Answer:
v₂₁ = 1.45 m/s
Second runner is the winner.
Δs = 35.1 m
Explanation:
For the relative velocity, we use the formula:
v₂₁ = v₂ - v₁
where,
v₂₁ = relative velocity of second runner with respect to first runner = ?
v₁ = velocity of first runner = 3.4 m/s
v₂ = velocity of second runner = 4.85 m/s
Therefore,
v₂₁ = 4.85 m/s - 3.4 m/s
v₂₁ = 1.45 m/s
Now, for finding the winner, we calculate the time taken by both the runners to reach finish line, by using following equation:
s = vt
t = s/v
for first runner:
t₁ = (215 m)/(3.4 m/s)
t₁ = 63.23 s
for 2nd runner:
t₂ = (215 m + 41.6 m)/(4.85 m/s)
t₂ = 52.9 s
Since, t₂<t₁.
Therefore, second runner is the winner.
Now, for the difference between runners at the time of winning, we first calculate the distance covered by first runner at that time. Using second equation of motion:
s = (3.4 m/s)(52.9 s)
s = 179.9 m
So, the distance by which the second runner finishes ahead of the first runner is given as follows:
Δs = 215 m - 179.9 m
Δs = 35.1 m
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find
(a) the coefficient of static friction.
(b) the coefficient of kinetic friction between the block and the surface.
Answer:
(a) 0.31
(b) 0.245
Explanation:
(a)
F' = μ'mg.................... Equation 1
Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.
make μ' the subject of the equation above
μ' = F'/mg............. Equation 2
Given: F' = 75 N, m = 25 kg
constant: g = 9.8 m/s²
Substitute these values into equation 2
μ' = 75/(25×9.8)
μ' = 75/245
μ' = 0.31.
(b) Similarly,
F = μmg.................. Equation 3
Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.
make μ the subject of the equation
μ = F/mg.............. Equation 4
Given: F = 60 N, m = 25 kg, g = 9.8 m/s²
Substitute these values into equation 4
μ = 60/(25×9.8)
μ = 60/245
μ = 0.245
A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 N. 825 N. frictional force acting on the box as it moves to the right what is the net force in the Y direction
Answer:A
Explanation:
Explanation:
Given that,
Gravitational force = 600 N
Frictional force = 25 N
Pulled by the Force = 250 N
We know that,
The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.
The balance equation is along y-axis
The box will not move in y-axis therefore, the net force in the y-axis will be zero.
Hence, The net force in the y-direction will be zero.
A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
I2 = 3.076 A
Explanation:
In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:
[tex]B=\frac{\mu_oI}{2R}[/tex] (1)
I: current in the wire
R: radius of the wire
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:
[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex] (2)
I1: current of the first ring = 8A
R1: radius of the first ring = 0.078m
I2: current of the second ring = ?
R2: radius of the first second = 0.03m
To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:
[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]
The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.
Please help! Calculate velocity. Show all work!
Answer:
v = 23.66 m/s
Explanation:
recall that one of the equations of motion may be expressed:
v² = u² + 2as,
Where
v = final velocity (we are asked to find this)
u = initial velocity = 0 m/s since we are told that it starts from rest
a = acceleration = 0.56m/s²
s = distance traveled = given as 500m
Simply substitute the known values into the equation:
v² = u² + 2as
v² = 0 + 2(0.56)(500)
v² = 560
v = √560
v = 23.66 m/s
A ball drops some distance and gains 30 J of kinetic energy. Do NOT ignore air resistance. How much gravitational potential energy did the ball lose? Group of answer choices
Answer:
Greater than 30 J
Explanation:
We know that if the resistance air is nothing then the potential energy is directly translated into the energy of the kinetic i.e kinetic energy also there is second condition is arises If the resistance of air is used to operate towards the air resistance then the potential energy is greater then the kinetic energy .
In the given mention question the ball is gaining the 30 J of the kinetic energy on drooping it on the air resistance it simply means that the ball is losing the potential energy of 30 J.Therefore the gravitational potential energy of the ball is greater then than 30 J
The gravitational potential energy where the ball loses should be Greater than 30 J
Translation of the potential energy to the kinetic energy:In the case when the resistance air should be nothing so this transformation could be done. Also there should be the second condition when the air resistance should be used for operating other than the potential energy that should be more than the kinetic energy.
Since in the given situation, the ball should be gained the 30 J of the kinetic energy so here the ball should lose the potential enery of 30 J.
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Light of wavelength 575 nm falls on two double slits spaced 0.30 mm apart. What is the required distance from the slits to a screen if the spacing between the first and second dark fringe is to be 4.00 mm
Answer:
L = 2.1 m
Explanation:
From Young's Double Slit Experiment, the formula for the distance between two consecutive dark or bright fringes, called fringe spacing, is derived as:
Δx = λL/d
where,
Δx = distance between first and second dark fringe = 4 mm = 4 x 10⁻³ m
λ = wavelength of light = 575 nm = 5.75 x 10⁻⁷ m
d = distance between the slits = 0.3 mm = 3 x 10⁻⁴ m
L = Distance from slits to screen = ?
Therefore,
4 x 10⁻³ m = (5.75 x 10⁻⁷ m)(L)/(3 x 10⁻⁴ m)
L = (4 x 10⁻³ m)/(1.92 x 10⁻³)
L = 2.1 m
a straight wire that is 0.60m long and carrying a current of 2.0a is placed at an angle with respect to a magnetic field of strenght 0.30t if the wire experiences a force of magnitude 0.18 n what angle does the wire
Answer:
The angle is [tex]\theta = 30^o[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]l = 0.60 \ m[/tex]
The current is [tex]I = 2.0 \ A[/tex]
The magnetic field strength is [tex]B = 0.30 \ T[/tex]
The magnitude of the magnetic force is [tex]F _b = 0.18 \ N[/tex]
Generally the magnetic force exerted on the wire is mathematically represented as
[tex]F_b = I * l * B * sin \theta[/tex]
Making [tex]\theta[/tex] the subject
[tex]\theta =sin^{-1} [ \frac{ F_b }{I * l * B } ][/tex]
substituting values
[tex]\theta =sin^{-1} [ \frac{ 0.18 }{ 2.0 * 0.6 * 0.3 } ][/tex]
[tex]\theta = 30^o[/tex]
g Doppler Radar gathers information about precipitation by sending out pulses of ______ energy that is reflected back by the precipitation towards the radar. Group of answer choices
Answer:
Doppler Radar gathers information about precipitation by sending out pulses of ___Radio wave___ energy
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. What is the direction of the electric field vector?
Answer:
the electric field is pointing horizontal direction and in south direction
Explanation:
In an electromagnetic wave, the magnetic field and electrical field are perpendicular to each other and these are perpendicular to the direction of the waves.
Determining the Mass of a Mystery Mystery object Object distance (m) Brick mass (kg) Brick distance (m) Brick torque (Nm) Object mass (kg) Fobject (N) Fbricks (N) Fpivot point (N) A. 1.00 20 B. 1.00 20 C. 1.00 20 D. 1.00 20
Answer:
Explanation:
according to resultant of two parallel forces,
Fpivot = Fobject + Fbricks
so that, the net force is zero
what is the largest star in our night sky
The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field with a solenoid of the same size, carrying a current of 1.8 A , how many turns of wire would you need
Answer:
The number of turns of the solenoid is 3536 turns
Explanation:
Given;
magnetic field of the solenoid, B = 0.1 T
current in the solenoid, I = 1.8 A
length of the solenoid, L = 8cm = 0.08m
The magnetic field near the center of the solenoid is given by;
B = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
n is number of turns per length
I is the current in the coil
The number of turns per length is calculated as;
n = B / μ₀I
n = (0.1 ) / (4π x 10⁻⁷ x 1.8)
n = 44203.95 turns/m
The number of turns is calculated as;
N = nL
N = (44203.95)(0.08)
N = 3536 turns
Therefore, the number of turns of the solenoid is 3536 turns
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V
A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?
C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.
Answer:
Explanation:
A )
electric field E = V / d where V is potential difference between plates separated by distance d .
putting the given values
E = 370 / .040 V / m
= 9250 V / m
B )
Force on charged particle of charge q in electric field E
F = q E
F = 2.4 x 10⁻⁹ x 9250
= 22200 x 10⁻⁹
= 222 x 10⁻⁷ N .
C ) since field is uniform , force will be constant
work done by electric field putting up this force
= force x displacement
= 222 x 10⁻⁷ x 40 x 10⁻³
= 888 x 10⁻⁹ J
D )
change in potential energy
= q ( V₁ - V₂ )
= 2.40 X 10⁻⁹ x 370
= 888 x 10⁻⁹ J .
(a) The magnitude of electric field in the region between the plates is 9,250 V/m.
(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is [tex]8.88 \times 10^{-7} \ J[/tex].
(d) the change of the potential energy is [tex]8.88 \times 10^{-7} \ J[/tex].
The given parameters;
distance between the two metal plates, d = 40 mmpotential difference between the plates, V = 370 V(a) The magnitude of electric field in the region between the plates is calculated as;
[tex]E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m[/tex]
(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;
F = Eq
F = 9,250 x 2.4 x 10⁻⁹
F = 2.22 x 10⁻⁵ N
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;
[tex]W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J[/tex]
(d) the change of the potential energy is calculated as;
[tex]\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J[/tex]
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A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 105 Pa for the atmospheric pressure and 1200 kg/m3 for the density of the sauce, find the absolute pressure in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.
Answer:
(a) P = 103064 Pa = 103.064 KPa
(b) P = 102476 Pa = 102.476 KPa
Explanation:
(a)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)
Pg = 1764 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1764 Pa
P = 103064 Pa = 103.064 KPa
(b)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)
Pg = 1176 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1176 Pa
P = 102476 Pa = 102.476 KPa
The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.
Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.
In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.
The absolute pressure in the bulb can be calculated using the following formula:
P = P₀ + ρgh
where:
P is the absolute pressure in the bulb,
P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),
ρ is the density of the sauce (1200 kg/m³),
g is the acceleration due to gravity (9.8 m/s²), and
h is the height of the sauce in the tube.
(a) When h = 0.15 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)
P ≈ 1.013 x 10⁵ Pa + 1764 Pa
P ≈ 1.031 x 10⁵ Pa
(b) When h = 0.10 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)
P ≈ 1.013 x 10⁵ Pa + 1176 Pa
P ≈ 1.025 x 10⁵ Pa
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A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk.
(a) What is the new angular velocity of the disk?
(b) What is the change in the kinetic energy of the system?
(c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk then?
(d) What is the new kinetic energy of the system? (e) What is the cause of the increase and decrease of kinetic energy?
Answer:
Id = 1/2 Md * R^2 = 1/2 * .1 * .1^2 = .0005 kg m^2 inertia of disk
Ib = Mb * R^2 = .02 * .1^2 = .0002 kg m^2 inertia of bug at edge
(Id + Ib) w1 = Id w2 conservation of angular momentum
w2 = .0007 / .0005 * 10 = 14 /sec angular speed with bug at center
KE1 = 1/2 I1 * w1^2 = 1/2 * .0007 * 10^2 = .035 kg m^2 / s^2
KE2 = 1/2 * I2 w2*2 = (.0005 / 2 ) ^ 14^2 = .049 kg m^2 / s^2
The bug has to exert radial force on the disk to maintain its
centripetal acceleration. As the bug crawls to the center of the disk it
does work against this centripetal force which appears as an increase
of rotational energy of the disk. As the the bug crawls back to the edge
of the disk, the disk does work on the bug and loses KE.
A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?
Answer:
the car have travelled 0.31 mile during that time
Explanation:
Applying the Equation of motion;
s = 0.5(u+v)t
Where;
s = distance travelled
u = initial speed = 0 mph
v = Final speed = 50 mph
t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour
Substituting the given values into the equation;
s = 0.5(0+50)×(1/80)
s = 0.3125 miles
s ~= 0.31 mile
the car have travelled 0.31 mile during that time
A piece of tape is pulled from a spool and lowered toward a 100-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.
Requried:
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance.
Answer:
The magnitude of the electric force is [tex]F_e = 0.00098 \ N[/tex]
Explanation:
From the question we are told that
The mass of the paper is [tex]m= 100 mg = 100 *10^{-6} \ kg[/tex]
The position is [tex]d = 8.0\ mm = 0.008 \ m[/tex]
Generally the magnitude of the electric force at the point of equilibrium between the electric force and the gravitational force is mathematically represented as
[tex]F_e = F_g = mg[/tex]
Where [tex]F_g[/tex] is gravitational force
substituting values
[tex]F_e = 100 *10^{-6} * 9.8[/tex]
[tex]F_e = 0.00098 \ N[/tex]
Now generally the gravitational force acts downward (negative y axis ) hence the reason the electric force is same magnitude but opposite in direction (upward + y - axis )
Which pieces of information does the National Weather Service produce?
Answer:
1. It collects weather data as part of a network around the country.
2. its territories, adjacent waters and ocean areas for the protection of life and property and the enhancement of the national economy.
Answer:
Maps with isotherms
Explanation:
Unpolarized light is incident upon two polarization filters that do not have their transmission axes aligned. If 38 % of the light passes through this combination of filters, what is the angle between the transmission axes of the filters
Answer:
64°
Explanation:
See attached file
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.659 kJ of heat. It shrinks on cooling, and the atmosphere does 385 J of work on the balloon.
-310J
Explanation:
The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e
ΔE = Q + W ----------------------(i)
If heat is released by the system, Q is negative. Else it is positive.
If work is done on the system, W is positive. Else it is negative.
In this case, the system is the balloon and;
Q = -0.659kJ = -695J [Q is negative because heat is removed from the system(balloon)]
W = +385J [W is positive because work is done on the system (balloon)]
Substitute these values into equation (i) as follows;
ΔE = -695 + 385
ΔE = -310J
Therefore, the change in internal energy is -310J
PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.
A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva so you can better study its flow pattern in the laboratory. Your model must be able to move at 50 cm/s and you will place the model in honey instead of water. Honey has a density of 1400 kg/m3 and a viscosity of 600 Pa-s.
Required:
How long should your model be?
Answer:
Explanation:
For the problem, we should have same reynolds number
ρvd/mu = constant
1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600
d = 25.66 cm
"Consider the Earth and the Moon as a two-particle system. (a) How far from the center of the Earth is the gravitational field of this two-particle system zero?"
a is the correct
Explanation:
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The electric field strength is 1.70 × 104 N/C inside a parallel-plate capacitor with a 0.800 m spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?
Answer:
Here, "v" is the velocity of electron and "V" is the potential.
Three flat layers of transparent material are stacked upon one another. The top layer has index of refraction n1, the middle has n2 and the bottom one has n3. If n1 > n2 > n3, and if a ray of light strikes the top layer at an angle of incidence, in which layer is the angle of refraction the greatest? Why?
a. the bottom layer
b. the top layer
c. Once the ray enters the touching layers, the angle of refraction remains constant.
d. the middle layer
Answer:
a. the bottom medium
Explanation:
it has the least index of refraction and hence most rarer.
Unpolarized light enters a polarizer with vertical polarization axis. The light that passes through passes another polarizer with transmission axis at 40 degrees to the horizontal. What is the intensity of the light after the second polarizer expressed as a fraction of the original intensity
Answer:
I = 0.2934 I₀
Explanation:
The expression that governs the transmission of polarization is
I = I₀ cos² θ
Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of
I₁ = I₀ / 2
this is the light that enters the second polarizer
I = I₁ cos² θ
we substitute
I = I₀ / 2 cos² 40
I = I₀ 0.2934
I = 0.2934 I₀
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.61 rad/sec. The moment of inertia of the student plus the stool is 6 kg m^2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.39 m from the rotation axis.
Required:
a. Calculate the final angular speed of the student. Answer in units of rad/s.
b. Calculate the change in kinetic energy of the system. Answer in units of J.
Answer:
a) the final angular speed is 0.738 rad/s
b) the change in kinetic energy = 0.3 J
Explanation:
the two 1 kg objects have a total mass of 2 x 1 = 2 kg
radius of rotation of the objects = 0.9 m
moment of inertial of the student and the chair = 6 kg-m^2
initial angular speed of rotation of the sitting student and object system ω1 = 0.61 rad/s
final angular speed of rotation of the sitting student and object system ω2 = ?
moment of inertia of the rotating object is
[tex]I = mr^{2}[/tex] = 2 x [tex]0.9^{2}[/tex] = 1.62 kg-m^2
total moment of inertia of sitting student and object system will be
==> 6 + 1.62 = 7.62 kg-m^2
The initial angular momentum of the sitting student and object system will be calculated from
==> Iω1 = 7.62 x 0.61 = 4.65 kg-rad/s-m^2
if the radius of rotation of the object is reduced to 0.39 m,
new moment of inertia of the rotating object will be
[tex]I = mr^{2}[/tex] = 2 x [tex]0.39^{2}[/tex] = 0.304 kg-m^2
new total moment of inertia of the sitting student and object system will be
==> 6 + 0.304 = 6.304 kg-m^2
The final momentum of the sitting student and object system will be calculated from
==> Iω2 = 6.304 x ω2 = 6.304ω2
According to conservation of angular momentum, initial momentum of the system must be equal to the final momentum of the system. Therefore,
4.65 = 6.304ω2
ω2 = 4.65/6.30 = 0.738 rad/s
b) Rotational kinetic energy of the system = [tex]\frac{1}{2} Iw^{2}[/tex]
for the initial conditions, kinetic energy is
==> [tex]\frac{1}{2} Iw1^{2}[/tex] = [tex]\frac{1}{2}* 7.62*0.61^{2}[/tex] = 1.417 J
for the final conditions, kinetic energy is
==> [tex]\frac{1}{2} Iw1^{2}[/tex] = [tex]\frac{1}{2}*6.304*0.738^{2}[/tex] = 1.717 J
change in kinetic energy = final KE - initial KE
==> 1.717 - 1.417 = 0.3 J
A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?
The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration
Complete question:
A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?
Answer:
the total units of insulin admistered each morning
= 22 units of qam and humulin
Explanation:
given
44 units and Humnlin N
with concentration 50/100 = 1/2 = 0.5
∴ 44 × 0.5 ≈ 22 units in the morning
regular insulin administered each day
(22 + 35)units of qam and humulin
= 57units
A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/s. What is the average power done by the torque
Answer:
128.61 WattsExplanation:
Average power done by the torque is expressed as the ratio of the workdone by the toque to time.
Power = Workdone by torque/time
Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]
I is the rotational inertia = 16kgm²
[tex]\theta = angular\ displacement[/tex]
[tex]\theta = 2 rev = 12.56 rad[/tex]
[tex]\alpha \ is \ the\ angular\ acceleration[/tex]
To get the angular acceleration, we will use the formula;
[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]
[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]
Workdone by the torque = 16 * 1.28 * 12.56
Workdone by the torque = 257.23 Joules
Average power done by the torque = Workdone by torque/time
= 257.23/2.0
= 128.61 Watts
A particle accelerator fires a proton into a region with a magnetic field that points in the x-direction. (a) If the proton is moving in the y-direction, what is the direction of the magnetic force on the proton
Answer:
The magnitude of the magnetic field will act in a direction towards me.
Explanation:
When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field. In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.
Complete Question
The diagram for this question is showed on the first uploaded image (reference homework solutions )
Answer:
The velocity at the bottom is [tex]v = 11.76 \ m/ s[/tex]
Explanation:
From the question we are told that
The total distance traveled is [tex]d = 1.2 \ m[/tex]
The mass of the block is [tex]m_b = 0.3 \ kg[/tex]
The height of the block from the ground is h = 0.60 m
According the law of energy
[tex]PE = KE[/tex]
Where PE is the potential energy which is mathematically represented as
[tex]PE = m * g * h[/tex]
substituting values
[tex]PE = 3 * 9.8 * 0.60[/tex]
[tex]PE = 17.64 \ J[/tex]
So
KE is the kinetic energy at the bottom which is mathematically represented as
[tex]KE = \frac{1}{2} * m v^2[/tex]
So
[tex]\frac{1}{2} * m* v ^2 = PE[/tex]
substituting values
=> [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]
=> [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]
=> [tex]v = 11.76 \ m/ s[/tex]