Answer:
When an object is balanced, about a pivot, the total clockwise moment must be equal to the total anticlockwise moment about that pivot.
Hope that helps.
When an object is balanced about a pivot, the total clockwise moment must be equal to the total anticlockwise moment.
What is balancing?
Clockwise involves turning to the right while going in the direction of a clock's hands. The moment is referred to as a clockwise moment if the force acting on a body rotates the body clockwise with regard to the axis of rotation.
On the other hand, the moment is referred to as an anti-clockwise moment if the force rotates the body in the opposite direction. The clock face rule can be used to determine this magnet's polarity.
The face of the loop will display the North Pole if the current is flowing counterclockwise. On the other hand, if the current is moving counterclockwise, the South Pole is visible on the loop's face.
Therefore, the total clockwise and total counterclockwise moments must match when an object is balanced about a pivot.
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A man pushes a box with a force of 75N across a 12M loading dock. How much work did he do on the box?
Once again, Have a Happy, Merry Christmas!! =)
Answer: 75n,a * (12m)
Explanation I'm trying to understand the question but this is what i got.
Happy Holidays!
We can use the following equation to calculate work done on an object.
W = F · d
Work (J) = Force (N) · displacement (m)
Plug in the given values into the equation:
W = 75 · 12 = 900 J
If the forces acting on an object at rest are ______,the object will remain at rest.
Answer:
friction. or gravity both ofbthem .....................
..
Both igneous and sedimentary rock can become metamorphic rock if enough ________________ are applied.
Something has a mass of 40 kg and a velocity of 1 m/s
How much kinetic energy does it have?
Part 2: If it doubles its mass, will its KE also double?
[tex]\\ \sf\Rrightarrow K.E=\dfrac{1}{2}mv^2[/tex]
[tex]\\ \sf\Rrightarrow K.E=\dfrac{1}{2}(40)(1)^2[/tex]
[tex]\\ \sf\Rrightarrow K.E=20J[/tex]
Part 2:-m=80kg[tex]\\ \sf\Rrightarrow K.E=\dfrac{1}{2}(80)(1)^2[/tex]
[tex]\\ \sf\Rrightarrow K.E=40J[/tex]
YesIf you were traveling 60 mph and not wearing a seatbelt and collided with a fixed object, how fast would you be traveling when you hit the windshield
Your speed when your car hits a fixed object is 60 mph.
The given parameters:
Your initial speed, v = 60 mphWhat is relative velocity?Relative velocity tells us how fast we are traveling from a fixed point or reference point.
When travel at 60 mph in a car, you are moving at the same rate with the car.If you are not wearing a seat belt, when you hit a fixed object, you will move forward at the same rate as the car's speed.Thus, we can conclude that your speed when your car hits a fixed object is 60 mph.
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Block A has a mass of 2kg and a speed of 50 m/s along the positive x axis.
Block B has a mass of 4 kg and a speed of 25 m/s along the negative x axis.
The two blocks collide head-on in a perfectly elastic collision.
Determine the velocity of each mass after the collision.
Momentum is conserved, so
(2 kg) (50 m/s) + (4 kg) (-25 m/s) = (2 kg) v₁' + (4 kg) v₂'
where v₁' and v₂' are the velocities of the two blocks after the collision. Simplifying this gives
100 kg•m/s - 100 kg•m/s = (2 kg) v₁' + (4 kg) v₂'
0 = (2 kg) v₁' + (4 kg) v₂'
v₁' = -2v₂'
Energy is also conserved, so
1/2 (2 kg) (50 m/s)² + 1/2 (4 kg) (-25 m/s)² = 1/2 (2 kg) (v₁')² + 1/2 (4 kg) (v₂')²
Simplifying yields
2500 J + 1250 J = (1 kg) (v₁')² + (2 kg) (v₂')²
3750 J = (1 kg) (v₁')² + (2 kg) (v₂')²
Substitute v₁' = -2v₂' and solve for v₂' :
3750 J = (1 kg) (-2v₂')² + (2 kg) (v₂')²
3750 J = (6 kg) (v₂')²
(v₂')² = 625 J/kg = 625 m²/s²
v₂' = 25 m/s
Then the first block has final velocity
v₁' = -2 (25 m/s)
v₁' = -50 m/s
A pitcher throws a softball
Answer:
Did you forget the answers
Explanation:
if so leave in comments and i will try to figure out
Answer:
what is the question and the answers??
As best you can, describe how we can tell the age of the Earth and rocks found on it.
Answer:
We can determine the age of the Earth and the rocks found on Earth with techniques such as measuring ice cores and digging and analysing fossils.
Explanation:
Hopefully this helped!
A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?
A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?
0.57 N
32 N
98 N
0.00005 N
From Newton's second law of motion, the average force on the ball by the racket is 98 Newtons. The correct answer is option C
Given that a 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. And the time for contact with the racket is 0.04 s, that is,
mass m = 0.07 kg
velocity v = 56 m/s
time t = 0.04 s
force f = ?
To calculate the average force on the ball by the racket, let us apply Newton's second law of motion.
Impulse = change in momentum
ft = mv
Substitute all the parameters into the equation above
0.04f = 0.07 x 56
make f the subject of the formula
f = 3.92 / 0.04
f = 98 N
Therefore, the average force on the ball by the racket is 98 Newtons. The correct answer is option C
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The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
From the question, we are to determine the average force on the ball by the racket.
From the formula,
[tex]F = \frac{mv}{t}[/tex]
Where F is the force
m is the mass
v is the velocity
and t is the time
From the given information
m = 0.07 kg
v = 56 m/s
t = 0.04 s
Putting the parameters into the formula,
we get
[tex]F = \frac{0.07 \times 56}{0.04}[/tex]
[tex]F = \frac{3.92}{0.04}[/tex]
F = 98 N
Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
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explain why an astronaut on moon seems to be attracted only to the moon
Answer: the gravitational attraction on the Moon is much less than it is here on Earth, and a person weighs less on the Moon.
Explanation:
[URGENT] A swimmer wants to end up at a dock due north of her starting position on the south shore of a river. In still water her maximum speed is 1.25m/s. The river has a current flowing 0.35m/s [E].
a.) calculate the direction that she must aim herself through the water to arrive at the dock.
b.) how long will it take for the swimmer to cross the river, if the distance from the starting point to the dock is 300 m?
Please show all your work, thanks
Explanation:
As you can see in the picture, we want the swimmer to go on a straight line, so the speed of the water must be equal to the speed of the swimmer along the x-axis. We also know the value of v, so we can calculate the of the cosine of the angle (alpha) between Vx and V. Thanks to the fundamental relation of gioniometry (cos^2(x) + sin^2(x) = 1) we can find the sine of alpha and calculate Vy. With Vy we can calculate the time that the swimmer will use for reaching the dock: s = Vy * t => t = s/(Vy).
I'll let you do all the calculations, you just have to plug in values.
a)The direction that she must aim herself through the water to arrive at the dock will be 73.7°.
b)The time it takes for the swimmer to cross the river will be 240 seconds.
What is speed?Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.
Given data;
The maximum speed of the swimmer In still water,v=1.25m/s.
Speed of river = 0.35m/s [E]
α is the direction that she must aim herself through the water to arrive at the dock.
t is the time it takes for the swimmer to cross the river
The distance from the starting point to the dock is,s = 300 m
If the swimmer were to swim straight, the water's speed along the x-axis would have to match that of the swimmer.
[tex]\rm v_w = v_x = 0.35 \ m/sec[/tex]
α is the angle between vx and v
The value of the
[tex]\rm cos \alpha = \frac{v_x}{v} \\\\ \alpha = cos ^{-1}(\frac{0.35}{1.25} )\\\\ \alpha = 73.7 ^0[/tex]
The time that the swimmer will use for reaching the dock;
s = V×t
t = s/(v)
t=300 m / 1.25 m/sec
t = 240 seconds
Hence, the direction that she must aim herself through the water to arrive at the dock. and it takes for the swimmer to cross the river will be 73.7° and 240 seconds respectively.
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the breaks on a 15,680 N car exert a stopping force of 640 N.
A block of mass 1.0 kg rests on a trolley of mass 4.0 kg. The coefficient of dynamic friction between the block and the trolley is 0.30. A horizontal force f = 5.0 n acts on the block. The block slides over the trolley. What is the acceleration of the trolley?
Answer:
the acceleration of the trolley is the acceleration of the trolley is 0.75 m/s^-2
Explanation:
friction force between the block and the trolley
= 1×0.3×9.8
= 2.94 N
(F = ma)
the acceleration of the trolley
= friction force/mass
= 2.94/4
= 0.735 m/s^-2
= 0.75 m/s^-2
The acceleration of the trolley will be 0.75 m/s².Acceleration is the ratio of the force to the mass.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
The friction force between the block and the trolley is found as;
[tex]\rm F_f = \mu N \\\\ \rm F_f = \mu \times mg \\\\ \rm F_f = 0.30 \times 1 \timess 9.81 \\\\ \rm F_f= 2.94 \ N[/tex]
Mechanical force is equal to the product of the mass and the acceleration which is equal to the frictional force;
[tex]\rm F_f =F=ma \\\\ \rm a = \frac{F}{m} \\\\ \rm a = \frac{2.94}{4} \\\\ a=0.735 \ ms^{-2}[/tex]
Hence the acceleration of the trolley will be 0.75 m/s².Acceleration is the ratio of the force to the mass.
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If the energy in a mechanical wave increases or decreases, what is also going to increase or decrease? A wavelength B amplitude с frequency d refraction
Answer:
i think wave lengh not 100% sure tho
If the energy of the mechanical wave will increase or decrease, then the wavelength will also change accordingly. Hence, option A is correct.
What is Wavelength?The distance between two identical locations (adjacent crests) in successive cycles is known as the wavelength, and it is used to describe waveform signals that are transmitted over wires or into space. Typically, in wireless systems, this length is specified in meters (m), centimeters (cm), or millimeters (mm). The wavelength is more frequently described in nanometers (nm), which are units of 10⁻⁹ m, or angstroms, which are units of 10⁻¹⁰ m, for infrared (IR), visible light (UV), and gamma radiation.
Frequency, which is defined as the number of wave cycles per second, and wavelength have an inverse relationship. The wavelength of a signal decreases with increasing frequency, or we can say that wavelength and refractive index are inversely proportional to each other.
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PLEASE HELP!!
5. A 700 kg race car travels
around the track at 65 m/s. The track has a radius of 75 m.
a. What is the centripetal acceleration?
b. What is the centripetal force?
c. Is the net force on the car less than, equal to, or greater than the centripetal force? Why?
Hi there!
a.
The equation for centripetal acceleration is as follows:
[tex]\large\boxed{a_c = \frac{v^2}{r}}}[/tex]
Plug in the given values to solve:
[tex]\large\boxed{a_c = \frac{(65)^2}{75} = 56.33 m/s^2}[/tex]
b.
According to Newton's Second Law:
[tex]\large\boxed{\Sigma F = ma}}[/tex]
The acceleration is v²/r, so the net force is:
[tex]\large\boxed{\Sigma F = m(\frac{v^2}{r})}}[/tex]
Multiply by the given mass:
[tex]\large\boxed{\Sigma F = 700(56.33) = 39433.33 N}}[/tex]
c.
There is NO net force in the vertical direction since the object is NOT accelerating in the vertical direction (normal force and weight cancel).
Thus, the ONLY net force experienced by the object is in the horizontal direction and is EQUAL to the centripetal force.
The Earth currently has an orbital period of 1 year and a semi-major
axis length of 1 AU. If the Earth's orbit changed and the semi-major
axis length increased to 1.1 AU, how would the Earth's orbital period change?
A: It would increase by 0.15 years
B: It would increase by 0.032 years
C: It would decrease by 0.15 years
D: It would decrease by 0.032 years
The Earth's orbital period change is A: It would increase by 0.15 years
Using Kepler's third law which states that the square of the orbital period of the planet is directly proportional to the cube of its distance from the sun.
So, T² ∝ R³
T'²/T² = R'³/R³ where
T = orbital period at R = 1 yearR = initial axis length = 1 AUT' = orbital period at R'R' = final axis length = 1.1 AU.So, making T' subject of the formula, we have
T' = [√(R'/R)³]T
T' = [√(1.1 AU/1 AU)³] × 1 year
T' = [√(1.1)³] × 1 year
T' = √1.331 × 1 year
T' = 1.15 × 1 year
T' = 1.15 years.
So, the change in the Earth's orbital period ΔT = T' - T
= 1.15 years - 1 year
= 0.15 years
Since this is positive, the orbital period increases by 0.15 years.
So, the Earth's orbital period change is A: It would increase by 0.15 years
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light of wavelength 633 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away.
Answer:
because of the gravity of the earth
describe how visible light is used with solar photovoltaic panels to produce electricity
Photovoltaic panels are used to produce alternating current using sunlight. It uses solar cells which are photoactive and produce a photoelectric effect and thereby a potential is created.
What is photoelectric effect?When photons from the light energy is hit on a metallic surface the photon collides with the electrons of the metals and electrons are ejected from the metal surface. This phenomenon is called photoelectric effect.
In a solar photovoltaic panel, photovoltaic cells layers of semi-conducting materials, such as silicon, are sandwiched photovoltaic cells. When photons from sunshine strike each layer, its unique electronic properties energise and produce an electric field.
The current required to generate electricity is produced through what is known as the photoelectric effect. Electricity produced by solar panels is in the direct current form. This is then transformed into an alternating current by passing via an inverter.
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anyone to help with this question
Answer:
i can
Explanation:
i know what's the answer
A 25 N force acts on a box moving along a plane at a distance of 20 meters. What is the work it does with this force?
Help!!
Answer:
500 J.
Explanation:
Work done = force * distance
= 25 * 20
= 500 Joules.
Name a device or gadget that converts
a) Heat energy to mechanical energy b) heat energy to light energy
c)electrical energy to sound energy
Answer:
a heat engine converts heat to mechanical energy.
electric generator
Help it’s multiple choice 11 through 15 please!
1. • Here, force of gravity on the block = 20 N.
• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]
• The coefficient
[tex] u_{k} = 0.4[/tex]
• Force of friction =
[tex]u_{k} \times \: normal \: \: \: force \\ = 0.4 \times 20N \\ = 8N[/tex]
• Hence, the force of sliding friction between the block and the ground is 8 N.
• So, it is option c. 8 N
2. The answer is option d. continue in the same direction with no change in speed.
We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.
3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.
Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.
So, 1 N = 1 Kg × 1 m/s^2.
4. It is d. not zero.
Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.
5. Force = 2 N
Acceleration of the object A = 2 m/s^2.
Acceleration of the object B = 1 m/s^2.
Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg
And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg
So, the mass of object B is greater than that of object A.
Hence, the answer is option c. Object B has more mass.
Hope you could get an idea from here.
Doubt clarification - use comment section.
when did humans learn that the earth is not the center of the universe?
Answer:
When did humans learn that the Earth is not the center of the universe?
Answer
1
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4 Answers
Asked in 3 Spaces


Science - Next Generation
Alexander Somm
, Consultant, Investor Relations at Novelpharm AG (2015-present)
Answered Oct 16
What, it isn’t?!
Sorry, I had to.
As far as I have read and understood, the Sumerians and later the Babylonians both had astronomical calendars that already differentiated planets and stars. Earth was not the center to them, the Sun likely was. That was around 2,200 - 1,600 BC.
After that, Greek philosopher Aristarchus of Samos (310 - 230 BC) was the first (recorded) to have believed the solar system was organized around the Sun, rather than the Earth. His heliocentric model was unpopular during Aristarchus’ lifetime, although it would inspire astronomers centuries later, such as Copernicus and Galileo.
Now, there are numerous archeological findings (cave paintings) and studies, that all suggest an understanding of complex astronomy in prehistoric times dating back as far as 40,000 years. This also explains how early, prehistoric migrants may have navigated the seas.
Explanation:
hope it helps
have a good day
Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su
densidad es 1.10 3 kg/m 3 y su módulo de comprensibilidad vale 2,16x10 9 N/m 2 ¿Cuál
es la longitud de onda de las ondas sonoras en el agua si su frecuencia es de 1000hz?
1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua ([tex]v[/tex]), en metros por segundo:
[tex]v = \sqrt{\frac{K}{\rho} }[/tex] (1)
Donde:
[tex]K[/tex] - Módulo de compresibilidad, en newtons por metro cuadrado.[tex]\rho[/tex] - Densidad del agua, en kilogramos por metro cúbico.Si sabemos que [tex]\rho = 1\times 10^{3}\,\frac{kg}{m^{3}}[/tex] y [tex]K = 2,16\times 10^{9}\,\frac{N}{m^{2}}[/tex], entonces la velocidad de las ondas sonoras es:
[tex]v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }[/tex]
[tex]v\approx 1469,694\,\frac{m}{s}[/tex]
La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
2) Luego, determinamos la longitud de onda ([tex]\lambda[/tex]), en metros, mediante la siguiente fórmula:
[tex]\lambda = \frac{v}{f}[/tex] (2)
Donde [tex]f[/tex] es la frecuencia de las ondas sonoras, en hertz.
Si sabemos que [tex]v\approx 1469,694\,\frac{m}{s}[/tex] y [tex]f = 1000\,hz[/tex], entonces la longitud de onda de las ondas sonoras es:
[tex]\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}[/tex]
[tex]\lambda = 1,470\,m[/tex]
La longitud de onda de las ondas sonoras es 1,470 metros.
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A 100 g mass of tungsten at 100.0°C is placed in 200 g of water at 20.0°C. The
mixture reaches equilibrium at 21.6°C. Calculate the specific heat of tungsten
The specific heat capacity of tungsten given the data from the question is 0.17 J/gºC
Data obtained from the questionMass of tungsten (M) = 100 g
Temperature of tungsten (T) = 100 °C
Mass of water (Mᵥᵥ) = 200 g
Temperature warm water (Tᵥᵥ) = 20 °C
Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
Equilibrium temperature (Tₑ) = 21.6 °C
Specific heat capacity of tungsten (C) = ?
How to determine the specific heat capacity of tungstenHeat loss = Heat gain
MC(T – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
100 × C × (100 – 21.6) = 200 × 4.184(21.6 – 20)
100 × C × 78.4 = 1338.88
Divide both sides by 100 × 78.4
C = 1338.88 / (100 × 78.4 )
C = 0.17 J/gºC
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Which statement describes the magnetic field inside a bar magnet? It points from north to south. It points from south to north. It forms loops inside the magnet. There is no field within the magnet.
Inside the bar magnet, the magnetic field points from north to south. Statement A is correct.
Magnetic Field:
It is defined as a vector field or the influence of the magnet on the electric current, charges and ferromagnetic substance.
The strength of magnetic field is depends up on the numbers of magnetic field lines per unit area.
Magnetic field lines emerge from the North pole and end in the South pole of a bar magnet. Inside the magnet are also present inside the bar magnet and never intersect at any point.Therefore, inside the bar magnet, the magnetic field points from north to south.
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Answer:b
Explanation:
. A teacher pushes a desk across the floor for a distance of 5 meters. She exerted a horizontal force of 20N. How much work was done?
Answer:
100 J work was done.
Hope you could get an idea from here.
Doubt clarification - use comment section.
A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal. What is the vertical component of the velocity?
a 40 m/s
b 34.6 m/s
c 0 m/s
d 20 m/s
Vertical component = 40 m/s x sin(60) = 34.6 m/s
Answer: b. 34.6 m/s
John does 3.2J of work to lower the window shade in her bedroom a distance of 0.8m. How much force must John exert on the window shade?
Answer:
[tex]\boxed {\boxed {\sf 4 \ Newtons}}}[/tex]
Explanation:
Work is energy transferred when an object is moved. It is the product of force and distance.
[tex]W= F*d[/tex]
John did 3.2 Joules of work or 3.2 Newton meters of work (the units are equivalent). The window shade was lowered a distance of 0.8 meters.
W=3.2 N*m d= 0.8 mSubstitute the values into the formula.
[tex]3.2 \ N *m = F * 0.8 \ m[/tex]
We are solving for the force, so we must isolate the variable F. It is being multiplied by 0.8 meters. The inverse operation of multiplication is division, so divide both sides of the equation by 0.8 m.
[tex]\frac {3.2 \ N*m}{0.8 \ m}= \frac{ F* 0.8 \ m}{0.8 \ m}[/tex]
The units of meters cancel.
[tex]\frac {3.2 \ N}{0.8 }=F[/tex]
[tex]4 \ N = F[/tex]
John must exert 4 Newtons of force.
why does rhododendron not grow well in the terai region of nepal give reason
Answer:
Tge Rhododendrob is the national flower of Nepal. The hills and mountain sides ranges of Nepal are decorated with diff colours and shapes