Answer:
Magnitude of the electrostatic force on the +32 µC charge, [tex]F_{net} = 12 N[/tex]
Explanation:
Let q₁ = +32 µC, x₁ = 0
q₂ = +20 µC, x₂ = 40 cm = 0.4 m
q₃ = -60 µC, x₃ = 60 cm = 0.6 m
Let magnitude of the electrostatic force on the +32 µC charge due to the + 20 µC charge = F₁ (i.e force on q₁ due to q₂)
[tex]F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }[/tex]
[tex]F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N[/tex]
Let magnitude of the electrostatic force on the +32 µC charge due to the -60 µC charge = F₂ (i.e force on q₁ due to q₃)
[tex]F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }[/tex]
[tex]F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N[/tex]
The electrostatic force on the 32 µC charge, [tex]F_{net} = |F_{2} + F_{3}|[/tex]
[tex]F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N[/tex]
Electric motors convert electrical energy to mechanical energy. When the current-carrying coil is placed between the magnetic poles, a force acts on it that causes it to rotate. The image below shows a simple electric motor. The motor is used to lift metal boxes. How can the motor be changed to be able to lift a heavier box? A. rotate the coil in a counter-clockwise direction B. add more loops of wire between the magnets C. change the polarity of the magnet D. decrease the size of the magnets
Answer:
B. add more loops of wire between the magnets
Explanation:
this would increase the magnetic force acting on the rod therefore increasing
By adding more loops of wire between the magnets the motor is changed to be able to lift a heavier box.
What is an electric motor?An electric motor is a mechanism that turns electricity into mechanical energy.
The interaction between the motor's magnetic field and electric current in a wire winding generates force in the form of torque imparted to the motor's shaft in most electric motors.
An electric generator is physically equivalent to an electric motor, but it converts mechanical energy into electrical energy using a reversed flow of power.
The load capacity in the motor can be increased by increasing the no of loops. So by adding more loops of wire between the magnets the motor is changed to be able to lift a heavier box.
Hence option B is correct.
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When a switch is closed to complete a DC series RL circuit which has a large time constant, Group of answer choices the electric field in the wires gradually increases to a maximum value. the rate of change of the electric and magnetic fields is greatest at the instant when the switch is closed. all of the above are true. the magnetic field outside the wires gradually increases to a maximum value. only (a) and (c) above are true.
Answer:
The electric and magnetic fields gradually increase.
Explanation:
Time constant in RL circuits, denoted by τ, is equal to the value of L / R which is the value of the inductor over the resistor. It is used to calculate the point when the current will reach the maximum value in the steady state of the circuit. Because of this behavior of the circuit, the magnetic field and the electric field gradually increase to their maximum values.
I hope this answer helps.
A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the circuit?
Answer:0.00125 watts
Explanation:
resistance=50 ohms
Current=5 milliamps
Current=5/1000 milliamps
Current =0.005 amps
power=(current)^2 x (resistance)
Power=(0.005)^2 x 50
Power=0.005 x 0.005 x 50
Power=0.00125 watts
The power in the circuit is 1.25 mW.
What is Ohm's law?
Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature and other physical conditions remain constant.
In other words, the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R):
I = V/R.
This relationship is often written as V = IR or R = V/I. Ohm's law is named after Georg Simon Ohm, a German physicist who first formulated the relationship in the early 19th century.
Here in the Question,
Using Ohm's law, we can find the voltage in the circuit as:
V = IR
V = (5 mA)(50 Ω) = 0.25 V
Using the formula for power, we can find the power in the circuit as:
P = IV
P = (5 mA)(0.25 V) = 0.00125 W or 1.25 mW
Therefore, the power in the circuit is 1.25 mW.
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6. The two ends of an iron rod are maintained at different temperatures. The amount of heat thatflows through the rod by conduction during a given time interval does notdepend uponA) the length of the iron rod.B) the thermal conductivity of iron.C) the temperature difference between the ends of the rod.D) the mass of the iron rod.E) the duration of the time interval.Ans: DDifficulty: MediumSectionDef: Section 13-27. The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twiceas long and twice the diameter conduct heat between the same two temperatures
Answer:
20cal/s
Explanation:
Question:
There are two questions. The first one has been answered:
From the formular, Power = Q/t = (kA∆T)/l
the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.
The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).
Second question:
The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?
Solution:
Power = 10cal/s
Power = energy per unit time = Q/t
Where Q = energy
Power = (kA∆T)/l
k = thermal conductivity of iron
A = area
Area = πr^2
r = radius
Diameter = d = 2r
r = d/2
Area = (πd^2)/4
Length = l
∆T = change in temperature
10 = (kA∆T)/l
For a steel rod with length doubled and diameter doubled:
Let Length (L) = 2l
Diameter (D)= 2d
Area = π [(2d)^2]/4 = (π4d^2)/4
Area = 4(πd^2)/4
Using the formula Power = (kA∆T)/l, insert the new values for A and l
Power = [k × 4(πd^2)/4 × ∆T]/2l
Power = [4k((πd^2)/4) ∆T]/2l
Power = [(4/2)×k((πd^2)/4) ∆T]/l
Power = [2k(A) ×∆T]/l = 2(kA∆T)/l
Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l
Recall initial Power = (kA∆T)/l = 10cal/s
And ∆T is the same
2[(kA∆T)/l] = 2 × 10
Power of a steel that has its length doubled and diameter doubled = 20cal/s
A particle with charge q = +1e and mass m = 5.5×10^-26 kg is injected horizontally with speed 2.4×10^6 m/s into the region between two parallel horizontal plates. The plates are 22 cm long and 1.2 cm apart. The particle is injected midway between the top and bottom plates. The top plate is negatively charged and the bottom plate is positively charged, so that there is an upward-directed electric field between the plates, of magnitude E = 33 kN/C. Ignore the weight of the particle.
(a) How long, in seconds, does it take for the particle to pass through the region between the plates?
(b) When the particle exits the region between the plates, what will be the magnitude of its vertical displacement from its entry height, in millimeters?
Answer:
a) t = 9.16*10^{-18} s
b) y = 0.402 mm
Explanation:
(a) To find the time that the particle takes to pass trough the region between parallel plates, you take into account that the horizontal component of the velocity is constant in all trajectory of the particle. Then, you use the following formula:
[tex]t=\frac{x}{v_x}\\\\[/tex]
x: length of the sides of the plates = 0.22m
v_x: horizontal component of the velocity = 2.4*10^6 m/s
[tex]t=\frac{0.22m}{2.4*10^6m/s}\\\\t=9.16*10^{-8}s=91.6ns[/tex]
(b) To find the vertical displacement of the particle you first calculate the acceleration of the particle generated by the electric force:
[tex]F_e=ma\\\\qE=ma\\\\a=\frac{qE}{m}\\\\a=\frac{(1.6*10^{-19}C)(33*10^3N/C)}{5.5*10^{-26}C}=9.6*10^{10}\frac{m}{s^2}[/tex]
where you have used that the charge is 1.6*10^-19 C (charge of an electron).
With the values of the acceleration and time you use the following kinematic equation to calculate the vertical displacement:
[tex]y=v_{oy}t+\frac{1}{2}at^2\\\\v_{oy}=0m/s\\\\y=\frac{1}{2}(9.6*10^{10}\frac{m}{s^2})(9.16*10^{-8}s)^2=4.02*10^{-4}m=0.402mm[/tex]
Following are the solution to the given points:
When calculating the time required the particle will pass through the zone between plates, remember that such a horizontal component of the particle's velocity is fixed throughout the particle's route. The following formula is then used:
[tex]\to t=\frac{x}{v_x}[/tex]
[tex]\to x:[/tex] plate length = 0.22m
[tex]\to v_x :[/tex]velocity's horizontal component[tex]= 2.4\times 10^{6} \ \frac{m}{s}\\\\[/tex]
[tex]\to t=\frac{0.22\ m}{ 2.4 \times 10^6 \frac{m}{s}}= 9.16 \times 10^{-8} \ s = 91.6 \ ns[/tex]
To determine the particle's vertical displacement, firstly compute the particle's acceleration due to electric force:[tex]\to F_e=ma\\\\\to qE=ma\\\\\to a=\frac{qE}{m}\\\\[/tex]
[tex]=\frac{1.6 \times 10^{-19}\ C \times 33 \times 10^3 \frac{N}{C}}{5.5 \times 10^{-26} C} \\\\ =\frac{1.6 \times 10^{-16} \times 33 \times 10^{26}}{5.5} \\\\ =\frac{1.6 \times 33 \times 10^{10}}{5.5} \\\\ =\frac{16 \times 33 \times 10^{10}}{55} \\\\ = 9.6 \times 10^{10} \ \frac{m}{s^2}\\\\[/tex]
when you have used that charge [tex]1.6\times 10^{-19}\ C[/tex]electronic charge).The following kinematic formula is used to determine vertical displacement using the values of acceleration and time:[tex]\to y=v_{oy} t+\frac{1}{2} at^2\\\\\to v_{oy}= 0 \ \frac{m}{s}\\\\\to y=\frac{1}{2} (9.6 \times 10^{10} \frac{m}{s^2}) (9.6 \times 10^{-8}\ s)^2[/tex]
[tex]= 4.02 \times 10^{-2} \ m \\\\ = 0.402 \ mm\\\\[/tex]
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When a high‑energy photon passes near a heavy nucleus, a process known as pair production can occur. As a result, an electron and a positron (the electron's antiparticle) are produced. In one such occurrence, a researcher notes that the electron and positron fly off in opposite directions after being produced, each traveling at speed 0.941c. The researcher records the time that it takes for the electron to travel from one position to another within the detector as 15.7 ns. How much time would it take for the electron to move between the same two positions as measured by an observer moving along with the positron?
Answer:
1.47*10^{-8}s
Explanation:
You first calculate the distance traveled by the electron:
[tex]x=vt\\\\x=(0.941(3*10^8m/s))(15.7*10^{-9}s)=4.43m[/tex]
Next, you calculate the relative speed as measure by an observer in the positron, of the electron:
[tex]u'=\frac{u+v}{1+\frac{uv}{c^2}}\\\\u'=\frac{0.941c+0.941c}{1+\frac{(0.941)^2c^2}{c^2}}\\\\u'=0.99c[/tex]
with this relative velocity you calculate the time:
[tex]t=\frac{x}{u'}\\\\t=\frac{4.43m}{0.99c}=1.47*10^{-8}s[/tex]
A 20-kilogram box is sitting on an inclined plane with a 30-degree slope. If the box is at rest, what is the force
on the box due to friction?
Answer:
Explanation:
The box is at rest because , the component of weight along the surface downwards is equal to frictional force .
frictional force = mgsinθ
= 20 x 9.8 sin 30
= 98 N .
Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of
Answer:
18
Explanation:
got it right on edge
the speed of sound is 343m/s. dezeirey is positioned 5m behind her. how many seconds will it take for the echo from the wall to reach her
Answer:
t = 0.029 s
Explanation:
We have,
Speed of sound is 343 m/s.
Dezeirey is positioned 5 m behind her.
It is required to find the time taken for the echo from the wall to reach her. The total distance covered by the echo when it reaches her is 2d or 10 m.
Time taken,
[tex]t=\dfrac{d}{v}\\\\t=\dfrac{10\ m}{343\ m/s}\\\\t=0.029\ s[/tex]
So, it will take 0.029 seconds for the echo from the wall to reach her.
Which of the following characteristics of Earth's crust is likely to contribute to geological events?
A. can be broken into many plates
B. has has uniform thickness throughout
C. cysts on top of the solid rock of the lower mantle
D. is composed of a single layer that surrounds Earth
Answer:A
Explanation:
The crust Of The earth's has plates boundaries,when the layers that forms to boundaries glide on each other, vibrations occur which are known as earthquakes
Answer:
A) Can be Broken Into Many Plates
What types of mediums are involved in the energy transfer
Answer:
In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.
Forests have been completely wiped out in some parts the world. In areas that are economically depressed, people use the cut wood to fuel fires for cooking and for making shelters or homes. How would deforestation most likely impact Earth's biogeochemical cycles?
Group of answer choices
Less carbon dioxide will be present in the atmosphere because the people living in the area still breath in oxygen.
An increase in carbon dioxide will occur in the atmosphere because there are less trees to use the gas in the process of photosynthesis.
Carbon dioxide levels will not change, since plants do not have a role in the carbon cycle.
More water will be released into the atmosphere because transpiration and condensation will produce more precipitation.
Answer: An increase in carbon dioxide will occur in the atmosphere because there are less trees to use the gas in the process of photosynthesis.
Explanation: Photosynthesis can be simply defined as a process where green plants uses minerals from the soil, water and carbon dioxide to produce their own food (energy-rich organic compound/starch) while oxygen is released into the environment.
On the other hand, Human beings and other animals inhales the oxygen released into the environment by green plants and breath out carbon dioxide.
Deforestation is the permanent removal trees either for commercial or household use or in some cases, to make room or space for other activities or development.
Now when these trees (green plants) are permanently cut down without replacement, humans and other animals continuously exhale carbon dioxide and there will be little or no green plants (trees) available to convert this carbon dioxide into oxygen.
Hence, there will be an increased amount of carbon dioxide in the atmosphere as their are little or no tress to use the gas (carbon dioxide) in the process of photosynthesis.
As SCUBA divers go deeper underwater, the pressure from the weight of all the water above them increases tremendously which compresses the gases in their blood. What happens to the volume of gas in their blood as the diver rises quickly to the surface?
Answer:
The volume of gas in an early diving bell full of air at sea level is halved at 10 m according to Boyle’s law
Explanation:
;at 20 m pressure is 300 kPa absolute and the gas is compressed into one third the volume.
HOPE THIS HELPS i did this before
What do you think will be different about cars in the future? Describe a change that is already being developed or that you think should be invented.
Answer:
Flying cars.
Explanation:
Which term defines the distance from crest to crest
Answer:
The horizontal distance between two adjacent crests or troughs is known as the wavelength.
Answer: Wavelength
Explanation:
From crest to crest, it is one full wavelength
Which of these parameters is directly related to sound frequency?
Answer:Velocity
Explanation:
Velocity is directly proportional to the frequency of a wave.
Velocity=frequency x wavelength
Speed is the rate of change of position expressed as____ Traveled per unit of time
Answer:
distance
Explanation:
Speed is the rate of change of position expressed as distance travelled per unit of time
1. A 5 Ohm resistor is connected to a 9 Volt battery. How many Joules of thermal energy are produced in 7 minutes?
2. The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?
3. If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?
4. A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the circuit?
5. How many Joules of electric energy are delivered to a 60 Watt lightbulb if the bulb is left on for 2.5 hours?
Answer:
Question 1: 189 joules
Question 2: 2.25 watts
Question 3: 405 joules
Question 4: 0.00125 watts
Question 5: 450000 joules
Explanation:
question 1:
Volt=1.5v
Resistance =5 ohms
Time=7 minutes
Time=7 x 60
Time=420 seconds
Current =voltage ➗ resistance
Current =1.5 ➗ 5
Current =0.3 amps
Energy=current x voltage x time
Energy=0.3 x 1.5 x 420
Energy =189 joules
Question 2:
Current =0.5 amps
Voltage =4.5v
Power=current x voltage
Power=0.5 x 4.5
Power=2.25 watts
Question 3:
Current=0.5 amps
Voltage=4.5v
Time =3 minutes
Time =3x60
Time =180 seconds
Energy=current x voltage x time
Energy=0.5 x 4.5 x 180
Energy =405 joules
Question 4:
Resistance=50 ohms
Current =5 milliamps
Current =5/1000
Current =0.005 amps
Power =current x current x resistance
Power=0.005 x 0.005 x 50
Power=0.00125 watts
Question 5:
Power =50 watts
Time=2.5 hour
Time=2.5 x 60 x 60
Time =9000 seconds
Energy=power x time
Energy=50x9000
Energy=450000 joules
Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have albedo values of 0.15 and 0.75, and observed surface temperatures of approximately 220 K and 700 K, respectively. The average distance of Mars from the sun is 2.28 x 108 km, and the average distance of Venus is 1.08 x 108 km. Given that the radius of the sun is 7 x 108 m, and the energy flux at the surface of the sun is 6.28 x 107 W/m2 , what is the extent of the greenhouse effect for Mars
Answer:
The extent of greenhouse effect on mars is [tex]G_m = 87 K[/tex]
Explanation:
From the question we are told that
The albedo value of Mars is [tex]A_1 = 0.15[/tex]
The albedo value of Mars is [tex]A_2 = 0.15[/tex]
The surface temperature of Mars is [tex]T_1 = 220 K[/tex]
The surface temperature of Venus is [tex]T_2 = 700 K[/tex]
The distance of Mars from the sun is [tex]d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m[/tex]
The distance of Venus from sun is [tex]d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m[/tex]
The radius of the sun is [tex]R = 7*10^{8} \ m[/tex]
The energy flux is [tex]E = 6.28 * 10^{7} W/m^2[/tex]
The solar constant for Mars is mathematically represented as
[tex]T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ][/tex]
Where [tex]\sigma[/tex] is the Stefan's constant with a value [tex]\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}[/tex]
So substituting values
[tex]T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}[/tex]
[tex]T = 307K[/tex]
So the greenhouse effect on Mars is
[tex]G_m = T - T_1[/tex]
[tex]G_m = 307 - 220[/tex]
[tex]G_m = 87 K[/tex]
A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed uniformly over the cross section of the cylinder. A cylindrical shell with radius b = 21 cm is concentric with the solid cylinder and carries a current I2 = 3 A coming out of the page. 1)Calculate the y component of the magnetic field By at point P, which lies on the x axis a distance r = 41 cm from the center of the cylinders.
Answer:
Explanation:
We shall use Ampere's circuital law to find magnetic field at required point.
The point is outside the circumference of two given wires so whole current will be accounted for .
Ampere's circuital law
B = ∫ Bdl = μ₀ I
line integral will be over circular path of radius r = 41 cm .
Total current I = 5A -3A = 2A .
∫ Bdl = μ₀ I
2π r B = μ₀ I
2π x .41 B = 4π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ x 2 / .41
= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.
A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object of mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
a) Make sure that the length of the string is not too long.
b) Make sure that the mass of the pendulum is not too large.
c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
d) Make sure that the experiment is conducted in an environment that has minimal wind resistance.
Answer:
the answers the correct one is cη
Explanation:
In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so
sin θ = θ
with this approach the equation will be surveyed
d² θ / dt² = - g / L sin θ
It is reduced to
d² θ / dt² = - g / L θ
in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%
The angle is related to the height of the pendulum
sin θ = h / L
h = L sin θ.
Therefore the student must be careful that the height is small.
When reviewing the answers the correct one is cη
Considering the approximation of simple harmonic motion, the correct option is:
(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
Simple Harmonic MotionAccording to Newton's second law in case of rotational motion, we have;
[tex]\tau = I \alpha[/tex]
Applying this, in the case of a simple pendulum, we get;
[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]
On, rearranging the above equation, we get;
[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]
Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.
Then, [tex]sin\, \theta \approx \theta[/tex]
[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]
This is now in the form of the equation of a simple harmonic motion.
[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]
Comparing both these equations, we can say that;
[tex]\omega = \sqrt{\frac{g}{L}}[/tex]
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]
This relation for the time period can only be obtained if the angular displacement is very less.
So, the correct option is;
Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.
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A bicycle coasting downhill reaches its maximum speed at the bottom of the
hill.
This speed would be even greater if some of the bike's
energy had
not been transformed into
energy
A) kinetic; heat
OB) heat; potential
C) kinetic; potential
OD) potential; kinetic
OB
mmnjnjlkdhfutydjfyiudtkcgvyftdcgvjyiluftgyiuyu ( had to do that cuz it wouldn't let through)
a girl pushes an 18.15 kg wagon with a force of 3.63 N. what is the acceleration?
A. 0.06 m/s2
B. 5 m/s2
C. 0.2 m/s2
Answer:0.2 m/s^2
Explanation:
mass=18.15kg
Force=3.63N
Acceleration=force ➗ mass
Acceleration=3.63 ➗ 18.15
Acceleration=0.2 m/s^2
A girl pushes a wagon of mass 18.15 kg with a force of 3.63 N, so the acceleration of the wagon will be 0.2 m/s².
What is acceleration?In mechanics, acceleration is the measure of how rapidly an object's velocity changes over time. Accelerations as a vector quantity. An object's acceleration depends on the direction of the net force exerted on it.
A vector quantity, acceleration, is something that has both a magnitude and a direction. As a vector quantity, velocity is also. The ratio of the velocity vector change over a time interval to that interval is the definition of acceleration.
Mass, m =18.15 kg
Force, f = 3.63 N
Force = m × a
a= f / m
a =3.63 / 18.15
a = 0.2 m/s²
Hence, the acceleration of the wagon will be 0.2 m/s².
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A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 14.9 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 3h0 above the spring, what would be the maximum compression of the spring?
Answer:
a)5.88J
b)-5.88J
c)0.78m
d)0.24m
Explanation:
a) W by the block on spring is given by
W= [tex]\frac{1}{2}[/tex]kx² = [tex]\frac{1}{2}[/tex](530)(0.149)² = 5.88 J
b) Workdone by the spring = - Workdone by the block = -5.88J
c) Taking x = 0 at the contact point we have U top = U bottom
So, mg[tex]h_o[/tex] = [tex]\frac{1}{2}[/tex]kx² - mgx
And, [tex]h_o[/tex]= ( [tex]\frac{1}{2}[/tex]kx² - mgx )/(mg) = [tex][\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)[/tex]]/(0.645x9.8)
[tex]h_o[/tex]= 0.78m
d) Now, if the initial initial height of block is 3[tex]h_o[/tex]
[tex]h_o[/tex] = 3 x 0.78 = 2.34m
then, [tex]\frac{1}{2}[/tex]kx² - mgx - mg[tex]h_o[/tex] =0
[tex]\frac{1}{2}[/tex](530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0
265x² - 6.321x - 14.8 = 0
a=265
b=-6.321
c=-14.8
By using quadratic eq. formula, we'll have the roots
x= 0.24 or x=-0.225
Considering only positive root:
x= 0.24m (maximum compression of the spring)
How do most black holes form?
Has anyone read the book Third level
In order to get going fast, eagles will use a technique called stooping, in which they dive nearly straight down and tuck in their wings to reduce their surface area. While stooping, a 6- kg golden eagle can reach speeds of up to 53 m/s . While golden eagles are not very vocal, they sometimes make a weak, high-pitched sound. Suppose that while traveling at maximum speed, a golden eagle heads directly towards a pigeon while emitting a sound at 1.1 kHz. The emitted sound has a sound intensity level of 30 dB when heard at a distance of 5 m .A) Model this stooping golden eagle as an object moving at terminal velocity. The eagle’s drag coefficient is 0.5 and the density of air is 1.2 kg/m 3 . What is the effective cross-sectional area of the eagle’s body while stooping?B) What is the doppler-shifted frequency that the pigeon will hear coming from the eagle?C) Consider the moment when the pigeon is 5 m away from the eagle. At the pigeon’s position, what is the intensity (in W/m^2 ) of the sound the eagle makes?D) The golden eagle slams into the 250- g pigeon, which is initially moving at 10 m/s in the opposite direction (toward the eagle). The eagle grabs the pigeon in its talons, and they move off together in a perfectly inelastic collision. How fast do they move after the collision?
Answer:
Check the explanation
Explanation:
Part A
F = CA
this drag force balances the weight = 6X 9.8
so
6X9.8 = 0.5 X A X0.5 X 1.2 X 532
A= 0.069 m2
Part B
here the sorce is moving and the observer is at rest
so f= f(- 1 - 1
f = 1.1X10 343 343 – 53
f' = 1.3 KHz
Part C:
given the intensity = 30 dB
we know that I dB = 10 log (I(W/m2))
so we get I (W/m2) = 1000
Part D : The catch
Given that U1 = 53 M1 = 6 kg
U2 =-10 M2=0.25
V1=V2
now conserving momentum
6 X 53 -0.25 X10 =(6+0.25)V
V= 50.48 m/sec
A radiator rests snugly on the floor of a room when the temperature is 10 oC. The radiator is connected to the furnace in the basement by a pipe that is 15 m long. How far off the floor will the radiator be lifted when it is filled with steam at 102 oC? The iron expands 1.0 * 10-5 / oC.
In witch water environment would holdfast be most useful
Answer: Holdfast is a root- like structure by which an aquatic sessile algae are attached to a substrate. Its primary function is to secure the organism to the sea floor, i.e anchorage.
Mutations that neither benefit nor harm the organisms have blanks effect on the organisms survival
Answer:
I would say that the mutation has no effect on the organism, as it doesn't help or harm it.
hope this helps :)
