Answer:
person that you know and like (not a member of your family), and who likes you
two 100 ohm resistors are connected inparallel and one identical resister in series. The maximum power that can be delivered to any one resistor is 25W. What is the maximum voltage that can be applied between the terminals A and B ?
A. 50V
B. 75V
C. 100V
D. 125V
SOLVED DOWN BELOW
Explanation:
In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law
I = 25/200
I= .125 amps or 125 ma
__________
R= r1 * r2 / r1 +r2
R= 100 * 100 / 100 + 100
R= 10000 / 200
R= 50 ohms
Which describes the greenhouse effect?
a. an artificial process
b. a dangerous process
c. a natural process
d. new process
c. a natural process
It is a natural process
In a double-slit arrangement, the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) Calculate the angular separation, !, in radians between the central maximum and the 1st order maximum
Solution :
The conditions for the maximum in the Young's experiment is :
d sin θ = m λ, where m = 0, 1, 2, 3, .....
The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,
d sin θ = λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]
Given : d = 100 λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]
[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]
[tex]$=0.573^\circ$[/tex]
= 0.01 rad
Two gerbils run in place with a linear speed of 0.55 m/s on an exercise wheel that is shaped like a hoop. Find the rotational kinetic energy of the wheel if the exercise wheel has a radius of 9.5 cm and a mass of 5.0 g. Each gerbil has a mass of 0.02 kg if you think that is important.
Answer:
K = 7.56 10⁻⁴ J
Explanation:
The rotational kinetic energy is
K = ½ I w²
They ask us for the kinetic energy of the wheel, which can be approximated as a thin ring, its moment of inertia is
I = M r²
the linear speed of the gerbils is equal to the linear speed of the wheel. The linear and rotational variables are related
v = w r
w = v / r
we substitute
K = ½ (M r²) v² / r²
K = ½ M v²
let's calculate
K = ½ 5 10⁻³ 0.55²
K = 7.56 10⁻⁴ J
A 40-kg crate is being lowered with a downward acceleration is 2.0 m/s2 by means of a rope. (a) What is the magnitude of the force exerted by the rope on the crate
Answer:
F = 312 N
Explanation:
Given that,
The mass of a crate, m = 40 kg
Acceleration of the crate, a = 2 m/s²
As the carte is falling downward, the net force exerted by the rope on the carte is given by :
F = m(g-a)
Put all the values,
F = 40(9.8-2)
F = 312 N
Hence, the required force exerted by the rope on the crate is equal to 312 N.
A 1.2 kg mass is suspended from the ceiling by a string. A second horizontal string holds the mass at rest next to the wall. The angle between the string and the ceiling is 65o. What is the tension force in the horizontal rope
Answer:
The tension in the horizontal string is 25.2 N.
Explanation:
mass, m = 1.2 kg
Angle, A = 65 degree
The tension in the string is T.
So, T cos A = m g
T x cos 65 = 1.2 x 9.8
T x 0.4226 = 11.76
T = 27.83 N
Tension in the horizontal string, T' = T sin A = 27.83 x sin 65 = 25.22 N
Answer:
Explanation:
See the figure attached . T₁ and T₂ are tension in the inclined and horizontal string .
The vertical component of T₁ will balance weight and horizontal component will balance the tension T₂.
T₁ sin 65 = mg and T₁ cos 65 = T₂
Dividing ,
Tan 65 = mg / T₂
T₂ = mg / tan 65
= 1.2 x 9.8 / 2.1445
= 5.5 N
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
A.
69.58 newtons
B.
45.05 newtons
C.
25 newtons
D.
98.26 newtons
Answer:
Option A (69.56 newtons) is the appropriate solution.
Explanation:
According to the question,
On the X-axis,
⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]
or,
[tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]
On substituting the values, we get
[tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]
[tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)
On the Y-axis,
⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]
[tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]
[tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]
From equation 1, we get
[tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]
[tex]T_1+3T_1=278.4 \ N[/tex]
[tex]4T_1=278.4 \ N[/tex]
[tex]T_1=\frac{278.4}{4}[/tex]
[tex]=69.6 \ N[/tex]
Answer:
69.58
Explanation:
When thrust is doubled, pressure is______.
Answer:
doubled
Explanation:
When thrust is double so will the pressure I hope this helps
enjoy
A 4.9 A current is set up in a circuit for 4.7 min by a rechargeable battery with a 12 V emf. By how much is the chemical energy of the battery reduced
Answer:
E = 16581.6 J
Explanation:
Given that,
Current, I = 4.9 A
Time for which the current is set up, I = 4.7 min = 282 s
The voltage of the battery, V = 12 V
We need to find how much chemical energy of the battery reduced. Let It is E. We know that,
E = P t
Where
P is power of battery, P = VI
So,
[tex]E=VIt[/tex]
Put all the values,
[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]
So, 16581.6 J of chemical energy of the battery is reduced.
Please helppppppp!!!!!!
Answer:
The resulting force on the first object is 800 N.
Explanation:
Given;
force exerted on one of the objects, F₁ = 400 N
let the first charge = q₁
let the second charge = q₂
The force of repulsion between the objects is calculated using Coulomb's law;
[tex]F =\frac{kq_1q_2}{r^2} \\\\\frac{k}{r^2} = \frac{F}{q_1q_2} = \frac{F_1}{q_1\times2q_2} \\\\F_1 = \frac{F(q_1\times2q_2)}{q_1q_2} \\\\F_1 = 2F\\\\F_1 = 2(400 \ N)\\\\F_1 = 800 \ N[/tex]
Therefore, the resulting force on the first object is 800 N.
Coherent monochromatic light falls perpendicularly on two slits (each of width 0.10 mm) separated by 0.50 mm. In the resulting interference pattern on a screen 2.80 m away, adjacent bright fringes are separated by 2.80 mm. (a)What is the wavelength of the light that falls on the slits
Answer:
The correct answer is "[tex]0.5\times 10^{-6} \ m[/tex]".
Explanation:
Given:
[tex]\frac{\lambda D}{d} =2.8\times 10^{-3}[/tex]
[tex]d = 0.5\times 10^{-3}[/tex]
[tex]D = 2.80[/tex]
Now,
The wavelength will be:
⇒ [tex]\lambda = 2.8\times 10^{-3}\times \frac{d}{D}[/tex]
By putting the values, we get
⇒ [tex]=\frac{2.8\times 10^{-3}\times 0.5\times 10^{-3}}{2.8}[/tex]
⇒ [tex]=\frac{1.4\times 10^{-6}}{2.8}[/tex]
⇒ [tex]=0.5\times 10^{-6} \ m[/tex]
What is the relationship between the density of the equipotential lines, the density of the electric field lines and the strength of the electric field?
Answer:
I dont. understand the question, maybe insert the picture?
After turning on the power source connected to your two electrodes, we expect to see the microbeads moving through the solution. What forces are acting on the microbeads as they move (ignore vertical forces)
Answer:
the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
Explanation:
The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
F = q E
____ is the study of things getting faster as they move.
A. Anatomy
B. Force
C. Physics
D. Dynamics
Answer: b force
Explanation:
yes because the world comin g up with more technique
People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm. a. What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.5 cm
This question is incomplete, the complete question is;
People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.
a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.) __ mm
Answer:
1) the power in D of the eyes of a woman is 61.7647 D
2) the size in mm of an image of a 8.00 mm object is -1.882 mm
Explanation:
Given the data in the question;
a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,
so
u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m
v = 2.00 cm = ( 2 / 100 )m = 0.02 m
Now, we know that power of lens p = 1 / u + 1 / v
so we substitute
p = ( 1 / 0.085 ) + ( 1 / 0.02 )
p = 11.7647 + 50
p = 61.7647 D
Therefore, the power in D of the eyes of a woman is 61.7647 D
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)
we know that;
m = -v / u
we substitute
m = -0.02 / 0.085
m = -0.2353
since H₀ = 8.0 mm
H[tex]_i[/tex] = m × H₀
H[tex]_i[/tex] = -0.2353 × 8.0
H[tex]_i[/tex] = -1.882 mm
the size in mm of an image of a 8.00 mm object is -1.882 mm
Since when was the light we see now emanating from the quasar? Note that the distance between the Earth and the quasar is 598 Mpc
a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(16N )and f2 =-(12.0N)+(8.0N) find the third force
By Newton's second law, the net force on the object is
∑ F = m a
∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N
Let f be the unknown force. Then
∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f
=> f = (-2.0 i - 12.0 j ) N
Explain how muscles are effected by space travel
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force between them?
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
The function s(t)s(t) describes the position of a particle moving along a coordinate line, where ss is in feet and tt is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, specd, and acceleration at time t
Answer:
Explanation:
From the given information:
Let's assume that the missing function is:
s(t) = t³ - 6t², t ≥ 0
From part (b), we are to find the given required terms when time t = 2
So; from the function s(t) = t³ - 6t², t ≥ 0
[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]
[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]
[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]
[tex]acceleration\ a(t) = 6t - 12[/tex]
At time t = 2
The position; S(2) = (2)² - 6(2)²
S(2) = 8 - 6(4)
S(2) = 8 - 24
S(2) = - 16 ft
v(2) = 3(2)² - 12 (2)
v(2) = 3(4) - 24
v(2) = 12 - 24
v(2) = - 12 ft/s
speed = |v(2)|
|v(2)| = |(-12)|
|v(2)| = 12 ft/s
acceleration = 6t - 12
acceleration = 6(2) - 12
acceleration = 12 - 12
acceleration = 0 ft/s²
an animal which is known as an ascendant of man
Answer:
spirit animal?
Explanation:
a train is traveling at 50km/h average .what is the displacement of the train per second?
A cart weighing 40 pounds is placed on a ramp incline 15 degrees to the horizon. The cart is held in place by a rope inclined 60 degrees to the horizontal. find the force that the rope must exert on the cart to keep it from rolling down the ramp.
Answer: [tex]14.64\ N[/tex]
Explanation:
Given
Inclination of ramp is [tex]\theta=15^{\circ}[/tex]
Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal
Weight of cart [tex]W=40\ lb[/tex]
from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp
Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position
[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]
Answer this
a) which ink is likely to be pure? Why?
b) What does the chromatography tell us about ink Y
c) Why are the three different spots separated out from ink Y found at different heights?
Answer:
a) Ink X is likely to be pure because it only contain 1 spot.
b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.
c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.
The different spots from Y are found at various heights because they represent different compounds.
What is chromatography?The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.
We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.
Learn more about chromatography:https://brainly.com/question/26491567?
#SPJ6
Two workers are sliding 450 kg kg crate across the floor. One worker pushes forward on the crate with a force of 380 NN while the other pulls in the same direction with a force of 230 NN using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
Answer:
The coefficient of kinetic friction on the floor is 0.138
Explanation:
Given;
mass of the crate, m = 450 kg
force applied by the first worker, F₁ = 380 N
force applied by the second worker in the same direction as the first worker, F₁ = 230 N
frictional force opposing the motion of the box = -[tex]F_k[/tex]
Apply Newton's second law of motion;
∑F = ma
[tex]F_1 + F_2 - F_k = ma[/tex]
If the crate slides with constant speed, acceleration is zero (0).
[tex]F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138[/tex]
Therefore, the coefficient of kinetic friction on the floor is 0.138
Cuando Daniel hace oscilar un péndulo, este realiza 30,6 ciclos (completos) en 9 [s].
¿Cuál es la frecuencia del péndulo?
A )3,4 [Hz].
B )4,3 [Hz].
C )30 [Hz].
D )5 [Hz]
(10) The use of Doppler radar for speed detection and enforcement on the roads is very common and has been in use for a long time. Suppose a 10 GHz radar (also called radar gun or speed gun) measures the speed of a car at 120 km/h moving towards the radar gun. a. What is the change in the frequency of the reflected wave due to the speed of the car b. Calculate the sensitivity of the device in [Hz/km].
Answer:
The sensitivity of the device = 1.234 Hz per km
Explanation:
Given
Frequency (f) = 10 gHz
Speed of the car = 120 Km/h
As per the doppler’s effect
V = (change in frequency /frequency) *(c/2)
Substituting the given values, we get –
Change in frequency = {(2*10^9*120)/(3*10^8)} * (1000/3600)
Change in frequency = 37.03 Hz
b) speed of light = wavelength * frequency
3*10^8 = wavelength * 10*10^9
Wavelength = 0.03 m
Sensitivity = change in frequency /wavelength = 37.03/0.03 = 1234 Hz/m
1.234 Hz per km
Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of planets from earliest to latest.
a. Small clumps of matter stick together via the process of accrection to form plantesimals a few hundred kilometers in diameter
b. Dust keeps matter inside the disk cool long enough for planet formation to start
c. Planetisimals begin to accrete, forming protoplanets
d. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter
e. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
Answer: See explanation
Explanation:
The evolutionary stages for the formation of planets from earliest to latest will be:
1. Dust keeps matter inside the disk cool enough for planet formation to start
2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.
3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.
4. Planetesimals begin to accrete, forming protoplanets.
5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
Danny is riding his bike at 12m/s he tries to pop a wheelie but he fails hits a curb flies through the air and comes to a complete stop in 30 seconds what is Danny's deceleration
Answer:
a = -0.4m/s²
Explanation:
v_f = v_I + (a)(t)
a(t) = v_f-v_I
a = (v_f-v_I)/t
a = (0m/s-12m/s)/30s
a = -0.4m/s²
Which of these is NOT an effect of humor?
strengthened immune system
reduced stress levels
reduced feelings of anxiety
feelings of jealousy and envy