Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
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5. The path length difference for the waves exiting the two slits of the double slit experiment must be equal to _____ for a bright fringe to appear.
Answer:
An integral or whole multiple of the wavelength λ
or d sin θ = mλ,
for m = 0, 1, −1, 2, −2, ...,
Explanation:
In the double slit interference pattern, if we consider how two waves travel from the slits to the screen, we'll see that each slit is a different distance from a given point on the screen hence, they posses different wavelengths. Waves in a double slit experiment will be in phase if they interfere constructively by starting out crest to crest, or trough to trough. If the waves arrive crest to trough, they will interfere destructively, and arrive out of phase. A constructive interference occurs when the path length difference of the waves exiting the two slits forms an integral multiple of wavelength at the screen. A destructive interference occurs if the path length differs by half a wavelength. Constructive interference forms the bright fringes, while the dark fringes are formed by destructive interference.
why does a hot-air ballon rise? a the volume of the air dispalced by the balloon is less than the volume of the balloon. b the weight of the air dispaced by the ballon is less than the volue of the ballon. c the wight of the balloon is less than the weight of the air displced by the balloon.
Answer:
the answer is c
Explanation:
the balloon is lighter than the air around displaced by it
Question 5
A fidget spinner that is 4 inches in diameter is spinning clockwise. The spinner spins at 3000
revolutions per minute.
At t = 0, consider the point A on the outer edge of the spinner that is along the positive horizontal
axis. Let h(t) be the vertical position of A in inches. Suppose t is measured in minutes. Find a
sinusoidal function that models h(t).
h(t) =
Can someone please help me for this question?!!!!! ASAP?!!!!
Answer:
h = 4 sin (314.15 t)
Explanation:
This is a kinematics exercise, as the system is rotating at a constant speed.
w = θ / t
θ = w t
in angular motion all angles are measured in radians, which is defined
θ = s / R
we substitute
s / R = w t
s = w R t
let's reduce the magnitude to the SI system
w = 3000 rev / min (2π rad / 1rev) (1min / 60 s) = 314.16 rad / s
let's calculate
s = 314.16 4 t
s = 1,256.6 t
this is the value of the arc
Let's find the function of this system, let's use trigonometry to find the projection on the x axis
cos θ = x / R
x = R cos θ
x = R cos wt
projection onto the y-axis is
sin θ = y / R
how is a uniform movement
θ = w t
y = R sin wt
In the case y = h
h = R sin wt
h = 4 sin (314.15 t)
if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
Answer:
neither will happen
Explanation:
cause the water is already defreezed
A particle with a charge of 4.0 μC has a mass of 5.0 × 10 -3 kg. What electric field directed upward will exactly balance the weight of the particle?
Answer:
E = 12.25 x 10³ N/C = 12.25 KN/C
Explanation:
In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,
Electrostatic Force = Weight
E q = mg
where,
E = Electric Field = ?
m = Mass of the Charge = 5 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C
Therefore,
E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)
E = 0.049 N/4 x 10⁻⁶ C
E = 12.25 x 10³ N/C = 12.25 KN/C
The voltage across a membrane forming a cell wall is 84.0 mV and the membrane is 9.40 nm thick. What is the electric field strength (in V/m)
Answer:
8.9*10^6 V/m
Explanation:
The expression for electric field strength E is given as
[tex]E=\frac{V}{d}[/tex]
where V= voltage
d= distance of separation
Given data
[tex]voltage= 84 mV= 84*10^-^3\\distance= 9.4*10^-^9[/tex]
substituting our given data into the electric field strength formula we have
[tex]E= \frac{84*10^-^3}{ 9.4*10^-^9} \\\\E= \frac{84}{9.4} *10^-^3^-^(^-^9^)\\\\E= \frac{84}{9.4} *10^-^3^+^9\\\\E= \frac{84}{9.4} *10^6\\\\E=8.9*10^6 V/m[/tex]
A capacitor is to be constructed to have a capacitance of 100uF.The area of the plates is 6.om by 0.030m and the relative permitivityof dielectric is 7.0 Find the necessary separation of the plates and the electric field strength if a potential difference of 12V is applied across the capacitor.
Answer:
The answer is below
Explanation:
Given that:
The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²
the relative permittivity of dielectric (εr) is 7.0
Permittivity of free space (εo) = 8.854 × 10^(-12)
capacitance of 100uF
potential difference (V) of 12V
d = separation between plate
The capacitance (C) of a capacitor is given by:
[tex]C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m[/tex]
The electric field between plates is given as:
E = V /d
[tex]E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m[/tex]
Gun was fired with a muzzle velocity of 350m/s, mounted at an angle of 45’ above the ground. Neglecting air resistance, compute for the following;
*Maximum height reached
*Range of the projectile
*Total time of flight
Answer:
Maximum height, h = 3062.5m
Total time of flight, T = 49.49secs or 50secs approx.
Range, R = 12250m
Explanation:
Given data:
U = 350m/s
Angle = 45°
Assume g = 10m/s
At the greatest height, v = 0
Therefore,
V^2 = U^2 sin^2 × angle - 2×g×h
Substituting values:
0^2 = 350^2 sin^2 (45) - 2 × 10 × h
Let h = maximum height reached
Rearranging gives:
350^2 sin^2(45) = 2 x 10 x h
h = 350^2 sin^2(45)/2×10
h = 122500 x 0.5/20
h = 61250/20
h = 3062.5m
2)Total time of flight, T
T = 2U sin(angle)/g
= 2x350 sin(45)/10
= 494.9747/10
= 49.49secs or 50sec approx.
3) Range of projectile, R
R = U^2 sin2(angle)
= 350^2 sin2 (45)
= 122500 x 1/10
= 12250m
Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface
Answer:
(a) 4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface, (b) The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
Explanation:
The complete statement is: "Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface, (b) How far has the rocket traveled when half the total work has occurred?"
(a) The work required to propel the rocket is given by the change in gravitational potential energy, whose expression derives is described below:
[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]
Where:
[tex]U_{g,o}[/tex], [tex]U_{g,f}[/tex] - Initial and final gravitational potential energies, measured in joules.
[tex]m[/tex], [tex]M[/tex] - Masses of the rocket and planet Earth, measured in kilograms.
[tex]G[/tex] - Universal gravitation constant, measured in newton-square meters per square kilogram.
[tex]r_{o}[/tex], [tex]r_{f}[/tex] - Initial and final distances of the rocket with respect to the center of the Earth, measured in meters.
The initial distance and rocket mass are converted to meters and kilograms, respectively:
[tex]r_{o} = (4000\,mi)\cdot \left(1609.34\,\frac{m}{mi} \right)[/tex]
[tex]r_{o} = 6,437,360\,m[/tex]
[tex]m = (7\,ton)\cdot \left(1000\,\frac{kg}{ton} \right)[/tex]
[tex]m = 7000\,kg[/tex]
Given that [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]r_{f} \rightarrow +\infty[/tex], the work equation is reduced to this form:
[tex]U_{g,f} - U_{g,o} = \frac{G\cdot m \cdot M}{r_{o}}[/tex]
[tex]U_{g,f} - U_{g,o} = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)}{6,437,360\,m}[/tex]
[tex]U_{g,f} - U_{g,o} = 4.334\times 10^{11}\,J[/tex]
4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface.
(b) The needed change in gravitational potential energy is:
[tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex]
The expression for the change in gravitational potential energy is now modified by clearing the final distance with respect to the center of Earth:
[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]
[tex]\frac{U_{g,o}-U_{g,f}}{G\cdot M \cdot m} = \frac{1}{r_{f}} - \frac{1}{r_{o}}[/tex]
[tex]\frac{1}{r_{f}} = \frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m}[/tex]
[tex]r_{f} = \left(\frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m} \right)^{-1}[/tex]
If [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex], then:
[tex]r_{f} = \left[\frac{1}{6,437,360\,m}-\frac{2.167\times 10^{11}\,J}{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)} \right]^{-1}[/tex]
[tex]r_{f} \approx 12,874,502.49\,m[/tex]
The final distance with respect to the center of the Earth in miles is:
[tex]r_{f} = (12,874,502.49\,m)\cdot \left(\frac{1}{1609.34}\,\frac{mi}{m} \right)[/tex]
[tex]r_{f} = 7999.865\,mi[/tex]
The distance travelled by the rocket is: ([tex]r_{f} = 7999.865\,mi[/tex], [tex]r_{o} = 4000\,mi[/tex])
[tex]\Delta r = r_{f}-r_{o}[/tex]
[tex]\Delta r = 7999.865\,mi - 4000\,mi[/tex]
[tex]\Delta r = 3999.865\,mi[/tex]
The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
radiation transfers energy through___. a metal. b liquid. c touch. d waves.
Answer:
Radiation is transferred through electromagnetic waves so D.
Explanation:
Answer:
D. Waves
Explanation:
a and b don't make much sense, conduction is transfer of energy through touch
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
A wave travels at a consent speed. how does the frequency change if the wavelength is reduced by a factor of 4?
Answer:
The frequency increases by 4 because it is inversely proportional to the wavelength.
A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is _______.
Answer:
10 kgm/s
Explanation:
Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.
From the question,
ΔM = m(v-u)...................... Equation 1
Where ΔM = change in momentum, u = initial velocity, v = final velocity.
Note: Let upward direction be negative, and downward direction be positive.
Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s
Substitute into equation 1
ΔM = 0.2(-20-30)
ΔM = 0.2(-50)
ΔM = -10 kgm/s.
The negative sign shows that the change in momentum is Upward
The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
Given data:
The mass of rubber ball is, m = 0.2 kg.
The initial speed of ball is, u = 30 m/s.
The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)
The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,
[tex]p= m ( v-u)[/tex]
Solving as,
[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]
Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
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which of the following is true about scientific models?
A. models are used to simplify the study of things
B.computer models are the most reliable kind of model
C. models explain past, present and future information
D.a model is accurate if it does not change over time
Answer:
A- to simplify the study of things
Explanation:
a visual reference
BIO A trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. A 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms; these are all typical values. At what speed does it leave the ground
Answer:
Final velocity (v) = 0.509 m/s (Approx)
Explanation:
Ant use impulse power
Given:
Mass of ant = 12 mg = 12 × 10⁻⁶ kg
Average force = 47 mN = 47 × 10⁻³ N
Initial velocity(u) = 0
Time taken = 0.13 ms = 0.13 × 10⁻³ s
Find:
Final velocity (v)
Computation:
Force × Time = change in momentum
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = mv - mu
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = m(v - u)
6.11 × 10⁻⁶ = 12 × 10⁻⁶(v - 0)
6.11 = 12 v
Final velocity (v) = 0.509 m/s (Approx)
1) What is the highest atomic number element a red dwarf star can produce in its core? a. Carbon b. Oxygen c. Helium d. Iron
2) What is the highest atomic number element that can be produced in the cores of the largest stars?a. Helium b. Oxygen c. Iron d. Carbon
3) If formed at the same time, a red dwarf star is likely to become a white dwarf faster than a Sun-like star would. a) True b) False
Answer:
1) c. Helium
2) Iron
3) False.
Explanation:
1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.
2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.
3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.
Inductance is usually denoted by L and is measured in SI units of henries (also written henrys, and abbreviated H), named after Joseph Henry, a contemporary of Michael Faraday. The EMF E produced in a coil with inductance L is, according to Faraday's law, given by
E=−LΔIΔt.
Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?
Answer:
Explanation:
E= −L ΔI / Δt.
L = E Δt / ΔI
Hence the unit of inductance may be V s A⁻¹
or volt s per ampere .
In the given case
change in current ΔI = - 2.5 A
change in time = .015 s
L = .56 H
E = − L ΔI / Δt.
= .56 x 2.5 / .015
= 93.33 V .
If you have completely polarized light of intensity 125 W/m2, what will its intensity be after passing through a polarizing filter with its axis at an 89.5° angle to the light's polarization direction?
Answer:
When we have completely polarized light with intensity I0, and it passes through a polarizing filter with its axis at an angle θ with respect to the light's polarization direction, the new intensity of the light will be:
I = I0*cos^2(θ)
This is called the "Malus' law".
in this case, we have:
I0 = 125 W/m^2
θ = 89.5°
then:
I = (125 W/m^2)*cos^2(89.9°) = 0.00038 W/m^2
A brass ring of diameter 10.00 cm at 20.0 ^ { \circ } \mathrm { C }20.0 ∘C is heated and slipped over an aluminum rod of diameter 10.01 cm at 20.0 ^ { \circ } \mathrm { C }20.0 ∘C. Assuming the average coefficients of linear expansion are constant, What if the aluminum rod were 10.02 cm in diameter?
Answer:
b)using
DT = (LAl - LBr) / (LBr aBr - LAl aAl)
DT = (10.02-10)/(10*19x10-6 –10.02*24x10-6)
DT = -396 C°
20°C + -396 C° < -273.15 °C;
So the temperature will be -396° this is unattainable because we can’t go below absolute zero
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
The movement of a car on a road is represented in this figure. Between t = 0 and t = 0.6 hrs, what is the displacement made by the car?
1). 4.0 km.
2). 0.0 km.
3). -4.0 km. 4
). 8.0 km.
Answer:
0
Explanation:
On a graph, if there is plot of time vs velocity, then area of velocity plot gives the displacement.
also we can see area of plot is velocity* time which is equal to formula of displacement.
area for this plot is velocity * time
Thus,
from
t =0 to t = 0.2
v = 20
t = 0.2 - 0 = 0.2
thus, displacement till 0.2 seconds = 20*0.2 = 4 Km
____________________________________________________
from
t =0 to t = 0.4
v = 0
t = 0.4 - 0.2 = 0.2
thus, displacement from 0.2 seconds to 0.4 seconds = 0*0.2 = 0 Km
____________________________________________________
from
t =0.4 to t = 0.6
v = -20
t = 0.6 - 0.4 = 0.2
thus, displacement from 0.4 seconds to 0.6 seconds = -20*0.2 = -4 Km
Thus, total displacement = 4+0 -4 = 0
Thus, net displacement made by car is 0.
A 1300-turn coil of wire that is 2.2 cm in diameter is in a magnetic field that drops from 0.11 T to 0 {\rm T} in 12 ms. The axis of the coil is parallel to the field What is the emf of the coil?
Answer:
The induced voltage is [tex]\epsilon = 4.53 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1300 \ turns[/tex]
The diameter is [tex]d = 2.2 \ cm =0.022 \ m[/tex]
The initial magnetic field is [tex]B_i = 0.11 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
The time taken is [tex]t = 12 \ ms = 12*10^{-3} \ s[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.022}{2}[/tex]
[tex]r = 0.011 \ m[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{d\phi}{dt}[/tex]
Where [tex]d\phi[/tex] is the change in magnetic flux of the wire which is mathematically represented as
[tex]d \phi = dB* A * cos \theta[/tex]
=> [tex]d \phi = (B_f - B_i )* A * cos \theta[/tex]
Here [tex]\theta = 0[/tex]
since the axis of the coil is parallel to the field
Where A is the cross-sectional area of the coil which is mathematically represented as
[tex]A = \pi * r^2[/tex]
[tex]A = 3.142 * 0.011^2[/tex]
[tex]A = 3.80*10^{-4} \ m^2[/tex]
So the induced emf
[tex]\epsilon = - 1300 * \frac{(0- 0.11) * 3.80*10^{-4}}{12*10^{-3}}[/tex] Here we substituted the values of [tex]d \phi[/tex]
[tex]\epsilon = 4.53 \ V[/tex]
The emf induced in the coil at the given magnetic field strength is 4.53 V.
The given parameters;
number of turns, N = 1300 turnsdiameter of the coil, d = 2.2 cminitial magnetic field, B₁ = 0.11 Tfinal magnetic field, B₂ 0time, t = 12 msThe area of the coil is calculated as follows;
[tex]A = \pi r^2 = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times 0.022^2}{4} = 0.00038 \ m^2[/tex]
The emf induced in the coil is calculated as follows;
[tex]emf = -N\frac{d\phi}{dt} \\\\emf = N (\frac{\phi_1 - \phi_2}{t} )\\\\emf = N(\frac{AB_1 - AB_2}{t} )\\\\emf = NA(\frac{B_1 - B_2}{t} )\\\\emf = 1300 \times 0.00038 (\frac{0.11 - 0}{12 \times 10^{-3}} )\\\\emf = 4.53 \ V[/tex]
Thus, the emf induced in the coil at the given magnetic field strength is 4.53 V.
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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.
Answer:
206.67NExplanation:
The sum of force along both components x and y is expressed as;
[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]
The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
[tex]a_x = \frac{d^2 x }{dt^2}[/tex]
[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]
Similarly,
[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]
[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]
[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride. Part A If a boat and its riders have a mass of 1200 kg and the boat drifts in at 1.3 m/s how much work does Sam do to stop it
Answer:
W = 1014 J = 1.014 KJ
Explanation:
As, Sam has to stop the boats in the log ride. Therefore, the work Sam needs to do, in order to stop a boat must be equal to the kinetic energy of the boat:
Work Done by Sam = Kinetic Energy of the Boat
W = K.E
W = (1/2)mv²
where,
m = mass of boat and its rider = 1200 kg
v = speed of the boat = 1.3 m/s
Therefore,
W = (1/2)(1200 kg)(1.3 m/s)²
W = 1014 J = 1.014 KJ
Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?
Answer:
0.11m
Explanation:
let's assume the boat is of uniform construction
Ignoring friction losses
Also assume the origin is at the end of the boat originally with the heavier person
the center of mass of the whole system will not change relative to the water when the two swap ends
Originally, the center of mass is
85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin
after the swap, the center of mass is
50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin
The center of mass has shifted
1.14-1.030 = 0.11m
as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat
Visible light of wavelength 589 nm is incident on a diffraction grating that has 3500 lines/cm. At what angle with respect to the central maximum is the fifth order maximum observed
Answer:
dsinФ=mΔ
d=1/N
d=1/3500*[tex]10^{-2} m\\[/tex]
d=2.8*[tex]10^{-6}[/tex]
NOw apply all values on formula
dsinФ=mΔ
2.8*[tex]10^{-6\\}[/tex]sinФ=5*589*[tex]10^{-9}[/tex]
sinФ=1.05 error
so due fifth maximum order it cannot be soved by this grating
To compensate for acidosis, the kidneys will
Answer:
Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.
In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.
A system loses 510 J of potential energy. In the process, it does 430 J of work on the environment and the thermal energy increases by 100 J.
Required:
Find the change in kinetic energy.
Answer:
20J
Explanation:
The computation of the change in kinetic energy is shown below:
As we know that
Work was done on system = change in potential energy + change in kinetic energy + change in thermal energy
-430J = -510J + change in kinetic energy + 100J
-430J = -410J + change in kinetic energy
So, the change in kinetic energy is 20J
We simply applied the above formula to find out the change in kinetic energy
A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant for this circuit
Answer:
The time constant is [tex]\tau = 1.265 s[/tex]
Explanation:
From the question we are told that
the time take to charge is [tex]t = 2.4 \ s[/tex]
The mathematically representation for voltage potential of a capacitor at different time is
[tex]V = V_o - e^{-\frac{t}{\tau} }[/tex]
Where [tex]\tau[/tex] is the time constant
[tex]V_o[/tex] is the potential of the capacitor when it is full
So the capacitor potential will be 100% when it is full thus [tex]V_o =[/tex]100% = 1
and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so
V = 0.85
Hence
[tex]0.85 = 1 - e^{-\frac{2.4}{\tau } }[/tex]
[tex]- {\frac{2.4}{\tau } } = ln0.15[/tex]
[tex]\tau = 1.265 s[/tex]
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in its secondary coil, and an input voltage of 120 V. Randomized Variables Δ 33%
Part (a) What is the voltage output Vs, in volts, of the transformer used for to charge the batteries? Grade Summar Deductions Potential sin tan) ( Submissions Attempts remain coso cotan) asin) acos() atan acotan)sinh( cosh)tanhcotanh0 % per attempt detailed view END Degrees Radians DEL CLEAR Submit Hint I give up! Hints:% deduction per hint. Hints remaining:I Feedback: 1% deduction per feedback. - 쇼 33%
Part (b) what input current ,. İn milliamps, is required to produce a 3.2 A output current? 33%
Part (c) What is the power input, in watts?
Answer:
a) 0.72 V
b) 19.2 mA
c) 2.304 Watts
Explanation:
A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.
number of primary turns = [tex]N_{p}[/tex] = 500 turns
input voltage = [tex]V_{p}[/tex] = 120 V
number of secondary turns = [tex]N_{s}[/tex] = 3 turns
output voltage = [tex]V_{s}[/tex] = ?
using the equation for a transformer
[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]
substituting values, we have
[tex]\frac{V_{s} }{120 } = \frac{3 }{500} }[/tex]
[tex]500V_{p} = 120*3\\500V_{p} = 360[/tex]
[tex]V_{p}[/tex] = 360/500 = 0.72 V
b) by law of energy conservation,
[tex]I_{P}V_{p} = I_{s}V_{s}[/tex]
where
[tex]I_{p}[/tex] = input current = ?
[tex]I_{s}[/tex] = output voltage = 3.2 A
[tex]V_{s}[/tex] = output voltage = 0.72 V
[tex]V_{p}[/tex] = input voltage = 120 V
substituting values, we have
120[tex]I_{p}[/tex] = 3.2 x 0.72
120[tex]I_{p}[/tex] = 2.304
[tex]I_{p}[/tex] = 2.304/120 = 0.0192 A
= 19.2 mA
c) power input = [tex]I_{p} V_{p}[/tex]
==> 0.0192 x 120 = 2.304 Watts