Answer:
M L1 = m L2 torques must be zero around the fulcrum
M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg
Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
Define universal gravitational constant.
The magnification produced by spherical mirror is + 1/4. State the type of spherical mirror. State 3 characteristics of the image formed by the mirror: -
Answer:
Convex mirror.
Explanation:
Image is real.
Image is inverted.
Image is magnified.
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: Find the work function of the irradiated metal in electron volts. work function:
Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E = [tex]\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}[/tex]
= [tex]4.59\times 10^{-19} \ J[/tex]
or,
= [tex]\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
= [tex]2.87 \ eV[/tex]
(b)
As we know,
⇒ [tex]Vq=\frac{hc}{\lambda}-\Phi_0[/tex]
By substituting the values, we get
⇒ [tex]1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0[/tex]
⇒ [tex]\Phi_0=2.3\times 10^{-19} \ J[/tex]
or,
⇒ [tex]=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
⇒ [tex]=1.4375 \ eV[/tex]
How do solar panels work with conduction, convection and radiation?
Answer:
In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.
Explanation:
Can someone please help me with this problem
Answer:
resultant is equal to the sum of A vector or B vector and draw resultant in order that the tail of resultant vector concides with tail of vector a and head of resultant concides with the head of vector b
Explanation:
The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 157 lb (71.0 kg ) person of height 5.91 ft(1.80 m ) would have a body surface area of approximately 1.90 m2 .
Reqiuired:
a. What is the net amount of heat this person could radiate per second into a room at 19.0 ∘C (about 66.2∘F) if his skin's surface temperature is 31.0 ∘C? (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is 1.00, regardless of the amount of pigment.)
b. Normally, 80.0 % of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part A to find this person's basal metabolic rate(BMR).
Answer:
A. Net amount of heat radiated = 109.2W
B. Person's basal energy = 136.5
Explanation:
Part A:
Area of person, A = 1.90 m^2
Temperature of person , T = 31 C
T = 304 K
Temperature of surroundings , To = 19 C
To = 282 K
Now, net amount of heat radiated = e*A*sigma *(T^4 - To^4)
Net amount of heat radiated = 1 * 1.8 * 5.6703 *10^-8 *(304^4 - 294^4)
Net amount of heat radiated = 109.2 W
The net amount of heat radiated is 109.2 W
Part B:
Person's basal energy = net amount of heat radiated /(0.80)
Person's basal energy = 109.2/0.80
Person's basal energy = 136.5 W
Person's basal energy is 136.5 W
Effective sex education must engage _____ more than _____.
Answer:
pregnant
Explanation:
no interest at school
A cell phone weighs about 28x10" pounds. Which value of n is most reasonable?
Answer:
[tex] {3}^{n} [/tex]
Explanation:
https://www.doubtnut.com/question-answers/a-cellphone-weighs-about-28x10n-pounds-which-value-of-n-is-most-reasonable-a-3-b-2-c-0-d-1-433477
Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?
Answer:8pi
Explanat:Omega =2pi/T
When you are standing without moving, you exert a force on the ground. Why doesn't Earth slowly start accelerating downwards?
Explanation:
You would think it should. But remember the Force is also determined by mass. The mass of the earth markes our mass like the smallest part of a mosquito leg. The earth will go on it its merry way without cosidering us at all.
How many molecules do we have for Na2Co3?
105.9888 g/mol is the mass as far as i know, Don't know the amount of molecules tho.
mark me brainliestt :))
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
Required:
At what rate is the magnetic field changing?
This question is incomplete, the complete question is;
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
At what rate is the magnetic field changing?
Answer:
the magnetic field changing at the rate of 9.33 m T/s
Explanation:
Given the data in the question;
Electric field E = 7 mV/m
radius r = 1.5 m
Now, from Faraday law of induction;
∫E.dl = d∅/dt
E∫dl = A( dB/dt )
E( 2πr ) = πr² ( dB/dt )
( 0.007 ) = (r/2) ( dB/dt )
( 0.007 ) = 0.75 ( dB/dt )
dB/dt = 0.007 / 0.75
dB/dt = 0.00933 T/s
dB/dt = ( 0.00933 × 1000) m T/s
dB/dt = 9.33 m T/s
Therefore, the magnetic field changing at the rate of 9.33 m T/s
A person is driving a car down a straight road. The instantaneous acceleration is constant and in the direction of the car's motion. 1) The speed of the car is increasing. decreasing. constant. increasing but will eventually decrease. decreasing but will eventually increase.
Answer:Increasing
Explanation:
Given
Car is driven on the straight road with instantaneous acceleration in the direction of car's motion.
If instanateneous acceleration is constant then speed of car is increasing at a constant pace. As there are no turns on the road, therefore speed of car is increasing.
The speed of the car is "decreasing". A further description is provided in the below paragraph.
It's because the individual would be in a straightforward fashion. This same acceleration inclination comes contrary to the movement of the automobile. It indicates that it exerts pressure against the movement of the automobile. So, when it moves forward, the speed of the automobile decreases.
Thus the above answer is correct.
Learn more about the speed here:
https://brainly.com/question/5053192
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
what is the power of an electrical device which operates with a current of 12.4 A and a potential difference of 12 V
148.8 Watts
Explanation:
P = VI
= (12 V)(12.4 A)
= 148.8 Watts
How are hypotheses tested?
Answer:
by making observation hope it's helpful
A 10,000J battery is depleted in 2h. What power consumption is this? *
A) 5000W
B) 3W
C) 1.4W
D) 20000W
show your work please
Answer:
P = 1.4 W
Explanation:
Given that,
The work done or the energy of the battery, E = 10,000 J
Time, t = 2 h
We need to find the power consumption. Let it is P. Power is the rate of doing work. So,
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{10,000}{2\times 3600}\\\\P=1.38\ W[/tex]
or
P = 1.4 W
So, the power of the battery is 1.4 W.
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passes over the pulley as shown. Masses M1 and M3 lies on a 30o incline plane which slides down the plane. The coefficient of kinetic friction on the incline plane is 0.28. A. Draw a free body diagram of all the forces acting in the masses M1 and M2. B. Determine the tension in the string that connects M2 and M3.
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
∑Fᵧ = maᵧ T₁ - m₂g = m₂aᵧNote that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
T₁ = m₂aᵧ + m₂gT₁ = m₂(a + g)We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
∑Fₓ = maₓ F_f - F_g sinΘ = maₓThe normal force is equal to the x-component of the force of gravity.
(F_n · μ_k) - m₁g sinΘ = m₁aₓ (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ [2.539595871] - [-58.0962595] = 6aₓ 60.63585537 = 6aₓ aₓ = 10.1059759 m/s²Now let's go back to this equation:
T₁ = m₂(a + g)We have 3 known variables and we can solve for the tension force.
T = 2(10.1059759 + 9.8)T = 2(19.9059759)T = 39.8119518 NThe tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
A student must use an object attached to a string to graphically determine the gravitational field strength near Earth's surface. The student attaches the free end of the string to the ceiling and pulls the object-string system so that the string makes an angle of 5 degrees from the object's vertical hanging position. The student then releases the object from rest and uses a stopwatch to measure the time it takes for the object to make one complete oscillation. Which of the following is the next step that will allow the student to determine the gravitational field strength?
А) Repeat the experiment by adding additional mass to the object for multiple trials
B) Repeat the experiment by changing the length of the string for multiple trials
C) Repeat the experiment by changing the angle that the string makes with the object's vertical hanging position
D) Repeat the experiment by measuring the time it takes to make two oscillations, three oscillations, and additional oscillations for multiple trials
Answer:
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
Explanation:
The student is reacting a simple pendulum experiment where he can determine the value of the relationship of gravity with the expression
T = 2π [tex]\sqrt{\frac{L}{g} }[/tex]
let's analyze each statement
A) False. The mass is not a paramer of the period expression
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
C) False. The angle while it is small does not influence the period
D) True. By changing the number of oscillations the period does not change, so you can get the value of the pull of gravity.
We can see that the expressions B and d are true, the most exact value is obtained using procedure B since the graphs allow to reduce the errors
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
(a) v₁ = 51.96 km/h
(b) v₁ = 178 km/h
Explanation:
(a)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Cadillac = 2650 kg
v₂ = speed of Cadillac = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (2650\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{2650\ kg}{816\ kg})[/tex]
v₁ = 51.96 km/h
(b)
For having the same momentum:
m₁v₁ = m₂v₂
where,
m₁ = mass of Volkswagen = 816 kg
v₁ = speed of Volkswagen = ?
m₂ = mass of Truck = 9080 kg
v₂ = speed of Truck = 16 km/h
Therefore, using these values in the equation, we get:
[tex](816\ kg)v_1 = (9080\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{9080\ kg}{816\ kg})[/tex]
v₁ = 178 km/h
occurs when an air mass and its clouds encounter a mountain. This forces the air mass to move from a low elevation to a high elevation as it crosses over the mountain.
Frontal wedging
Orographic lifting
Localized convective lifting
Convergence
Jet streams
Answer:
Localized convective lifting
A 3-kg projectile is launched at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
1517.4 m
Step-by-step explanation:
Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity [tex]v_0[/tex] from the equation
[tex]\frac{1}{2}R= \dfrac{1}{2} \left(\dfrac{v_0^2}{g}\sin 2\theta_0 \right)[/tex]
where [tex]\theta_0 = 45°[/tex] and [tex]\frac{1}{2}R = 50\:\text{m}[/tex] so we will get [tex]v_0=31.3\:\text{m/s}[/tex]. Next, we can use the equation
[tex]v_y = v_0y - gt = v_0 \sin 45 - gt[/tex]
and since [tex]v_y=0[/tex] at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.
At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile, [tex]m_1[/tex] be the mass of the larger fragment (2 kg) and [tex]m_2[/tex] be the mass of the smaller fragment (1 kg). We can write the conservation law as
[tex]Mv_0x = m_1V_1 + m_2V_2[/tex]
where [tex]V_1\:\text{and}\:V_2[/tex] are the velocities of the fragments immediately after the break up. But we also know that [tex]V_1=0[/tex] so the velocity of [tex]m_2[/tex] can be calculated from the conservation law as
[tex]Mv_0 \cos 45° = m_2V_2[/tex]
or
[tex]V_2 = \dfrac{M}{m_2}v_0 \cos 45° = 66.4\:\text{m/s}[/tex]
Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is
[tex]x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}[/tex]
Therefore, the total distance traveled from the launch point is
[tex]D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}[/tex]
A light hollow tube of 2.00 cm diameter and 1.0 m length is filled with tiny beads of different density. The resulting density distribution is linear, with the left end having a density of 1.6 g/cm^3, and the right end having a density of 6.3 g/cm^3. How far from the left end will be the center of mass? (give answer in cm).
Answer:
i think c
Explanation:
cause
What is the internal resistance of a current source with an EMF of 12 V if, when a resistor with an unknown resistance is connected to it, a current of 2 A flows through the circuit? A voltmeter connected to the source terminals shows 8 V.
A charged particle accelerates as it moves from location A to location B. If VA = 260 V and VB = 210 V, what is the sign of the charged particle? positive negative (b) A electron loses electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
Answer:
(a) Positive
(b) Electron gains energy as it moves from A to B.
Explanation:
VA = 260 V
VB = 210 V
An electron moves from lower to higher potential which is negatively charged and a positively charged particle moves from higher to lower potential, so the charge particle is positive in nature.
(a) Positive
(b) No, electron gains energy as it moves from A to B.
I provided the question above.
Answer:
Explanation:
since it is connected in parallel combination
use this formula
[tex]\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2}[/tex]
[tex]\frac{1}{R} = \frac{1}{2} + \frac{1}{4}[/tex]
[tex]\frac{1}{R} = \frac{4+2}{2}[/tex]
[tex]\frac{1}{R} = \frac{6}{2}[/tex]
[tex]\frac{1}{R} = 3[/tex] ohm
therefore resistence = 3 ohm
then we should find power
P = VI
P = 12*3
P = 24 watt
now to find current use formula power = current * voltage
24 = current * 12
24/12 = current
2 = current
therefore current is 2 ampere (A).
to find potential difference (emf) use formula
V = IR
V = current * resistence
V = 2 * 3
V = 6 volt .
therefore potential difference is 6 volt.
To a man running east at the rate of 3m/s vain appears to fall vertically with a speed of 4m/s. Find the actual speed and direction of rain...
Answer:
The actual speed of the rain is 5 m/s and its direction is -53.13°
Explanation:
The actual speed of the rain V = speed of man, v + speed of rain relative to man, v'.
V = v + v'
We add these vectorially.
Since the man's speed is 3 m/s east, in the positive x - direction, we have v = 3i and the rain's speed is falling vertically at 4 m/s, in the negative y- direction, we have v' = -4j
So, V = v + v'
V = 3i + (-4j)
V = 3i - 4j
So, the magnitude of V which is its speed is V = √(3² + (-4)²) = √(9 + 16) = √25 = 5 m/s
The direction of V, Ф = tan⁻¹(vertical component/horizontal component) = tan⁻¹(-4/3) = tan⁻¹(-4/3) = tan⁻¹(-1.333) = -53.13°
So, the actual speed of the rain is 5 m/s and its direction is -53.13°
It takes a minimum distance of 98.26 m to stop a car moving at 17.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Answer:
x_f = 212.5m
Explanation:
t = (x_f-x_0)/(.5*(v_f-v_0))
t = (98.26m-0m)/(.5(0m/s-17m/s))
t = 11.56s
a = (v_f-v_0)/t
a = (0m/s-17m/s)/11.56s
a = -1.47m/s²
t = (v_f-v_0)/a
t = (0m/s-25m/s)/-1.47m/s²
t = 17s
x_f = x_0+(.5*(v_f-v_0))*t
x_f = 0m+(.5*(0m/s-25m/s))*17s
x_f = 212.5m
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
Explanation:
From the given information:
the car's momentum = momentum of the truck
∴
(a) 816 kg × v = 2650 kg × 16.0 km/h
v = (2650 kg × 16.0 km/h) / 816 kg
v = 51.96 km/hr
(b) 816 kg × v = 9080 kg × 16.0 km/h
v = (9080 kg × 16.0 km/h) / 816 kg
v = 178.04 km/hr