Answer:
c. decreasing.
Explanation:
Force produces acceleration or deceleration. Force is the product of a body's mass and its acceleration. When a force is applied to an object, the force tends to cause the body to move if the body was originally stagnant, cause the body to accelerate if applied in the direction of the body's velocity, or decelerate the body if applied in opposite direction to the velocity of the body. When the force that is exerted on a moving body is slowly reduced to zero, frictional forces between the body and the floor surface gradually decelerates the body. When this deceleration occurs, the velocity of the body gradually decreases t a stop.
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
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
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.
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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A stretched string is 120 cm long and has a linear density of 0.022 g/cm. What tension in the string will result in a second harmonic of 590 Hz
Answer:
T = 15,576 N
Explanation:
The speed of a wave on a string is given by
v = √ T /ρ rho
also the speed of the wave is given by the relationship
v = λ f
we substitute
λ f = √ T /ρ
T = (lam f)² ρ
let's find the wavelength in a string, fixed at the ends, the relation that gives the wavelength is
L= λ/2 n
λ= 2L / n
we substitute
T = (2L / n f)²ρ rho
let's calculate
T = (2 1.20 / 2 590) 0.022
T = 15,576 N
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.
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
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59-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when: the elevator is descending at a constant rate of 8 m/s
Answer:
578.2 N
Explanation:
The computation of reading on the scale is shown below:-
Data provided in the question
Weight of a girl = 59 kg
Constant rate = 8 m/s
Since the elevator is descended so the acceleration is zero
As we know that
Reading on the scale is
[tex]F = m\times g[/tex]
where, m = 59 kg
g [tex]= 59 \times 9.8 m/s^2[/tex]
So, the reading on the scale is
= 578.2 N
Therefore for computing the reading on the scale we simply applied the above formula.
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.
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.
The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
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!
Light of wavelength λ=0.01nm, is scattered at 1350 from a stationary electron. What is the kinetic energy of the recoiling electron?
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
how far do you think you would go in a car while sneezing for 2.5 seconds
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 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.
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
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.
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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What length (in m) should an oboe have to produce a fundamental frequency of 294 Hz on a day when the speed of sound is 350 m/s? It is open at both ends.
Answer:
L = 0.60 m
The length in metres should be 0.60 m
Explanation:
A pipe open at both ends can have a standing wave pattern with resonant frequency;
f = nv/2L ........1
Where;
v = velocity of sound
L = length of pipe
n = 1 for the fundamental frequency f1
Given;
Fundamental frequency f1 = 294 Hz
Velocity v = 350 m/s
n = 1
From equation 1;
Making L the subject of formula;
L = nv/2f1
Substituting the given values;
L = 1×350/(2×294)
L = 0.595238095238 m
L = 0.60 m
The length in metres should be 0.60 m
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
Answer:
q/6Eo
Explanation:
See attached file pls
symbol of science hhshsjsiwtwwisjzhJava
Answer:
is this a company name.? java is a computer software right..
A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to move this charge 50 cm along a path making an angle of 40
Answer:
The work required to move this charge is 0.657 J
Explanation:
Given;
magnitude of charge, q = 4.4 x 10⁻⁶ C
Electric field strength, E = 3.9 x 10⁵ N/C
distance moved by the charge, d = 50 cm = 0.5m
angle of the path, θ = 40°
Work done is given as;
W = Fd
W = FdCosθ
where;
F is the force on the charge;
According the coulomb's law;
F = Eq
F = 3.9 x 10⁵ x 4.4 x 10⁻⁶ = 1.716 N
W = FdCosθ
W = 1.716 x 0.5 x Cos40
W = 0.657 J
Therefore, the work required to move this charge is 0.657 J
Defination coulomb's law.
Answer:
a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
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?
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 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
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.
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.
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?
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]
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?
Answer:
(a) 62.3 N
(b) 1.89 N
(c) 0.430 kg m²
Explanation:
(a) Find the acceleration of block B.
Δy = v₀ t + ½ at²
1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²
a = 0.90 m/s²
Draw a free body diagram of block B. There are two forces:
Weight force mg pulling down,
and tension force Tb pulling up.
Sum of forces in the -y direction:
∑F = ma
mg − Tb = ma
Tb = m (g − a)
Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)
Tb = 62.3 N
(b) Draw a free body diagram of block A. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and tension force Ta pulling right.
Sum of forces in the +x direction:
∑F = ma
Ta = ma
Ta = (2.10 kg) (0.90 m/s²)
Ta = 1.89 N
(c) Draw a free body diagram of the pulley. There are two forces:
Tension force Tb pulling down,
and tension force Ta pulling left.
Sum of torques in the clockwise direction:
∑τ = Iα
Tb r − Ta r = Iα
(Tb − Ta) r = I (a/r)
I = (Tb − Ta) r² / a
I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)
I = 0.430 kg m²
The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].
Given :
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.a) First, determine the acceleration of the B block.
[tex]\rm s = ut + \dfrac{1}{2}at^2[/tex]
[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]
[tex]\rm a = 0.9\; m/sec^2[/tex]
Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.
[tex]\rm \sum F=ma[/tex]
[tex]\rm mg-T_b=ma[/tex]
[tex]\rm T_b = m(g-a)[/tex]
[tex]\rm T_b = 7\times (9.8-0.9)[/tex]
[tex]\rm T_b = 62.3\;N[/tex]
b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.
[tex]\rm \sum F=ma[/tex]
[tex]\rm T_a=ma[/tex]
[tex]\rm T_a = 2.1\times 0.9[/tex]
[tex]\rm T_a = 1.89\;N[/tex]
c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.
[tex]\rm \sum \tau = I\alpha[/tex]
[tex]\rm T_br-T_ar = I\alpha[/tex]
[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]
Now, substitute the values of the known terms in the above expression.
[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]
[tex]\rm I = 0.430 \; kg\;m^2[/tex]
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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)?
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
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
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 2.0-m wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the force it experiences
Answer:
The force experienced is 0.6 N
Explanation:
Given data
length of wire L= 2 m
current in wire I= 0.6 A
magnetic field B= 0.5
The force experienced can be represented as
[tex]F= BIL[/tex]
[tex]F= 0.5*0.6*2\\\F= 0.6 N[/tex]
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?
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
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!! In California, Clay is surfing on a wave that propels him toward the beach with a speed of 5.0 m/s. The wave crests are each 20 m apart. What is the frequency of the water wave? (please show your work and equation used)
Answer:
Frequency= 0.25m
Period= 4.0 secs
Explanation:
Clay is surfing on a wave with a speed of 5.0m/s
The wave crests are 20m apart
Therefore, the frequency of the wave can be calculated as follows
Frequency= wave speed/distance
= 5.0/20
= 0.25m
The period (T) can be calculated as follows
T= 1/frequency
T = 1/0.25
T= 4.0secs
Hence the frequency is 0.25m and the period is 4.0 secs
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)
Answer:
C is the correct answer
Explanation:
Find the change in the force of gravity between two planets when the distance between them becomes 10 times smaller. g
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.
Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec
Answer:
3g/(8π²) ≈ 0.372 m
Explanation:
Draw a free body diagram. There is a weight force at the center of the pendulum.
Sum of the torques about the pivot:
∑τ = Iα
mg (½ L sin θ) = (⅓ mL²) α
3g sin θ = 2L α
α = 3g/(2L) sin θ
For small θ, sin θ ≈ θ.
α = 3g/(2L) θ
θ" = 3g/(2L) θ
The solution of this differential equation is:
θ = θ₀ cos(√(3g/(2L)) t)
So the period is:
T = 2π / √(3g/(2L))
If the period is 1 second:
1 = 2π / √(3g/(2L))
√(3g/(2L)) = 2π
3g/(2L) = 4π²
L = 3g/(8π²)
L ≈ 0.372 m
The length of the pendulum rod is 0.37 m.
What is meant by time period of a pendulum ?The time period of a pendulum is defined as the time taken by the pendulum to complete one oscillation.
Here,
The mass of the pendulum, m = 2 kg
Time period of the pendulum, T = 1 s
Since, the pendulum is suspended from the pivot and is oscillating, at the position of the pendulum when it makes an angle θ with the pivot, there is a force of weight acting at the center of the pendulum.
The length of pendulum at that point = L/2
The perpendicular distance at that point, r = (L/2) sinθ
Therefore, the torque acting on the pendulum at that point,
τ = Iα
where I is the moment of inertia of the pendulum and α is the angular acceleration.
mg (L/2 sinθ) = (mL²/3)α
1/2 gsinθ = 1/3 Lα
Therefore,
α = (3g/2L) sinθ
For smaller values of θ, we can take sinθ = θ
So, α = (3g/2L) θ
We know that α = θ''
where θ is the angular displacement.
Therefore,
θ'' = (3g/2L) θ
So, ω = √3g/2L
Therefore, the equation of motion of the pendulum can be written as,
θ = θ₀ cos(ωt)
θ = θ₀ cos [(√3g/2L) t]
So, time period of the pendulum,
T = 2[tex]\pi[/tex]/ω
T = 2[tex]\pi[/tex]/(√3g/2L)
2[tex]\pi[/tex]/(√3g/2L) = 1
(√3g/2L) = 2[tex]\pi[/tex]
Therefore, length of the rod,
L = 3g/8[tex]\pi[/tex]²
L = 0.37 m
Hence,
The length of the pendulum rod is 0.37 m.
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Attaching the image file here.
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?
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