them. Then, find the magnitude and direction of the electric field there. potential: V magnitude of field: N/C The direction of field is toward one of the protons. another direction. undetermined.
Answers
Answer:
V = 2.95*10^-3 V
E = 0 N/C
Explanation:
In order to calculate the electric potential due to the two protons at the midpoint between them, you use the following:
[tex]V=V_1+V_2=k\frac{q}{r}+k\frac{q}{r}=2\frac{kq}{r}[/tex] (1)
where you have taken into account each contribution to the total electric potential, produced by each proton.
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q: charge of the proton = 1.6*10^{-19}C
r: distance from the point (at the midway distance between the protons) to one proton = 973nm = 973nm/2 = 486.5nm = 486.5*10^-9m
You replace the values of the parameters in the equation (1):
[tex]V=2\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{486.5*10^{-9}m}\\\\V=2.95*10^{-3}V[/tex]
The electric potential is 2.95*10^-3V
The electric field generated at the midpoint in between the protons is zero, because the electric field generated by each proton has the same magnitude but opposite direction.
E = 0N/C