Explanation:
The torque [tex]\tau[/tex] is given by
[tex]\tau=Fd = (65\:\text{lbs})(\frac{8}{12}\:\text{ft}) = 43.3\:\text{lbs-ft}[/tex]
Which of the following phenomenon odd called photoelectric effect?
A. High energy electrons impinge on a metallic Annie which emits electrons
B. A high energy photon emits photons as it slows down
C. A metal absorbs a quanta of light and then emits electrons
D. Two electrons are created from a quanta of light.
two electrons are created from a quanta of kight
what is effort arm
don't say the answer of gogle
Answer:
effort arm mean the use of any work by using your hand force motion or by hand power
help plzzzzzzzzzzzz ?
Explanation:
1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:
[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]
[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]
Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so
[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]
Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so
[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]
Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is
[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]
2. The total current in the circuit is
[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]
3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is
[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]
4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us
[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]
5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is
[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]
This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is
[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]
What must a plant need to produce 4 molecules of sugar?
What must a plant need to produce 4 molecules of sugar?
Answer:
To produce a single sugar molecule a plant must use a molecule of water (H2O) and carbon dioxide (CO2) and sunlight. To get 4 molecules, you simple use 4 molecules of water and 4 molecules of carbon dioxide
What does a model of a light wave tell us about
brightness and color?
Answer:
A wave model of light is useful for explaining brightness,color, and the frequency-dependent bending of light at a surface between media. However, because light can travel through space, it cannot be a matter wave, like sound or water waves.
Which of the following is a good example of a contact force?
ОА.
Earth revolving around the Sun
OB.
a bridge suspended by cables
OC.
a ball falling downward a few seconds after being thrown upward
OD. electrically charged hairs on your head repelling each other and standing up
Answer:
A bridge suspended by cables
Explanation:
Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.
A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1. Which of the following describes the electricfield inside this conductor?
a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.
2. The charge density inside theconductor is:
a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinite
Answer:pp
Explanation:
ii
If a rock displaces 7 ml of water, what is the volume of the rock?
Answer:
if i am not mistaken the volume is 7, because it only took that much space
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed into elastic potential energy of the bending board; if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the force. Ultimately the board breaks at an applied force of 870 N and a deflection of 1.4 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a distance of 1.2 cm, what is the average force on the hand?
Answer:
A. The hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand = 1025 N
Explanation:
A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.
Workdone = force × distance
Minimum force required = 870 N;
Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M
Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J
This work done = Kinetic energy of the hand
Kinetic energy = mv²/2 ; where m is mass and v is velocity
Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J
v² = 2 KE/m
v = √2KE/m
v = √(2 × 12.18/0.50)
v = 6.98 m/s
Therefore, the hand must move with a velocity of 6.98 m/s to break the board.
B. Average force on the hand
This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration
From the equation of motion, a = v² - u²/2s
At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m
a = 0² - 6.98²/ 2 × 0.012
a = -2030 m/s²
Force = 2030 m/s² × 0. 50 kg = 1015 N
Therefore, Average force on the hand = 1025 N
Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?
Explanatio
omega=2pi/T
Answer:
0
0000
Explanation:
Answer the following questions
1. Heat in liquid travels from
a) bottom to top
b) top to bottom
c) left to right
d) right to left
2. The direction of flow of heat is
a) always from a cooler body to a hotter body
b) always from a hotter body to cooler body
c) always from a body at a lower temperature to a body at a higher temperature
d) all the above
3. A cold steel spoon is dipped in a cup of hot milk. The steel spoon transfer the heat to its other end by the process of
a) convection
b) conduction
c) radiation
d) none of the above
who won the battle of Buxar
Answer:
British East India Company
The battle was fought at Buxar, a "small fortified town" within the territory of Bihar, located on the banks of the Ganga river about 130 kilometres (81 mi) west of Patna; it was a decisive victory for the British East India Company.
An iron nail floats in mercury and sinks in water. explain why?
Answer:
because density of iron is more than that of water but less then that of Mercury
hope it's helpful
hoW DO YOU SPELL THIS?
Answer:
Spell what?
Can u please mention that too?
Answer:
what
Explanation: how do you spell what please include this in your next post
When you release some of the paint from a spray paint can and the can remains at the same temperature, which gas law does this represent?
Answer:
Boyle's law.
Explanation:
Robert Boyle was an Irish chemist and is famously referred to as the first modern chemist. He was born on the 25th of January, 1627 in Lismore, Ireland and died on the 31st, December 1691, London, United Kingdom.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
[tex] PV = K[/tex]
[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]
Where;
P1 is the original pressure.
P2 is the final pressure.
V1 is the original volume.
V2 is the final volume.
Hence, when you release some of the paint from a spray paint can by applying an amount of pressure and the can remains at the same temperature, the gas law which this represent is Boyle's law.
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on Earth, and her friend Darien is stationed on the star where the race ends. Bob is in a rocket that can travel at 0.7c; whereas Charlie's rocket can reach a speed of 0.866c. Bob and Charlie start at the same time. Draw space- diagrams from each perspective.
Required:
Estimate how long it takes Bob and Charlie to finish the race from each perspective.
Solution :
The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years
But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.
For Bob,
Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]
So the distance appeared to Bob due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]
[tex]$=7.1 \ Ly$[/tex]
Therefore, the time required to finish the race by Bob is
[tex]$t_b = \frac{L_b}{V_b}$[/tex]
[tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]
= 10.143 year
For Charlie,
Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]
So the distance appeared to Charlie due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]
[tex]$=5 \ Ly$[/tex]
The time required to finish the race by Charlie is
[tex]$t_b = \frac{L_c}{V_c}$[/tex]
[tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]
= 5.77 year
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.6 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
a. How far from the wall is she?
b. What is the period of her up-and-down motion?
Answer:
a) [tex]d=11m[/tex]
b) [tex]T=4.68s[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=22m[/tex]
Velocity [tex]v=4.6m/s[/tex]
a)
Generally the equation for distance between her and the wall d is mathematically given by
Since
The First Anti node distance is [tex]\frac{\lambda}{2}[/tex]
Therefore
[tex]d= \frac{\22}{2}[/tex]
[tex]d=11m[/tex]
b)
Generally the equation for her up-and-down motion is mathematically given by
[tex]T=\frac{22}{4.7}[/tex]
[tex]T=4.68s[/tex]
The position of an object of
mass 5 kg as a function of
time is giving by r = (20m/s4)t4
i + (12 m/s3)t3 j. Find the
force acting on the object as a
function of time. Express the
force in unit vectors. Hint:
Remember that Newton's
second Law relates the force
to the acceleration
Answer:
[tex]F=5(240t^2i+72tj)\ N[/tex]
Explanation:
Given that,
The mass of the object, m = 5 kg
The position vector is, [tex]r=20t^4i+12t^3j[/tex]
Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]
Newton's second law of motion is given as follows:
F = ma
Put all the values,
[tex]F=5(240t^2i+72tj)\ N[/tex]
Hence, this is the required solution.
What is the magnitude of a vector that has the following components: x = 32 m y = -59 m
Answer:
Explanation:
Since the x and y components are given
The vectors Magnitude = √32²+(-59)²
=67.12m
40 ohms
1.2 A
40 ohms
12 V
Calculate the total energy developed in 5
minutes by the system above.
Answer:
17280 J and 1080 J
Explanation:
Given :
R= 40 ohm
I=1.2A
t= 5 min=60×5=300 sec
Now,
Total energy can be calculated as:
[tex]E=I^{2} Rt\\E=(1.2)^{2} *40*300\\E=17280 J[/tex]
Now,
V=12V
R=40 Ohm
[tex]E=\frac{V^{2} }{R} *t\\E=\frac{(12)^{2} }{40} *300\\E=1080 J[/tex]
Total energy is 17280 J and 1080 J
What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *
1 point
45 N
450 N
450 kg
10N
Answer:
450N
Explanation:
weight= m*g
weight=45*10
weight=450N
two groups of students carry out experiments to investigate the relationships between force and extenstion
Answer:
They investigated Hooke's law.
Explanation:
[tex]{ \tt{force \: \alpha \: extension }} \\ { \boxed{ \bf{F = kx}}} \\ x \: is \: extension \\ F \: is \: force \\ k \: is \: constant : k = 1[/tex]
A 2.7 kg mass is connected to a spring (k=159 N/m) and is sliding on a horizontal frictionless surface. The mass is given an initial displacement of +19 cm and released with an initial velocity of -13 cm/s. Determine the acceleration of the spring at t=3.4 seconds.
Answer:
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
Explanation:
The mass-spring system experiments a Simple Harmonic Motion, whose kinematic expression is the following:
[tex]x(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t + \phi\right)[/tex] (1)
Where:
[tex]x(t)[/tex] - Position, in centimeters.
[tex]A[/tex] - Amplitude, in centimeters.
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]m[/tex] - Mass, in kilograms.
[tex]\phi[/tex] - Phase, in radians.
By Differential Calculus, we derive expression for the velocity and acceleration of the mass-spring system:
[tex]v(t) = -\sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (2)
[tex]a(t) = -\frac{k\cdot A}{m}\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (3)
Where [tex]v(t)[/tex] and [tex]a(t)[/tex] are the velocity and acceleration of the system, in centimeters per second and centimeters per square second, respectively.
If we know [tex]m = 2.7\,kg[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]t = 0\,s[/tex], [tex]x(t) = 19\,cm[/tex] and [tex]v(t) = -13\,\frac{cm}{s}[/tex], then we have the following system of nonlinear equations:
[tex]A \cdot \cos \phi = 19[/tex] (1)
[tex]-7.674\cdot A \cdot \sin \phi = - 13[/tex] (2)
If we divide (2) by (1):
[tex]-7.674\cdot \tan \phi = -0.684[/tex]
[tex]\tan \phi = 0.089[/tex]
[tex]\phi = \tan^{-1} 0.089[/tex]
[tex]\phi \approx 0.028\,rad[/tex]
By (1), we get the value of the amplitude:
[tex]A = \frac{19}{\cos \phi}[/tex]
[tex]A = 19.075\,cm[/tex]
If we know that [tex]A = 19.075\,cm[/tex], [tex]k = 159\,\frac{N}{m}[/tex], [tex]m = 2.7\,kg[/tex], [tex]t = 3.4\,s[/tex] and [tex]\phi \approx 0.028\,rad[/tex], then the acceleration of the spring is:
[tex]a(t) = -29339.947\,\frac{cm}{s^{2}}[/tex]
The acceleration of the spring at [tex]t = 3.4\,s[/tex] is -29339.947 centimeters per square second.
Một ô tô khối lượng một tấn chuyển động trên một đường nằm ngang. Hệ số ma sát giữa bánh ô tô và mặt đường là 0,07. Gia tốc trọng trường g=9,8m/s2
a) vẽ và xác định tên các lực tác động lên vật. Viết phương trình chuyển động của vật.
b) nếu ô tô chuyển động đều, xuống dốc có độ dốc 5%. Tính lực kéo của ô tô.
c) nếu ô tô chuyển động đều. Lên dốc có độ dốc 5%. Tính lực kéo của động cơ ô tô
Record the lengths of the sides of ABC and ADE.
what is the resistance of a bulb of 40w connected in a line of 70v
Answer:
122.5 ohm
Explanation:
Given :
P=40 w
V= 70 V
R=?
Resistance can be calculated as :
[tex]P=\frac{V^{2} }{R} \\40=\frac{(70)^{2} }{R}\\40=\frac{4900}{R} \\R=\frac{4900}{40} \\R=122.5 ohm[/tex]
Therefore, resistance of the bulb will be 122.5 ohm
A 500 Kg block is attached with a rope of length 5m, having area 0.4× 10-4 m2. If final length is 6m, Calculate the Stress, Strain and Young's Modulus?
Answer:
stress = 1.225 x 10^8 N/m^2
strain = 1/5
Young's modulus = 6.125 x 10^8 N/m^2
Explanation:
mass, m = 500 kg
length, L = 5 m
Area, A = 0.4 x 10^-4 m^2
Final length, L' = 6 m
extension, x = L'-L= 6 - 5 = 1 m
Stress is defined as force per unit area.
[tex]stress =\frac{Force}{Area}\\\\stress =\frac{500\times 9.8}{0.4\times 10^{-4}}\\\\stress = 1.225\times 10^8 N/m^2[/tex]
Strain is defined as the ratio of change in length to the original length.
[tex]strain =\frac{x}{L}\\\\strain = \frac{1}{5}[/tex]
Young's modulus is given by the ratio of stress to the strain.
[tex]Y = \frac{1.225 \times 10^8}{\frac{1}{5}}\\\\Y = 6.125\times 10^8 N/m^2[/tex]
The lumberjack pulls on the sled with 40 N at an angle of 30 degrees, pulling so the sled moves at a constant velocity. 1) What is the x component of the applied force? 2) What is the y component of the applied force? 3) If the loaded sled has a mass of 65 kg, what is the magnitude of the force of gravity? 4) What is the magnitude of the normal force acting on the sled? 5) What is the coefficient of friction between the snow and the sled?
1) (40 N) cos(30°) ≈ 34.6 N
2) (40 N) sin(30°) = 20 N
3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N
4) The net force on the sled acting in the vertical direction is made up of
• the sled's weight, 585 N, pointing downward
• the vertical component of the applied force, 20 N, pointing upward
• the normal force, with magnitude n, also pointing upward
The sled does not move up or down, so by Newton's second law,
∑ F = n + 20 N - 585 N = 0 ==> n = 565 N
5) The net force in the horizontal direction consists of
• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)
• kinetic friction, with magnitude f, pointing in the opposite and negative direction
By Newton's second law,
∑ F = 34.6 N - f = 0 ==> f ≈ 34.6 N
Now if µ is the coefficient of kinetic friction, then
f = µn ==> µ = f/n = (34.6 N) / (565 N) ≈ 0.0613
The component of the force is the effective part of that force in that direction.
What is the component of a force?The component of the force is the effective part of that force in that direction.
1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N
2) The vertical component of the force = 40 N sin 30 degrees = 20 N
3) The magnitude of the gravitational force = mg cos 30 degrees = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N
4) The normal force = 551.7 N
5) The coefficient of friction = F/R = 40 N /551.7 N = 0.07
Learn more about component of a force:https://brainly.com/question/15529350
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Why do some "people" still believe that the earth is "flat".
I called my brother in Australia and it was day here, and night there, HOW CAN THE EARTH POSSIBLY BE FLAT, i did a zo,om meeting to prove that it was live. WHAT ARE PE,OPLE TH,INKING!!!
Answer: Members of the Flat Earth Society claim to believe the Earth is flat. Walking around on the planet's surface, it looks and feels flat, so they deem all evidence to the contrary, such as satellite photos of Earth as a sphere, to be fabrications of a "round Earth conspiracy" orchestrated by NASA and other government agencies.
♛ Mark me as Brainliest if I’m right
Got it never mind. The only reason I'm typing more is to fill out the required space
Answer:
hey. i dont know what you tryna say but if u replying to someone else, you should use the comments section. in that way you won't lose points.
