Answers
Complete Question
The complete question(reference (chegg)) is shown on the first uploaded image
Answer:
The magnitude of the resultant force is [tex]F = 199.64 \ N[/tex]
The direction of the resultant force is [tex]\theta = 4.1075^o[/tex] from the horizontal plane
Explanation:
Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be [tex]Fcos(\theta )[/tex] while if the force is moving away from the angle then the resolved force is [tex]Fsin (\theta )[/tex]
Now from the diagram let resolve the forces to their horizontal component
So
[tex]\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)[/tex]
[tex]\sum F_x = 199.128 \ N[/tex]
Now resolving these force into their vertical component can be mathematically evaluated as
[tex]\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)[/tex]
[tex]\sum F_{y} = 14.30[/tex]
Now the resultant force is mathematically evaluated as
[tex]F = \sqrt{F_x^2 + F_y^2}[/tex]
substituting values
[tex]F = \sqrt{199.128^2 + 14.3^2}[/tex]
[tex]F = 199.64 \ N[/tex]
The direction of the resultant force is evaluated as
[tex]\theta = tan^{-1}[\frac{F_y}{F_x} ][/tex]
substituting values
[tex]\theta = tan^{-1}[\frac{ 14.3}{199.128} ][/tex]
[tex]\theta = 4.1075^o[/tex] from the horizontal plane
Related Questions
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
symbol of science hhshsjsiwtwwisjzhJava
Answers
Answer:
is this a company name.? java is a computer software right..
A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon's radius increasing when the radius is 7ft
Answers
Answer:
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Explanation:
Given;
Rate of inflation dV = 140pift^3/min
Radius r = 7 ft
Change in radius = dr
Volume of a spherical balloon is;
V = (4/3)πr^3
The change in volume can be derived by differentiating both sides;
dV = (4πr^2)dr
Making dr the subject of formula;
dr = dV/(4πr^2)
Substituting the given values;
dr = 140π/(4π×7^2)
dr = 0.714285714285 ft/min
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
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.
If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of
Answers
Answer:
The object has an excess of [tex]10^{13}[/tex] electrons.
Explanation:
When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:
[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]
The object has an excess of [tex]10^{13}[/tex] electrons.
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
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]
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.
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.
The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answers
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
I hope it helps you!
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
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistance is 207.4 What is the temerature (in degree Celsius of this solution
Answers
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
T₂ = 95.56°C
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
Find the change in the force of gravity between two planets when the distance between them becomes 10 times smaller. g
Answers
Answer:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
Explanation:
The force of gravity between two planets, is given by the following formula:
F = Gm₁m₂/r² ----------- equation 1
where,
F = Force of gravity between two planets
G = Gravitational Constant
m₁ = Mass of one planet
m₂ = Mass of other plant
r = Distance between two planets
Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:
F' = Gm₁m₂/(4r)²
F' = (1/16) (Gm₁m₂/r²)
using equation 1:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
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 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]
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]
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You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.
Answers
Answer:
D
Explanation:
The power equation is P= V^2/R
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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.
What is a resistors?A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.
Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.
A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.
The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.
Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.
Thus, the correct option is D.
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Near the surface of Earth, an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (magnitude and sign) would have to be placed on a penny that has a mass of 3 g to cause it to rise into the air with an upward acceleration of 0.19 m/s2?
Answers
Answer:
q = 2.997*10^-4C
Explanation:
In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.
You use the second Newton law to sum both electrical and gravitational forces:
[tex]F_e-W=ma\\\\qE-mg=ma[/tex] (1)
Fe: electric force
W: weight of the penny
q: required charge = ?
m: mass of the penny = 3g = 0.003kg
E: magnitude of the electric field = 100N/C
g: gravitational acceleration = 9.8m/s^2
a: acceleration of the penny = 0.19m/s^2
You solve the equation (1) for q, and replace the values of the other parameters:
[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]
It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2
A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between the two ends of the wire.
Answers
Answer:
V = 0.45 Volts
Explanation:
First we need to find the total current passing through the wire. That can be given by:
Total Current = I = (Current Density)(Surface Area of Wire)
I = (Current Density)(2πrL)
where,
r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m
L = Length of Wire = 6.5 m
Therefore,
I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]
I = 1.25 x 10⁻⁴ A
Now, we need to find resistance of wire:
R = ρL/A
where,
ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm
A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²
Therefore,
R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)
R = 0.36 Ω
From Ohm's Law:
Voltage = V = IR
V = (1.25 x 10⁻⁴ A)(0.36 Ω)
V = 0.45 Volts
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
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
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]
Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Answers
Answer:
C is the correct answer
Explanation:
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.
a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?
Answers
Answer:
6.44 s
Explanation:
Given:
v₀ = 119 m/s
v = 233 m/s
a = 17.7 m/s²
Find: t
v = at + v₀
(233 m/s) = (17.7 m/s²) t + (119 m/s)
t = 6.44 s
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
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
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
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