Answer:
The correct option is a , c
Explanation:
Generally the fringe spacing is mathematically defined as
[tex]y = \frac{m \lambda D}{d}[/tex]
Where y is the fringe spacing
m is the order of the fringe
[tex]\lambda[/tex] is the wavelength
D is the distance between the slit and the screen
d is the distance between the slit
Now in order to increase the fringe spacing can do the following
Increase the wavelength increase the distance from the slit to the screen Decrease the distance between the slitThe following action would increase the fringe spacing because the from the question these parameters are directly proportional to the fringe spacing
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a great amount of angular momentum when rotating. A wind turbine has a total of 3 blades. Each blade has a mass of m = 5500 kg distributed uniformly along its length and extends a distance r = 46 m from the center of rotation. The turbine rotates with a frequency of f = 11 rpm.
Required:
a. Calculate the total moment of inertia of the wind turbine about its axis, in units of kilogram meters squared.
b. Calculate the angular momentum of the wind turbine, in units of kilogram meters squared per second.
Answer:
Explanation:
moment of inertia of each blade which is similar to rod rotating about its one end
= 1/3 ml²
moment of inertia of 3 blades = ml²
= 5500 x 46²
I = 11638 x 10³ kg m²
angular velocity = 2πn where n is rotation per second
n = 11 / 60
angular velocity = 2π x 11/60
= 1.1513 rad /s
angular momentum
= moment of inertia x angular velocity
= 11638 x 10³ x 1.1513
= 13399 x 10³ kg m² per second.
The shaft of a motor has an angular displacement θ that is a function of time given by the equation: θ(t) = 4.40 t 3 rad/s3 + 1.40 t2 rad/s2 . At time t = 0.00 s the wheel is at rest and is oriented at θ = 0.00 rad. a) Derive the equation that specifies the angular velocity of the shaft as a function of time. b) Derive the equation that specifies the angular acceleration as a function of time.
Answer:
a) [tex]\omega = 13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]
b) [tex]\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]
Explanation:
You have that the angular displacement is given by:
[tex]\theta=4.40t^3\frac{rad}{s^3}+1.40t^2\frac{rad}{s^2}[/tex]
a) the angular velocity is given by the derivative in time, of the angular displacement, that is:
[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[4.40 t^3 rad/s^3 + 1.40 t^2 rad/s^2]\\\\\omega=\frac{d\theta}{dt}=13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]
b) the angular acceleration is the derivative, in time, of the angular velocity:
[tex]\alpha=\frac{d\omega}{dt}=\frac{d}{dt}[13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}]\\\\\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Answer:
Force applied to stop the car = 1,250 N
Explanation:
Given:
Mass of car (M) = 1,000 kg
Initial velocity (U) = 20 m/s
Final velocity (V) = 0 m/s
Distance (S) = 160 m
Find:
Force applied to stop the car.
Computation:
[tex]v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N[/tex]
Force applied to stop the car = 1,250 N
Vocabulary Matching
The specialized equipment used to conduct research and repair
damaged equipment
Instruments
Space Station
Space Suit
Accomodations
Answer:
instruments
Explanation:
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 .
How the musculoskeletal and nervous system develop as a human grows
Answer:
Explanation:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
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)
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 solid cylinder of mass m and radius R rolls down a ramp, starting from rest at a height h above a nearby horizontal surface. The coefficients of kinetic and static friction and are non-zero, and sufficiently large that the cylinder rolls down the ramp without slipping. Assume that the coefficient of rolling friction is zero. As the cylinder leaves the ramp, it continues along a horizontal surface (with the same frictional coefficients as the ramp).
Required:
What is the speed V of the cylinder after it has traveled a distance D along the horizontal surface?
Answer:
the volocity is 50
Explanation:
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]
why can you see the path of light in a sunbeam?
Answer:
Sunbeams are seen because of light separated from water droplets and dust and smoke particles suspended in the air. If the cloud cover only has a few small holes in it, then separate rays of light will sprinkle light in every direction so you can see sunbeams.
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
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.
5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3 . The pressure is maintained at atmospheric pressure of 1.01 x 105 Pa. A moveable piston of negligible weight rests on the surface of the water. The water is then converted to steam by adding an additional amount of heat to the system. When all of the water is converted, the final volume of the steam is 8.50 m3 . The latent heat of vaporization of water is 2.26 x 106 J/kg. Calculate how much work is done and the change in the internal energy during this isothermal process.
Answer:
1.04 x 107 J.
Explanation:
We can use the following method to do the calculation
Total energy given to water to convert intosteam
dQ = m* l
dQ = 5.00* 2.26 * 106
= 1.13* 107 J
Work done at constantpressure dW = P* dV
Initialvolume V1 = 5.00kg / 958
= 5.22* 10-3 m3
Finalvolume = 8.50 m3
=> dW = 1.01* 105 * ( 8.50 - 5.22 * 10-3)
= 8.58* 105 J
First law of thermodynamicsis dQ = ΔU + dW
Change in internalenergy ΔU = 1.13* 107 - 8.58 *105
= 1.04 x 107 J as our answer
How is the particle displacement related to the direction of wave movement in a longitude wave?
Answer:
The displacement of particles is perpendicular to the direction of wave motion.
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 .
Einstein developed much of his understanding of relativity through the use of gedanken, or thought, experiments. In a gedanken experiment, Einstein would imagine an experiment that could not be performed because of technological limitations, and so he would perform the experiment in his head. By analyzing the results of these experiments, he was led to a deeper understanding of his theory. In each the following gedanken experiments, Albert is in the exact center of a glass-sided freight car speeding to the right at a very high speed vvv relative to you. Albert has a flashlight in each hand and directs them at the front and rear ends of the freight car. Albert switches the flashlights on at the same time.
In Albert's frame of reference, which beam of light travels at a greater speed, the one directed toward the front or the one toward the rear of the train, or do they travel at the same speed? Which beam travels faster in your frame of reference? Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light directed toward the front of the train travels at a greater speed and in your frame of reference the two beams travel at the same speed, then enter front,same.
Answer:
For eintein's frame of reference, both beam travel at the same speed.
For my own frame of reference, both beams travel at the same speed.
Explanation:
According to special relativity, the speed of light is the same in all direction on all reference frame. If not for this law we will assume the from beam will have a relative speed that will be the speed of light plus the speed of the fright car. This is not so and it violates the speed limit of light which according to the first law is the highest speed possible and nothing can go beyond that.
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.
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
10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave pulse moves from one end of the rope to another and find that it is 36 m/s. What frequency must they vibrate the rope at to create the second harmonic
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
f₂ = 3 Hz
does work done by the electric force depends on the path taken.
Answer:
yes but not sure
Explanation:
i don't know so much but I just guessed I'm not sure
A rectangular painting measures 1.0 m tall along the y' axis and 3.0 m wide along the
x' axis. The painting is hung on the side wall of a spaceship which is moving passed
the Earth at a speed of 0.9c. Assume that the spaceship is moving along the (x, x')
direction.
a) What are the dimensions of the picture according to the captain of the
spaceship?
b) What are the dimensions of the picture as seen by an observer on the Earth?
Answer:
a) 1 m tall, 3 m wide
b) 1 m tall, 1.31 m wide
Explanation:
According to the captain of the spaceship, the dimensions of the picture is the same i.e 1.0 m tall along the y' axis and 3.0 m wide along the x' axis.
b) The dimensions of the picture as seen by an observer on the Earth along the y axis will remain the same, 1.0 m tall, for the direction of the y axis is perpendicular to the spaceship movement.
The dimensions of the picture as seen by an observer on the Earth along the x axis will reduce if we are to go by the Lorentz contraction:
L(x) = L(x)' * √[1 - (v²/c²)]
where
L(x)' = the dimensions of the picture along the x axis on the spaceship,
v² = the speed of the spaceship and c² = the speed of light in the vacuum.
On substituting, we have
L(x) = 3 * √[1 - (0.81c²/c²)]
L(x) = 1.31 m
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
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 is the gravitational force related to the distance between two objects?
Answer:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.
answer: It is inversely proportional to the square of the distance.
Explanation:
for the team
What is an open circuit
Answer:An electrical circuit that is not complete.
Explanation:
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
g science is strictly limited to the study of natural phenomena (things that result as the outcome of natural laws like the speed of light. What is an example of a question that scientific studies cannot address? Question 3 options: 1) What is the purpose of life? 2) Where did an important battle take place? 3) What is the mean flight speed velocity of a sparrow? 4) How much energy is stored in a particular kind of covalent
Answer:
1) What is the purpose of life
Explanation:
This is an age long question that arises out of human curiosity about the beginning, existence and subsequently what happens to life after its gone. There exist no natural laws or methods currently that addresses this question.
A bike travels 15.0 km in 45.0 min. Its average speed in km/h is .
The average speed of a bus traveling a distance of 15.0 km in 45.0 min is 20 km/hour.
What is speed?The speed of an object, also known as v in kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time, making it a scalar quantity.
The distance travelled by an object in a time interval is divided by the length of the interval to determine its average speed.
Distance travelled by the bike = 15.0 km
Time taken by the object = 45.0 minute = (45.0 ÷ 60) hour = 0.75 hour
Hence, the average speed of the object = distance travelled / time taken
= 15.0 km/0.75 hour
= 20 km/hour,
Therefore, the average speed of a bus traveling a distance of15.0 km in 45.0 min is 20 km/hour.
Learn more about speed here:
https://brainly.com/question/28224010
#SPJ2
Part F A system experiences a change in internal energy of 14 kJkJ in a process that involves a transfer of 36 kJkJ of heat into the system. Simultaneously, which of the following is true? A system experiences a change in internal energy of 14 in a process that involves a transfer of 36 of heat into the system. Simultaneously, which of the following is true? 22 kJkJ of work is done by the system. 22 kJkJ of work is done on the system. 50 kJkJ of work is done by the system. 50 kJkJ of work is done on the system
Answer:
Explanation:
According to first law of thermodynamics :
Q = ΔE + W
Q is heat added , ΔE is increase in the internal energy of the system and W is work done by the system .
Here Q = 36 KJ
ΔE = 14 kJ
Putting the values in the equation
36 = 14 + W
W = 36 - 14
= 22 kJ .
Work done by gas or system = 22 kJ.
