Answer:
Both balls will hit the ground at the same time, because gravity is constant.Explanation:
Notice that both balls are being thrown at the same initial height.
It's important to know that these movements which depends of gravity (a constant acceleratio) they would fall at the same rythm, because the gravity is a constant.
Remember that gravity is an acceleration, which it's defined as the change of the velocity, so if both balls change their vertical velocity at the same rate, then they will fall at the same time, because they have the same initial height.
Additionally, when you throw a ball horizontally, it will bend down due to gravity, and this falling movement is the same as if you throw vertically as a free falling movement.
Therefore, both balls will hit the ground at the same time, because gravity is constant.
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. Long after contact is made with the battery (a) the voltage across the capacitor is A) equal to the battery's terminal voltage. B) less than the battery's terminal voltage, but greater than zero. C) zero. (b) the voltage across the resistor is A) equal to the battery's terminal voltage. B) less than the battery's terminal voltage, but greater than zero. C) zero.
Answer:
A) equal to the battery's terminal voltage.
Explanation:
When the capacitor is fully charged after long hours of charging , its potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and current becoming zero . There is no charging current when the capacitor is fully charged .
3. The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?
4.If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?
Answer:
Question 3: 2.25 watts
Question 4: 405 joules
Explanation:
question 3:
Current =0.5 amps
Voltage =4.5 volts
Power= current x voltage
Power=0.5 x 4.5
power=2.25 watts
Question 4
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
When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done by friction because: A. there is no friction present B. the angular velocity of the center of mass about the point of contact is zero C. the coefficient of kinetic friction is zero D. the linear velocity of the point of contact (relative to the inclined surface) is zero E. the coefficient of static and kinetic friction are equal
Answer:
D. the linear velocity of the point of contact (relative to the inclined surface) is zero
Explanation:
The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.
PIUDICITIS CONSECulvely and Circle your aliswers. Lilyo
proper significant digits.
53. When you turn on your CD player, the turntable accelerates from zero to 41.8 rad/s in
3.0 s. What is the angular acceleration?
or
Answer:
The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Explanation:
Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.
It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :
[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]
So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Which term BEST describes the movement of air from the ocean toward the land in the daytime? (AKS 4b DOK 1) *
1 point
Sea breeze
Land Breeze
Valley Breeze
Current Breeze
Answer:
Option A, Sea Breeze
Explanation:
Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.
Hence, option A is correct
A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.
Answer:
b) 20 kJ
Explanation:
Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .
T₁ = 270 + 273 = 543K
T₂ = 50 + 273 = 323 K
Putting the given values of temperatures
efficiency = (543 - 323) / 543
= .405
heat input = 50 KJ
efficiency = output work / input heat energy
.405 = output work / 50
output work = 20.25 KJ.
= 20 KJ .
A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The wrecking ball swings along the wall between her house and the neighbor’s house. In an effort to determine the length of the cable on the wrecking ball the student builds a pendulum using a random rock and a string. Her pendulum turns out to be 0.500m long. While she plays with her pendulum she realizes that the wrecking ball swings back and forth in the same amount of time that it takes the rock to complete 5 full oscillations. What is the length of the cable on the wrecking ball?
Answer:
The length of cable is 12.5 m
Explanation:
Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.
Therefore,
Time Period of Wrecking Ball = 5 (Time Period of Rock)
Since,
Time Period of Pendulum = 2π√(L/g)
Therefore,
2π√(L₁/g) = 5[2π√(L₂/g)]
√L₁ = 5√L₂
Squaring on both sides:
L₁ = 25 L₂
where,
L₁ = Length of Cable = ?
L₂ = Length of string = 0.5 m
Therefore,
L₁ = 25 (0.5 m)
L₁ = 12.5 m
Yellow light with wavelength 600 nm is travelling to the left (in the negative x direction) in vacuum. The light is polarized along the z direction. (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave. (c) If the wave were to represent blue light instead of yellow light, how would your pictures in parts a and b change? If there is no change, say so explicitly.
Answer: (a) and (b) => check attached file.
(c). Picture (a) and (b) will both remain the same.
Explanation:
IMPORTANT: The solution to the question (a) and (b) that is (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.
It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.
Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.
Hence, the solution to option C is given below;
(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500.0 N/m (assume that the spring is massless). The block is held in position such that the spring is compressed 4.00 cm shorter than its undisturbed length. The block is suddenly released and allowed to slide away on the frictionless surface. Find the speed the block will be traveling when it leaves the spring.
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]
[tex]kx^2 = mv^2[/tex]
[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]
[tex]7.2 =200*10^{-3}*v^{2}[/tex]
[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]
[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]
v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s
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.
Which is the correct representation of the right-hand rule for a current flowing to the right?
Answer:
The third image
Explanation:
The one with the thumb pointing to the right
Answer:
3, correct on Edge 2020
A mutation causes a dog to be born with a tail that is shorter than normal.
Which best describes this mutation?
Answer:
A mutation causes a dog to be born with a tail that is shorter than normal. Which best describes this mutation? It is harmful because it obviously affects the dog’s survival. It is harmful because it affects the dog’s physical appearance. It is neutral because it does not obviously affect the dog’s survival. It is beneficial because it affects the dog’s physical appearance.
Explanation:
Answer:
C
Explanation:
:)))
How are the elements in the same row similar
Answer:
All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.
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:
Now that you've done your research on the law of supply, you understand that it basically asserts that how much coffee you'd be willing to supply depends on how much money you can make for each cup.
Calculate potential energy of a 5 kg object sitting on 3 meter ledge
Answer:147 joules
Explanation:
Mass=m=5kg
Acceleration due to gravity=g=9.8m/s^2
Height=h=3 meter
Potential energy=m x g x h
Potential energy=5 x 9.8 x 3
Potential energy=147 joules
8. At temperature 15°C, aluminum rivets have a diameter of 0.501 cm, and holes drilled in a titanium sheet have a diameter of 0.500 cm. If both the aluminum rivets and the titanium sheet are cooled together, at what temperature will the rivets just fit into the appropriate holes in the titanium sheet? Use 25x10-6 (°C)-1 for the coefficient of linear expansion for aluminum, and 8.5x10-6 (°C)-1 for titanium
Answer:
The temperature is [tex]T = -106 ^oC[/tex]
Explanation:
From the question we are told that
The temperature is [tex]T_1 = T_t= T_a=15^oC[/tex]
The diameter is [tex]d_1 = 0.5001 cm[/tex]
The diameter of the hole [tex]d_2 = 0.500 \ cm[/tex]
The coefficient of linear expansion for aluminum is [tex]\alpha _1 = 25 *10^{-6} \ ^oC^{-1}[/tex]
The coefficient of linear expansion for titanium is [tex]\alpha _2 = 8.5 *10^{-6} \ ^o C^{-1}[/tex]
According to the law of linear expansion
[tex]d = d_o (1 + \alpha \Delta T )[/tex]
Where [tex]d_o[/tex] represents the original diameter
So for aluminum
[tex]d_a = d_1 (1 + \alpha_1 (T- T_a) )[/tex]
Where [tex]d_a[/tex] is the new diameter of aluminum
[tex]T_a[/tex] is the new temperature of the aluminum
So for titanium
[tex]d_t = d_2 (1 + \alpha_1 (T- T_t) )[/tex]
Where [tex]d_t[/tex] is the new diameter of titanium
[tex]T_t[/tex] is the new temperature of the aluminum
So for the aluminum rivets to fit into the holes
[tex]d_a = d_t[/tex]
=> [tex]d_1 (1 + \alpha_1 (T- T_a) ) = d_2 (1 + \alpha_2 (T- T_t) )[/tex]
Making T the subject of the formula
[tex]T = \frac{(d_1 - d_2 ) + (d_2 *\alpha_2 T_t) - d_1 \alpha_1 * T_a }{d_2 \alpha_2 - d_1 \alpha_1 }[/tex]
Substituting values
[tex]T = \frac{(0.501 - 0.500 ) + (0.500 *(8.5*10^{-6}) * 15) - 0.500* (25*10^{-6}) * 15 }{0.500 * (8.5 *10^{-6}) - 0.501 * (25 *10^{-6}) }[/tex]
[tex]T = -106 ^oC[/tex]
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.
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.
Learn more about simple harmonic motion here:
https://brainly.com/question/26114128
An astronaut visiting Jupiter's satellite Europa leaves a canister of 1.20 mol of nitrogen gas (28.0 g/mol) at 50.0 ∘C on the satellite's surface. Europa has no significant atmosphere, and the acceleration due to gravity at its surface is 1.30 m/s2. The canister springs a leak, allowing molecules to escape from a small hole. Neglect the interaction with surrounding atmosphere. (a) What is the maximum height (in km) above Europa's surface that is reached by a nitrogen molecule whose speed equals the rms speed? Assume that the molecule is shot straight up out of the hole in the canister, and ignore the variation in g with altitude. (b) The escape speed from Europa is 2025 m/s. Can any of the nitrogen molecules escape from Europa and into space?
Answer:
the answer is a
Explanation:
A convex mirror of focal length 33 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror.If the height of the image is 7.0 cm,where is the object located,and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.
Answer:
Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42
The image is a virtual and upright image.
Explanation:
The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.
The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.
Given f = -33.0cm, v = -19.0cm
using thr mirror formula to get the object distance u, we have;
[tex]\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm[/tex]
To calculate the image height, we will use the magnification formula
M = [tex]\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance} \\[/tex]
M = [tex]\frac{Hi}{HI}=\frac{v}{u}[/tex]
Given Hi = 7.0cm
v = 19.0cm
u = 44.8cm
HI = 7*44.8/19
HI = 16.5cm
The object height is 16.5cm
Magnification = v/u = 19.0/44.8 = 0.42
SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image
What must x be so that the handle end of the bat remains at rest as the bat begins to move? (Hint: Consider the motion of the center of mass and the rotation about the center of mass. Find x so that these two motions combine to give v=0 for the end of the bat just after the collision. Also, note that integration of equation ∑τ⃗ =dL⃗ dt gives ΔL=∫t1t2(∑τ)dt. )
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]
The bat length is [tex]L_b = 0.900 \ m[/tex]
The distance of the bat's center of mass to the handle end is [tex]z_c = 0.600 \ m[/tex]
The moment of inertia of the bat is [tex]I = 0.0530 \ kg \cdot m^2[/tex]
The objective of the solution is to find x which is the distance from the handle of the bat to the point where the baseball hit the bat
Generally the velocity change at the end of the bat is mathematically represented as
[tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]
Where [tex]\Delta v_c[/tex] is the velocity change at the center of the bat which is mathematically represented as
[tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]
We are told that the impulse is J so
[tex]\Delta v_c = \frac{J}{m_b }[/tex]
And [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as
[tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]
Now we have that
[tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero and the impulse will be 1
So
[tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
=> [tex]x = \frac{I}{m_b z_c} + m_b[/tex]
substituting values
[tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]
[tex]x = 0.710 \ m[/tex]
A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7 mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
L= 276.4 mm
Explanation:
Given that
E= 180 GPa
d= 3.7 mm
F= 1890 N
ΔL= 0.45 mm
We know that ,elongation due to load F in a cylindrical bar is given as follows
[tex]\Delta L =\dfrac{FL}{AE}[/tex]
[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]
Now by putting the values in the above equation we get
[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]
L=0.2764 m
L= 276.4 mm
Therefore the length of the specimen will be 276.4 mm
You could use an analytical or triple beam balance to determine a ___ called ____
A)
physical property; mass.
B)
chemical property, mass.
C)
physical property; weight.
D)
physical property; density.
Answer:
a and b are the correct answers
Explanation:
Answer:
A) physical property; mass.
Explanation:
took the test
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
water is pumped from a stream at the rate of 90kg every 30s and sprayed into a farm at a velocity of 15m/s. Calculate the power of the pump.
Answer:
340 W
Explanation:
Power = change in energy / change in time
P = ΔKE / Δt
P = ½ mv² / Δt
P = ½ (90 kg) (15 m/s)² / (30 s)
P = 337.5 W
Rounded to 2 significant figures, the power is 340 W.
To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Electric fields are MOST associated with ________.
A particle is projected at an angle 60 degrees to the horizontal with a speed of 20m/s. (i) calculate total time of flight of the particle. (i) speed of the particle at its maximum height
Answer:
Time of flight=3.5 seconds
Speed at maximum height is 0
Explanation:
Φ=60°
initial velocity=u=20m/s
Acceleration due to gravity=g=9.8 m/s^2
Total time of flight=T
Final speed=v
question 1:
T=(2 x u x sinΦ)/g
T=(2 x 20 x sin60)/9.8
T=(2 x 20 x 0.8660)/9.8
T=34.64/9.8
T=3.5 seconds
Question 2
Speed at maximum height is 0
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
