Answer:
a) v_{2f} = 12.0 m / s, b) v_{2f} = 9.6 m / s
Explanation:
This is an exercise in conservation of momentum, let's start by defining a system formed by the two balls, so that the forces during the collision have been internal and the moment is preserved.
Initial instant. Before the crash
p₀ = m v₁₀ + m v₂₀
where we use index 1 for the green ball and index 2 for the blue ball.
Final moment. After the crash
p_f = m v_{1f} + m v_{2f}
the moment is preserved
p₀ = p_f
m v₁₀ + m v₂₀ = m v_{1f} + m v_{2f}
they tell us that the blue ball is at rest before the crash
m v₁₀ = m v_{1f} + m v_{2f}
a) it is indicated that the green ball stops after the collision v1f = 0
m v₁₀ = m v_{2f}
v_{2f} = v₁₀
v_{2f} = 12.0 m / s
b) the speed of the green ball is v_{1f} = 2.4 m / s
m v₁₀ = m v_{1f} + m v_{2f}
v_{2f} = v_{1o}- v_{1f}
v_{2f} = 12.0 - 2.4
v_{2f} = 9.6 m / s
Arrange the following substances from the lightest to the heaviest:
Cl2, CH4; H20; NH3, N2
A. H2O<NH,< N2 CH4Cl2
B. CH< NH< H-0< < Cl2
C. Ny< Cl< H2O< CH«<NH3
D. NH;< CH«< Cl< H2O< N2
molecular weights are written in the picture.
CH4<NH3<H2O<Cl2
What is the magnitude of the resultant vector? Round
your answer to the nearest tenth.
m
R
5 m
13 m
Intro
Done
Answer: 13.9 m
Explanation:
Answer:
13.9m
Explanation:
Answer on Edge
A child swings with a small amplitude on a playground with a 2.5m long chain.
a) what is the period of the child’s motion
b) what is the frequency of vibration
a. The period of the child's motion is 3.171 seconds.
b. The frequency of the vibration is 0.315 Hz.
Calculation of the period & frequency:Since A child swings with a small amplitude on a playground with a 2.5m long chain.
a. So here the time period should be
[tex]2\times \ pi(L/g)^{1/2}\\\\2\times 3.14(2.5/9.81)^{1/2}[/tex]
T=3.171s
b) Now the frequency is
[tex]T=1\div f\\\\f=1\div 3.171[/tex]
f=0.315Hz
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ball A is dropped from a hot air balloon rising at a costant velocity of 14,7 m.s'1 at a height of 19,7 m above the ground.the ball took 1.5s to reach its maximum height and hits the ground after some time in air.ignore the effects of air resistanceUse the ground as zero reference.3.1.1calculate the maximum height reached by the ball above the ground
Answer:
this slow site thinks the answer is a link
Explanation:
this was a week ago so i dont know if u still need help
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first bounce and do NOT assume the total energy is the same as the total energy of the object before it hits the ground.) a. List the quantities that you need to know to determine the total energy of the object after it hits the ground. b. Record your measurements and describe how you measured them. c. Calculate the total energy of the object after it hit the ground. Your final answer: ______________ d. Determine whether or not the object’s energy was conserved when it hit the ground. If it was not conserved, explain where the energy went.
Answer:
a) quantity to be measured is the height to which the body rises
b) weighing the body , rule or fixed tape measure
c) Em₁ = m g h
d) deformation of the body or it is transformed into heat during the crash
Explanation:
In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.
a) What quantities must you know to calculate the energy after the bounce?
The quantity to be measured is the height to which the body rises, we assume negligible air resistance.
So let's use the conservation of energy
starting point. Soil
Em₀ = K = ½ m v²
final point. Higher
Em_f = U = mg h
Em₀ = Em_f
Em₀ = m g h₀
b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.
c) We use conservation of energy
starting point. Soil
Em₁ = K = ½ m v²
final point. Higher
Em_f = U = mg h
Em₁ = Em_f
Em₁ = m g h
d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.
There are two possibilities.
* that have been equal therefore energy is conserved
* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash
Match the following items.
1. Extremely small building blocks of matter.
2. Forming new matter from old matter.
3. Small bits of matter.
atom
molecule
chemical change
The surveillance camera on a satellite at 250 km above the earth is taking pictures of the earth surface. Suppose that the imaging wavelength is 550 nm and the diameter of the camera lens is 40 mm. (a) Calculate the angular resolution of the camera. (b) Suppose that the headlights of a car on the earth are 1.6 m apart, can the camera resolve them
Answer:
a) θ = 1.67 10⁻⁵ m, b) Consequently we must affirm that the vehicle headlights cannot resolve.
Explanation:
a) To find the resolution of the camera we use the Rayleigh criterion for diffraction
a sin θ = m λ
where m = 1 for the first zero of the slit
we must remember that the angles in the experiments are measured in radians and are very small
sin θ = θ
we substitute
θ = [tex]\frac{\lambda}{a}[/tex]
this expression is for a slit, in the case of circular openings the expression must be solved in polar coordinates giving
θ = [tex]1.22 \ \frac{\lambda}{D}[/tex]
where D diameters of the opening
let's calculate
θ = [tex]1.22 \ \frac{550 \ 10^{-9}}{ 40\ 10^{-3}}[/tex]
θ = 1.67 10⁻⁵ m
b) let's use trigonometry to find the separation distance on earth
tan θ = y / x
y = x tan θ
let's calculate
remember that the angles must be in radians
y = 250 10³ tan 1.67 10⁻0-5
y = 4.18 m
as they indicate that the separation of the headlights is y = 1.6m,
we see that this separation is greater than the separation distance separation.
Consequently we must affirm that the vehicle headlights cannot resolve.
What happens when a tennis racket hits
a ball?
A. The ball pushes back on the racket in the opposite
direction.
B. The ball pushes back on the racket in the same
direction.
C. The ball does not push back on the racket.
D. The ball pushes back on the racket perpendicularly.
Newton's third law of motion states that for every action, there is an equal and opposite reaction. When a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.
When a tennis racket hits a ball, the ball exerts a force on the racket, and according to Newton's third law of motion, the racket exerts an equal and opposite force on the ball. This means that the ball pushes back on the racket in the opposite direction to which the racket struck the ball.
This principle is often referred to as "action-reaction" or "equal and opposite forces." When the racket collides with the ball, the force applied by the racket causes the ball to accelerate in the opposite direction, leading to its movement away from the racket.
Therefore, when a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.
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The force of friction depends upon
Answer:
ask internet?
Explanation:
easy .....
..
In fatal crashes, more than __________% of passenger car occupants who were totally ejected from the vehicle were killed.
Answer:
it would be 83% in a fatal crash.
Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?
Answer:
* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
*E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
Explanation:
Let's start by finding the electric field of the charged ring
in the attachment we can see a diagram of the system. Due to circular symmetry, the electric field perpendicular to the axis is canceled and only the electric field remains parallel to the axis.
Eₓ = E cos θ (1)
E = k ∫ [tex]\frac{dq}{r^2}[/tex]
cos θ = x / r
using the Pythagorean theorem
r = [tex]\sqrt{x^2 + y^2}[/tex]
we substitute
Eₓ = k ∫ [tex]\frac{dq}{x^2+y^2} \ \frac{x}{\sqrt{ x^2+y^2} }[/tex]
Eₓ = [tex]k \frac{x}{(c^2+y^2)^{3/2} }[/tex] ∫ dq
Eₓ = k \frac{x}{(c^2+y^2)^{3/2} } Q
the ring's electric field
E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
Now let's find the electric field of the disk
The charge is distributed over the entire disk, so let's use the concept of charge density
σ = [tex]\frac{dq}{dA}[/tex]
Let's approximate the disk as a group of rings, the width of each ring is dr, the area is
dA = 2πr dr
we substitute
σ = [tex]\frac{1}{2\pi r} \ \frac{dq}{dr}[/tex]
dq = 2π σ r dr
we substitute in equation 1, where the electrioc field is of each ring
Eₓ = [tex]k \int\limits^R_0 \ { \frac{x}{(x^2+r^2)^{3/2} } \ 2\pi \sigma \ r } \, dr[/tex]
if we use a change of variable
dv = 2rdr
v = r²
Eₓ = [tex]k x \pi \sigma \int\limits^a_b { \frac{1}{(x^2+v)^{3/2} } } \, dv[/tex]
we integrate
Eₓ = k x π σ [tex][ \frac{ (x^2 + r^2)^{-1/2} }{-1/2} ][/tex]
we value in the limits from r = 0 to r = R
Eₓ = k π σ x (-2) [ [tex]\frac{1}{ \sqrt{x^2+R^2} } - \frac{1}{x}[/tex]]
Eₓ = 2π k σ ([tex]1 - \frac{x}{(x^2 + R^2 ) ^{1/2} }[/tex] )
σ = Q/πR²
substitute
Eₓ = 2 k Q/R² (1 - \frac{x}{(x^2 + R^2 ) ^{1/2} } )
E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
The two electric fields are
* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
*E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
we can see that the functional relationship of the two fields is different
Outline 3 disadvantages and advantage of water and alcohol as a template liquid
Answer:
Advantages of mercury as a thermometric liquid.
-It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.
-It does not wet (cling to the sides of) the tube.
-It has a high boiling point
-It expands uniformly (linear expansion) and responds quickly to temperature changes, hence is sensitive.
-It has a visible meniscus.
Disadvantages
-Mercury is very poisonous.
-its expansively is fairly low
-it is expensive
-It has a high freezing point therefore it cannot be used in places where the temperature gets very low.
Alcohol has a thermometric fluid
-Alcohol expands uniformly.
-It has a low freezing point (-115 degreecentigrade) therefore it is very suitable for place where the temperature gets very low.
-It has a large expansively
-It is an easily available cheap liquid, which is safe to use
Disadvantages of alcohol
-it wets the tube
-it has a low boiling point (cannot be used in places with high temperatures)
-it does not react quickly to changes in temperature
-It needs to be dyed, since it's colourless.
Disadvantages of water
-Water has high specific heat capacity. So it cannot be used for measuring small temperature differences.
- Water will wet the surface of the glass tube. It is a sticky substance.
- Water is transparent
Explanation:
Your heart pumps blood into your aorta (diameter 2.5 cm) with a maximum flow rate of about 500 cm^3/s. Assume that blood flow in the aorta is laminar (which is not a very accurate assumption) and that blood is a Newtonian fluid with a viscosity similar to that of water.
a. Find the pressure drop per unit length along the aorta. Compare the pressure drop along a 10 cm length of aorta to atmospheric pressure (105 Pa).
b. Estimate the power required for the heart to push blood along a 10 cm length of aorta, and compare to the basal metabolic rate of 100 W.
c. Determine and sketch the velocity profile across the aorta (assuming laminar flow). What is the velocity at the center
Answer:
a. i) The pressure drop per unit length is 52,151.89 Pa
ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta
b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts
ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta
c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel
ii) The velocity at the center is approximately 2.04 m/s
Explanation:
The given diameter of the aorta, D = 2.5 cm = 0.025 m
The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s
Assumptions;
The blood flow is laminar
The blood is a Newtonian fluid
The viscosity of water ≈ 0.01 poise = 1 cp
a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;
[tex]Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)[/tex]
Where;
Q = The flow rate = A·v
A = The cross sectional area
R = The radius = D/2
Δp/L = The pressure drop per unit length of the pipe
Therefore, we have;
[tex]\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89[/tex]
The pressure drop per unit length ΔP/L = 52,151.89 Pa
ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;
∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa
Given that the atmospheric pressure, [tex]P_{atm}[/tex] = 10⁵ Pa, we have;
[tex]P_{atm}[/tex]/ΔP = 10⁵/5,215.189 ≈ 19.175
Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta
b. i) The power, P = Q × ΔP
Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts
ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;
P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35
The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length
c. i) The velocity profile across the aorta is given as follows;
[tex]v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2[/tex]
Where;
[tex]v_m[/tex] = The velocity at the center
We get;
[tex]v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04[/tex]
The velocity at the center, [tex]v_m[/tex] ≈ 2.04 m/s
ii) The velocity profile, v(r), is given by the following formula;
[tex]v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right][/tex]
Therefore, we have;
[tex]v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2[/tex]
The velocity profile of the pipe is created with Microsoft Excel
ANSWER DIS ASAP FOR LUCK TMRW :}
List down ten situations that show how friction affects the movement of objects. I NEED THE ANSWER IN ONE MINUTE
Answer:
Walking –We can walk only if we apply frictional force. Friction is what holds your shoe to the ground. The friction present on the ice is very little, this is the reason why it is hard to walk on the slippery surface of the ice.
Writing – A frictional force is created when the tip of the pen comes in contact with the surface of the paper. Rolling friction is what comes into play while writing with a ballpoint pen while sliding friction arises when one writes with a pencil.
Skating – A thin film of water under the blade is necessary to make the skate slide. The heat generated by the skate blade rubbing against the surface of ice causes some of the ice to melt right below the blade where the skater glides over the ice. This water acts as a lubricant reducing friction.
Lighting a matchstick – When the head of the matchstick is rubbed against a rough surface, heat is generated and this heat converts red phosphorous to white phosphorous. White phosphorous is highly inflammable and the match stick ignites. Sometimes, matchsticks fail to ignite due to the presence of water. Water lowers friction.
Driving of the vehicle on a surface- While driving a vehicle, the engine generates a force on the driving wheels. This force initiates the vehicle to move forwards. Friction is the force that opposes the tyre rubber from sliding on the road surface. This friction avoids skidding of vehicles.
Applications of breaks in the vehicle to stop it- Friction braking is the most widely used braking method in vehicles. This process involves the conversion of kinetic energy to thermal energy by applying friction to the moving parts of a vehicle. The friction force resists the motion and in turn, generates heat. This conversion of energy eventually bringing the velocity to zero.
Flight of aeroplanes- Drag is the force that opposes the forward motion of the aeroplane. The friction which resists the motion of an object moving through a fluid or immobile in a moving fluid, as occurs when we fly a kite. The friction of the air is created as it meets and passes over an aeroplane and its components. Drag is generated by air impact force, skin friction and displacement of the air.
Drilling a nail into the wall- Friction is responsible for fixing of nails in a wall. As the nail is driven into the wall, the nearby material to the nail of the wall gets compressed. This exerts a force on the nail. This force is the friction that converts the normal force exerted by the compressed layers of the wall into the resisting shear force. In this manner the friction cause nails and screws to hold on to walls.
The dusting of the carpet by beating it with a stick- When the carpet is beaten with the stick, the dust comes out. The dust is carried off by the wind or falls on the floor. The carpet exhibits a little static friction that holds the dust to the carpet. When the carpet is beaten, it will overcome the friction and the carpet will move away from the dust making the carpet free from dust.
Sliding on a garden slide- We know that friction is a force that is present whenever two objects rub against each other. In case of a slide in the garden such as a slide and a person’s backside rub each other’s surface. Without friction, a slide would accelerate the rider too quickly, resulting in possible injury due to the fall. The friction reduces the velocity of the sliding person and makes him stop.
Hope that helps! :D Sorry if it's too lengthy...
Explanation:
Can you help me please here is the picture DUE NOW PLEASE.
they are formed when hot
it's dependent on the rate of cooling of the melt, slow cooling allows large crystals to form, fast cooling yields small crystals. They cool too quickly to form crystals.
Hope this helps! <3 :3
An important news announcement is transmitted by radio waves to people who are 46 km away, sitting next to their radios, and by sound waves to people sitting across the newsroom, 2.1 m from the newscaster. Take the speed of sound in air to be 348 m/s. What is the difference in time that the message is received
Answer:
[tex]0.005847\ \text{s}[/tex]
Explanation:
Radio waves travel at the speed of light
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
[tex]d_r[/tex] = Distance between two radios = 46 km
[tex]v[/tex] = Speed of sound in air = 348 m/s
[tex]d_s[/tex] = Distance sound travels across the newsroom = 2.1 m
Time taken by the radio signal to reach the required location is
[tex]t_r=\dfrac{d_r}{c}\\\Rightarrow t_r=\dfrac{46\times 10^3}{3\times 10^8}\\\Rightarrow t_r=0.000153\ \text{s}[/tex]
Time taken by sound to reach the required location is
[tex]t_s=\dfrac{d_s}{v}\\\Rightarrow t_s=\dfrac{2.1}{348}\\\Rightarrow t_s=0.006\ \text{s}[/tex]
The time difference is
[tex]t_s-t_r=0.006-0.000153=0.005847\ \text{s}[/tex]
The difference in time that the message is received is [tex]0.005847\ \text{s}[/tex].
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.80 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
10. A 50 kg bicyclist on a 10 kg bicycle speeds up from 5.0 m/s to 10 m/s.
(a) What was the total kinetic energy before accelerating? Full working out
Answer:
T.K.E = 750 Joules.
Explanation:
Given the following data;
Initial velocity, u = 5m/s
Final velocity, v = 10m/s
Mass of bicyclist = 50kg
Mass of bicycle = 10kg
Total mass, Tm = 50 + 10 = 60kg
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}mv^{2}[/tex]
Where;
K.E represents kinetic energy measured in Joules.
m represents mass measured in kilograms.
v represents velocity measured in metres per seconds square.
To find the total kinetic energy before accelerating simply means the kinetic energy due to the initial velocity and total mass;
[tex] T.K.E = \frac{1}{2}T_{m}U^{2}[/tex]
Substituting into the equation, we have;
[tex] T.K.E = \frac{1}{2}*60*5^{2}[/tex]
[tex] T.K.E = 30*25 [/tex]
T.K.E = 750 Joules.
Alternate current
Hello!
I need solve 2 questions, I tried but im so bad at physics :(
1. If the frequency in the network is 50Hz, what is the period of this voltage?
2. The effective voltage value in the network is 230V. What is its maximum value?
I would like an explanation with a calculation for my understanding!!
Thanks in advance you all!
sorry bro but i don't know his i am also bad
Explanation:
but maybe you can you the formula of volatage as well as
True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy.
If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy. _ This is false statement.
What is projectile motion?When a particle is hurled obliquely near the surface of the earth, it travels along a curved path while accelerating continuously in the direction of the planet's center (we assume the particle stays close to the surface of the globe). Such a particle's motion is known as projectile motion, and its route is referred to as a projectile.
In projectile motion total energy is conserved. Hence, when a rock is thrown into the air, the increase in the height would increase the rock’s potential energy, and then the increase in the velocity as it falls to the ground would increase its kinetic energy.
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The bus lay 40 km at a speed of 72 km / h, and then another 60 km at a speed of 30 m / s. Determine the average speed of the bus along the way.
Answer:
25 m/s
Explanation:
From the question,
Average speed = Total distance /total time.
S' = D/T........................... Equation 1
D = 40+60 = 100 Km = 100000 m.
T = t₁+t₂
t₁ = (40×3600/72) s = 2000 s
t₂ = 60000/30 = 2000 s
T = 2000+2000 = 4000 s.
SUbstitute the values of T and D into equation 1
S' = 100000/4000
S' = 25 m/s
5j-Tj=6j+3 Tj
Determine the value of T from the given vector
. Four railroad cars, each of mass 2.50 104 kg, are coupled together and coasting along horizontal tracks at a speed of vi toward the south. A very strong but foolish movie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to 4.00 m/s southward. The remaining three cars continue moving toward the south, now at 2.00 m/s. (a) Find the initial speed of the cars. (b) How much work did the actor do
Answer:
a) v₀ = 2.5 m / s, heading south.
b) W = 1,219 10⁵ J
Explanation:
a) For this exercise we can use the conservation of the moment, we create a system formed by all the 4 cars, in this case when the last one separates the forces are intense and the moment is conserved
initial instant. Before separation
p₀ = M v₀
final instant. When uncoupling the last car
p_f = 3m v₁ + m v₂
where they indicate that the speed of the wagons is v₁ = 2.00 m / s and the speed of the last wagon is v₂ = 4.00 m / s
the total mass is M = 4m
how the moment is preserved
p₀ = p_f
4m v₀ = 3m v₁ + mv₂
v₀ = ¾ v₁ + v₂ / 4
let's calculate
v₀ = ¾ 2 + ¼ 4
v₀ = 2.5 m / s
heading south.
b) work is equal to the change in kinetic energy
W = ΔK = K_f -K_o
W = ½ m v_f² - ½ m v₀²
W = ½ m (v_f² -v₀²)
W = ½ 2.50 10⁴ (4² - 2.5²2)
W = 1,219 10⁵ J
3. A backpack weighs 8.2 N and has a mass of 5 kg on the moon. What is t
strength of gravity on the moon?
8.2 N
describe how electrons create electricity
describe how electrons create electricity
Atoms are made up of even smaller elements, called protons, electrons and neutrons. When electrical and magnetic forces move electrons from one atom to another, an electrical current is formed
Electrons in atoms can act as our charge carrier, because every electron carries a negative charge. If we can free an electron from an atom and force it to move, we can create electricity. ... In its balanced state, copper has 29 protons in its nucleus and an equal number of electrons orbiting around it.
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What color is a carrot?
Answer:
reddish-orrange
Explanation:
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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.10 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m. The fisherman sees that the wave crests are spaced a horizontal distance of 6.10 m apart. A. How fast are the waves traveling?B. What is the amplitude of each wave? C. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling? D. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) v = 2.9 m / s, b) A = 0.350 m, c) v = 2.9 m / s, d) A = 1.00 m
Explanation:
The oscillatory motion is described by the expression
x = A cos (wt + Ф)
the wavelength which is the distance for the wave to repeat and the frequency which is the number of times a wave oscillates per unit of time
a) In this part they ask us for the speed of the wave.
Let's use the relationship between speed, wavelength and frequency
v = λ f
For the wavelength they indicate that the distance between two crest is 6.1 m
λ / 2 = 6.10
λ = 12.20 m
They give us the period of the wave is the time it takes to return to the same point, in this case they give half a period
A / 2 = 2.10
A = 4.20 me
f = 1 / t
f = ¼, 2
f = 0.238 Hz
let's calculate
v = 12.20 0.238
v = 2.9 m / s
b) the amplitude of the wave, is the distance from zero to some maximum
2A = 0.700
A = 0.350 m
c) the speed of the wave is not function of the amplitude, so the speed is the same
v = 2.9 m / s
d) the amplitude is
2A = 0.50
A = 1.00 m
Which is a primary energy source used by power plants to generate electricity?
coal
O wood
o gasoline
batteries
Answer:
Your answer is wood that is what they used
The primary energy source used by power plants to generate electricity is coal.
What are power plants?Power plants are plants which use fuels to generate electricity for use by homes and industries.
Power plants have different energy sources for their fuel.
The primary energy source used by power plants to generate electricity is coal.
Therefore, coal is a primary source of fuel for power plants.
Learn more about power plants at: https://brainly.com/question/16064039
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what is the magnitude of electric field at apoint 2 meter from a point charge q= 4nc
Which resource when used causes the most pollution?
solar
coal
wind
O water