Answer:
Visible light intensity at the surface of the bulb (I) = 331 W/m²
Explanation:
Given:
Energy = 75 W
Radius = 6 /2 = 3 cm = 3 × 10⁻² m
Energy goes to visible light = 5% = 0.05
Find:
Visible light intensity at the surface of the bulb (I)
Computation:
Visible light intensity at the surface of the bulb (I) = P / 4A
Visible light intensity at the surface of the bulb (I) = (0.05)(75) / 4π(3 × 10⁻²)²
Visible light intensity at the surface of the bulb (I) = 3.75 / 4π(9 × 10⁻⁴)
Visible light intensity at the surface of the bulb (I) = 331 W/m²
21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fricción μr =0.55. ¿Cuánto desciende antes de detenerse?
Answer:
y = 12.82 m
Explanation:
We can solve this exercise using the energy work theorem
W = ΔEm
friction force work is
W = fr . s = fr s cos θ
the friction force opposes the movement, therefore the angle is 180º
W = - fr s
we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular
N -Wy = 0
N = mg cos θ
the friction force remains
fr = μ N
fr = μ mg cos θ
work gives
W = - μ mg s cos θ
initial energy
Em₀ = ½ m v²
the final energy is zero, because it stops
we substitute
- μ m g s cos θ = 0 - ½ m v²
s = ½ v² / (μ g cos θ)
let's calculate
s = ½ 20² / (0.55 9.8 cos 20)
s = 39.49 m
this is the distance it travels along the plane, to find the vertical distance let's use trigonometry
sin 20 = y / s
y = s sin 20
y = 37.49 sin 20
y = 12.82 m
If you weigh 685 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 km
CHECK COMPLETE QUESTION BELOW
you weigh 685 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 kmkm ? Take the mass of the sun to be msmsm_s = 1.99×1030 kgkg , the gravitational constant to be GGG = 6.67×10−11 N⋅m2/kg2N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be ggg = 9.8 m/s2m/s2 .
Answer:
5.94×10^15N
Explanation:
the weight on the surface of a neutron star can be calculated by below expresion
W= Mg
W= weight of the person
m= mass of the person
g=gravity of the neutron star
But we need the mass which can be calculated as
m= W/g
m= 685/9.81
m= 69.83kg
From the gravitational law equation we have
F= GMm/r^2
G= gravitational constant = 6.67x10⁻¹¹
M= mass of the neutron star = 1.99x10³⁰ kg
r = distance between the person and the surface
Then r can be calculated as = 25/2 = 12.5 km , we divide by two because it's the distance between the person and the surface
g=gravity of the neutron star can be calculated as
g=(6.67×10^-11 ×1.99×10^30)/(12.5×10^3)^2
= 8.50×10^13m/s^2
Then from W= mg we can find our weight
W= 8.50×10^13m/s^2 × 69.83
= 5.94×10^15N
Therefore, weight on the surface of a neutron star is 5.94×10^15N
PLEASE ANSWER ASAP
What happens to the ocean water before the precipitation part of the water cycle? ANSWERS; A.The ocean water condenses into the clouds. B.The ocean water collects back in the ocean. C.The ocean water falls back to Earth's surface. D. The ocean water runs off Earth's surface.
Answer:
B.
Explanation:
The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.
The ocean water prior to the part of the water cycle should be option B.
Ocean water:The ocean water should be collected back in the ocean prior to the part of the water cycle.
Because this should be done when it is evaporated by the sun. When the evaporation is done so the water should be transformed into water vapor.
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(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
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Complete each of the statements
A. Lines of force are lines used to represent ________ an ________ electric field
B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge
C. The electric field is uniform if at any point in the field its _________ and ________ is the same
D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation
Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
A rope, under a tension of 153 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by . where at one end of the rope, is in meters, and is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?
Complete question is;
A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by
y = (0.15 m) sin[πx/3] sin[12π t].
where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?
Answer:
A) Length of rope = 4 m
B) v = 24 m/s
C) m = 1.0625 kg
D) T = 0.11 s
Explanation:
We are given;
T = 153 N
y = (0.15 m) sin[πx/3] sin[12πt]
Comparing this displacement equation with general waveform equation, we have;
k = 2π/λ = π/2 rad/m
ω = 2πf = 12π rad/s
Since, 2π/λ = π/2
Thus,wavelength; λ = 4 m
Since, 2πf = 12π
Frequency;f = 6 Hz
A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;
λ = 2L/n
Since second harmonic, n = 2 and λ = L = 4 m
Length of rope = 4 m
B) speed is given by the equation;
v = fλ = 6 × 4
v = 24 m/s
C) To calculate the mass, we will use;
v = √T/μ
Where μ = mass(m)/4
Thus;
v = √(T/(m/4))
Making m the subject;
m = 4T/v²
m = (4 × 153)/24²
m = 1.0625 kg
D) Now, the rope oscillates in a third harmonic.
So n = 3.
Using the formula f = 1/T = nv/2L
T = 2L/nv
T = (2 × 4)/(3 × 24)
T = 0.11 s
hi guys!!! i have no more points, can someone nice guess all of these for me? :)
1.What happens to the ocean water before the precipitation part of the water cycle
2.During which stage of the water cycle does water from the ocean form clouds?
3.what is a runoff??
4.Which statement about oceans is incorrect? A.Evaporation occurs when water is warmed by the sun. B.Most evaporation and precipitation occur over the ocean. C.97 percent of Earth's water is fresh water from the ocean. D.Water leaves the ocean by the process of evaporation
5.How does most ocean water return to the ocean in the water cycle
tysm to u who answers :)
1. The ocean water collects back in the ocean.
2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
3. an excessive amount of water flowing from downslope along earths surface
4. A.Evaporation occurs when water is warmed by the sun.
5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.
Hope this answer correct ✌️
We've seen that for thermal radiation, the energy is of the form AVT4, where A is a universal constant, V is volume, and T is temperature. 1) The heat capacity CV also is proportional to a power of T, Tx. What is x
Answer:
this raise the temperature is x = 3
Explanation:
Heat capacity is the relationship between heat and temperature change
C = Q / ΔT
if the heat in the system is given by the change in energy and we carry this differential formulas
[tex]c_{v}[/tex] = dE / dT
In this problem we are told that the energy of thermal radiation is
E = A V T⁴
Let's look for the specific heat
c_{v} = AV 4 T³
the power to which this raise the temperature is x = 3
Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.
Answer:
1.125×10⁻⁹ J
Explanation:
Applying,
E = 1/2CV²................... Equation 1
Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.
Given; C = 1.0 nF, = 1.0×10⁻⁹ F, V = 1.5 V
Substitute into equation 1
E = 1/2(1.0×10⁻⁹×1.5²)
E = 1.125×10⁻⁹ J
Hence the energy stored by the capacitor is 1.125×10⁻⁹ J
If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.
Answer:
D. A convex lens in air
Explanation:
This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens
I WILL GIVE BRAINLIEST Identify two types of motion where an object's speed remains the same while it continues to change direction
Answer:
velocity and acceleration
Answer:
Hey there!
Centripetal (Circular Motion) and Oscillating Motion.
Let me know if this helps :)
A ball is thrown upward from a height of 432 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Answer;
A)S(t)=96t-16t² +432
B)it will take 9 seconds for the ball to reach the ground.
C)864feet
Explanation:
We were given an initial height of 432 feet.
And v(t)= 96-32t
A) we are to Find s(t), the function giving the height of the ball at time t
The position, or heigth, is the integrative of the velocity. So
S(t)= ∫(96-32)dt
S(t)=96t-16t² +K
S(t)=96t-16t² +432
In which the constant of integration K is the initial height, so K= 432
b) we need to know how long will the ball take to reach the ground
This is t when S(t)= 0
S(t)=96t-16t² +432
-16t² +96t +432=0
This is quadratic equation, if you solve using factorization method we have
t= -3 or t= 9
Therefore, , t is the instant of time and it must be a positive value.
So it will take 9 seconds for the ball to reach the ground.
C)V=s/t
Velocity= distance/ time
=96=s/9sec
S=96×9
=864feet
By applying the integrations,
(a) [tex]S = 96t-16t^2+432[/tex]
(b) Time will be "t = 9".
(c) Height will be "576"
Given:
Height,
423 feetInitial velocity,
96 feet/secAccording to the question,
(a)
Integrate v:
[tex]S = 96t-16t^2+C[/tex]Initial Condition,
→ [tex]S = 96t-16t^2+432[/tex]
(b)
Hits the ground when,
S = 0→ [tex]0=96t-16t^2+432[/tex]
→ [tex]t =9[/tex]
(c)
Maximum height when,
v = 0→ [tex]0 = 96-32 t[/tex]
→ [tex]t = 3[/tex]
Now,
→ [tex]S = 96\times 3-16\times 3^2+432[/tex]
[tex]= 576[/tex]
Thus the answer above is correct.
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Two protons moving with same speed in same direction repel each other but what about two protons moving with different speed in the same direction?
Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.
In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force
Explanation:
Explanation:
Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.
which is example of radiation
Answer:
Ultraviolet light from the sun.
Explanation:
This is an example of radiation.
Answer:
X-Ray
Explanation:
x-Ray is an example of radiation.
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?
Answer:
Kindly check explanation
Explanation:
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.
Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
The calculation is as follows;
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
[tex](1093.613 \times x) = 45[/tex]
[tex]x = 45 \div 1093.613[/tex]
x = 0.0411480 km
here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.
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how many stars are in our solar system?
Answer:
there are over 100 billion stars in our galaxy.
At a department store, you adjust the mirrors in the dressing room so that they are parallel and 6.2 ft apart. You stand 1.8 ft from one mirror and face it. You see an infinite number of reflections of your front and back.(a) How far from you is the first "front" image? ft (b) How far from you is the first "back" image? ft
Answer:
a) 3.6 ft
b) 12.4 ft
Explanation:
Distance between mirrors = 6.2 ft
difference from from the mirror you face = 1.8 ft
a) you stand 1.8 ft in front of the mirror you face.
According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,
your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft
b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.
the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,
the first image of your back will be 4.4 ft into the back mirror,
therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft
An interference pattern is produced by light with a wavelength 590 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.580 mm .
Required:
a. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
b. What would be the angular position of the second-order, two-slit, interference maxima in this case?
Answer:
a. 0.058°
b. 0.117°
Explanation:
a. The angular position of the first-order is:
[tex] d*sin(\theta) = m\lambda [/tex]
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]
Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.
b. The angular position of the second-order is:
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]
Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.
I hope it helps you!
In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.
Answer:
a. to the west.
Explanation:
An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.
In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.
A small glass bead charged to 5.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 910 N. What is the total charge on the rod?
Answer:
Explanation:
Let B= bead
Q = rod
the electric field at the glass bead pocation is
(Gauss theorem)
E = Q / (2 π d L εo)
the force is
F = q E = q Q / (2 π d L εo)
then
Q = 2 π d L εo F / q
Q = 2*3.14*4x10^-2*10^-1*8.85x10^-12*910x10^-4 / 5x10^-9 = 2.87x10^-8 C = 40.5 nC
I MIND TRICK PLZ HELP LOL
Troy and Abed are running in a race. Troy finishes the race in 12 minutes. Abed finishes the race in 7 minutes and 30 seconds. If Troy is running at an average speed of 3 miles per hour and speed varies inversely with time, what is Abed’s average speed for the race?
Answer:
Explanation:
Let the race be of a fixed distance x
[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]
Troy's Average speed = 3 miles/hr = x / 0.2 hr
x = 0.6 miles
Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr
If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop across this same diode be if the bias current is increased to 1mA? g
Answer:
the voltage drop across this same diode will be 760 mV
Explanation:
Given that:
Temperature T = 300°K
current [tex]I_1[/tex] = 100 μA
current [tex]I_2[/tex] = 1 mA
forward voltage [tex]V_r[/tex] = 700 mV = 0.7 V
To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.
Using the formula:
[tex]I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
where;
[tex]V_r[/tex] = 0.7
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}[/tex]
[tex]\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
[tex]\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
Suppose n = 1
[tex]V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV[/tex]
Then;
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}[/tex]
[tex]{e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =29.37[/tex]
[tex]V_r'=29.37 \times nV_T[/tex]
[tex]V_r'=29.37 \times 25.86[/tex]
[tex]V_r'=759.5 \ mV[/tex]
[tex]Vr' \simeq[/tex] 760 mV
Thus, the voltage drop across this same diode will be 760 mV
The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?
Answer:
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Explanation:
The transformer turns ratio is given by the following equation:
V₁/V₂ = N₁/N₂
where,
V₁ = Voltage of outlet = 120 V
V₂ = Device Voltage = 3 V
N₁ = No. of turns on outlet side = ?
N₂ = No. of turns on side of device = 45
Therefore,
120 V/3 V = N₁/45
N₁ = (40)(45)
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
The block moves up an incline with constant speed. What is the total work WtotalWtotalW_total done on the block by all forces as the block moves a distance LLL
Answer:
External force W₁ = F L
Friction force W₂ = - fr L
weight component W₃ = - mg sin θ L
Y Axis Force W=0
Explanation:
When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.
For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular
let's write the equations of translational equilibrium in given exercise
X axis
F - fr -Wₓ = 0
F = fr + Wₓ
the components of the weight can be found using trigonometry
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
let's look for the work of these three forces
W = F x cos θ
External force
W₁ = F L
since the displacement and the force have the same direction
Friction force
W₂ = - fr L
since the friction force is in the opposite direction to the displacement
For the weight component
W₃ = - mg sin θ L
because the weight component is contrary to displacement
Y Axis
N- Wy = 0
in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0
therefore work is worth zero
) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms
Explanation:
Hey, there!
Here, In question given that,
potential difference (V)= 8V
resistance (R)= 10 ohm
Now,
According to the Ohm's law,
V= R×I { where I = current}
or, I = V/R
or, I = 8/10
Therefore, current is 4/5 A or 0.8 A.
(A= ampere = unit of current).
Hope it helps...
A student wants to create a 6.0V DC battery from a 1.5V DC battery. Can this be done using a transformer alone
Answer:
Therefore, we need an invert, and a rectifier, along with the transformer to do the job.
Explanation:
A transformer, alone, can not be used to convert a DC voltage to another DC voltage. If we apply a DC voltage to the primary coil of the transformer, it will act as short circuit due to low resistance. It will cause overflow of current through winding, resulting in overheating pf the transformer.
Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.
So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.
Therefore, we need an invert, and a rectifier, along with the transformer to do the job.
PLEASE HELP WILL GIVE BRAINLIEST In an experiment, the hypothesis is that if leaf color is related to temperature, then exposing the plant to low temperatures will result in a leaf color change . This hypothesis is _____. 1. testable 2.falsifiable 3.a and b above 4.none of the above
Answer:
testable
Explanation:
high heat can cause browning and or welting
Answer:
A
Explanation:
If it is tested, it will really changed when the plant into low temperature
Which statement belongs to Dalton’s atomic theory? Atoms have a massive, positively charged center. Atoms cannot be created or destroyed. Atoms can be broken down into smaller pieces. Electrons are located in energy levels outside of the nucleus.
Answer:
the correct statement is
* atoms cannot be created or destroyed
Explanation:
The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.
Based on the previous descriptive, the correct statement is
* atoms cannot be created or destroyed
Answer:
the Answer is b hope it help
Explanation:
A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston
Answer:
The cross-sectional area of the larger piston is 392 cm²
Explanation:
Given;
output mass of the piston, m₀ = 2000 kg
input force of the piston, F₁ = 500 N
input area of the piston, A₁ = 10 cm² = 0.001 m²
The output force is given by;
F₀ = m₀g
F₀ = 2000 x 9.8
F₀ = 19600 N
The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;
[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]
Therefore, the cross-sectional area of the larger piston is 392 cm²
The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is
Answer:
E = -8.23 10⁻¹⁷ N / C
Explanation:
In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is
E = k q₁ q₂ / r²
in this case
q₁ is the charge of the proton and q₂ the charge of the electron
E = - k e² / a₀²
let's calculate
E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²
E = -8.23 10⁻¹⁷ N / C
