We can calculate the width/thickness of a coin using a screw gauge by placing it in between the teeth of the gauge.
How we can calculate the thickness using screw gauge?We can determine the width/thickness of a coin by using a screw gauge because screw gauge is an instrument that measures thickness of an object that are very thin. First we can placed the coin in between the teeth of the gauge and then move the gauge until the coin can be held tightly. After that note the reading on the scale.
So we can conclude that we can calculate the width/thickness of a coin using a screw gauge by placing it in between the teeth of the gauge.
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list five advantage of mecury over alcohol as a thermoelectric liquid
Explanation:
Advantages
It is a good conductor of heat and hence, can measure even high temperatures.
It gives results quickly – has a fast response time.
It expands linearly and less than alcohol and any other liquid.
It is more durable than alcohol thermometer because mercury does not evaporate easily.
Which point shows the prevailing westerlies?
1
2
3
4
Answer:
2
Explanation:edu
with detailed explaniation
#A
[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=20sin^237[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=7.2m[/tex]
#B
[tex]\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}[/tex]
[tex]\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}[/tex]
[tex]\\ \rm\Rrightarrow R=40sin74[/tex]
[tex]\\ \rm\Rrightarrow R=38.5m[/tex]
#C
[tex]\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}[/tex]
[tex]\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}[/tex]
[tex]\\ \rm\Rrightarrow T=4sin37[/tex]
[tex]\\ \rm\Rrightarrow T=2.4s[/tex]
Now
[tex]\\ \rm\Rrightarrow v=u-gt[/tex]
[tex]\\ \rm\Rrightarrow v=20-10(2.4)[/tex]
[tex]\\ \rm\Rrightarrow v=20-24[/tex]
[tex]\\ \rm\Rrightarrow v=-4m/s[/tex]
Uranium-235 Fission
Given: energy released = about 200 MeV per individual reaction (mass = 235 amu)
Part A
Find the balanced nuclear reaction.
In the balanced nuclear reaction of uranium 235, we have thorium 231, helium atom and energy released at the right of the equation
Balanced nuclear reaction of UraniumThe balanced nuclear reaction of uranium is determined as follows;
[tex]^{235}_{92}U\ --- > \ ^4_2He\ + \ ^{231}_{90}Th\ + \ energy[/tex]
Thus, in the balanced nuclear reaction of uranium 235, we have thorium 231, helium atom and energy released at the right of the equation while uranium 235 is on the left hand side of the nuclear reaction equation.
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X-ray production occurs in which process?
Answer:
Explanation:
X-rays are produced due to sudden deceleration of fast-moving electrons when they collide and interact with the target anode. In this process of deceleration, more than 99% of the electron energy is converted into heat and less than 1% of energy is converted into x-rays
an inclined plane has a velocity ratio of 2 and efficiency of 95%. it is used to raise a load of 400newtons. determine mechanical advantage? effort required?
Answer:
1.9
Explanation:
Efficiency=mechanical advantage/velocity ratio×100
95=M.A/2×100
95=50M.A
M.A=95/50=1.9
The mechanical advantage of the inclined plane is approximately 2.1053, and the effort required to raise the load of 400 newtons is approximately 189.78 newtons.
To determine the mechanical advantage and effort required for the inclined plane, we can use the formula:
Mechanical Advantage (MA) = Velocity Ratio (VR) / Efficiency
Effort Required = Load / Mechanical Advantage
Given:
Velocity Ratio (VR) = 2
Efficiency = 95% = 0.95
Load = 400 newtons
Calculate the Mechanical Advantage (MA):
MA = VR / Efficiency
MA = 2 / 0.95
MA ≈ 2.1053
Calculate the Effort Required:
Effort Required = Load / MA
Effort Required = 400 / 2.1053
Effort Required ≈ 189.78 newtons
So, the mechanical advantage of the inclined plane is approximately 2.1053, and the effort required to raise the load of 400 newtons is approximately 189.78 newtons.
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at what temperature does the fahrenheit scale read quartered the celsius scale ? what is the corresponding reading in kelvin scale ?
The temperature at which the Fahrenheit scale reads quartered the celsius scale is : -5.16 ⁰F, -20.64⁰C
Reading in kelvin scale = 252.5 K
Determine the temperature at which the Fahrenheit scale reads quartered the celsius scaleThe relationship between Fahrenheit and celsius can be expressed as
F = (9/5)(C) + 32 ---- ( 1 )
Therefore the temperature at which Fahrenheit scale quarters the celsius scale is :
F = 1/4 C --- ( 2 )
Back to equation ( 1 )
1/4 C = (9/5)(C) + 32
1/4C - 9/5C = 32
C ( 1/4 - 9/5 ) = 32
C = 32 / ( 1/4 - 9/5 )
= -20.64⁰C
F = 1/4 ( -20.64 ) = -5.16 ⁰F
Hence we can conclude that The temperature at which the Fahrenheit scale reads quartered the celsius scale is : -5.16 ⁰F, -20.64⁰C. Reading in kelvin scale = 252.5 K
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while standing at the edge of a cliff 45m high, you drop a ball. When the ball has fallen 5m, you throw a second ball straight down. What initial speed must you give the second ball if they are both to reach the ground at the same time?
The initial speed of the second ball if they are both reach the ground at the same time is 19.8 m/s.
Speed of the balls
The initial speed of the second ball is calculated as follows;
v = √2gh
v = √(2 x 9.8 x 45)
v = 29.7 m/s
Time for the ball to travel 5 meterst = √(2h/g)
t = √(2 x 5/9.8)
t = 1.01 s
Initial speed of the first ballvf = vi + gt
29.7 = vi + 9.8(1.01)
29.7 = vi + 9.9
vi = 19.8 m/s
Thus, the initial speed of the second ball if they are both reach the ground at the same time is 19.8 m/s.
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What happens to an object that is moving at 10 m/s if only balanced forces act on it?
Heelp me! and please explain...
If only balanced forces act on an object it will keep moving in the same speed and direction.
What is Vector?
This is defined as the type of quantity which has both magnitude and direction.
\Having balanced forces will bring about equilibrium which is why the obect will move in the same speed and direction.
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(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of the forces to assist you with the calculation.)
(8)Figure 5 shows three charges arranged in a right angled formation.
(8.1)Draw a free body diagram of the forces that act on the -0,03 uC charge.
(8.2)Calculate each force that acts on the -0,03 uC charge.
(8.3) Find the magnitude and direction of the net force that acts on the 0,03 μC charge with the aid of a diagram and by calculations.
Help Please.
Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is
[tex]F = \dfrac{kq_1q_2}{r^2}[/tex]
where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.
8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.
8.2. First convert everything to base SI units:
0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C
0,03 µC = 3 × 10⁻⁸ C
0,04 µC = 4 × 10⁻⁸ C
300 mm = 300 × 10⁻³ m = 0,3 m
600 mm = 0,6 m
Force due to Q₁ :
[tex]F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}[/tex]
Force due to Q₃ :
[tex]F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}[/tex]
8.3. The net force on the particle at Q₂ is the vector
[tex]\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}[/tex]
Its magnitude is
[tex]\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}[/tex]
and makes an angle θ with the positive horizontal axis (pointing to the right) such that
[tex]\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}[/tex]
where we subtract 180° because [tex]\vec F[/tex] terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.
Rajani had bought a new bottle of pickle from the market. She tried to open the
metal cap but could not do so. She then took a bowl of hot water and immersed
the upper end of the bottle in it for five minutes. She could easily open the
bottle now. Why?
BEST ANSWER WILL BE MARKED
Answer:
The pickle bottle cap on dipping it in hot water expanded.
Explanation: Hope it helps you:))))
Have a good day
When you throw a ball into the air, its kinetic energy _
Answer:
increases
Explanation:
The ball's kinetic energy is the energy it has due to its motion. When you throw the ball into the air, it gains kinetic energy from the force of your throw.
Look at this picture of people standing in a line. What could they do to create a model of a longitudinal wave?
Two identical masses are attached to frictionless pulleys by very light strings wrapped around the rim of the pulley and are released from rest. Both pulleys have the same mass and same diameter, but one is solid and the other is a hoop As the masses fall in which case is the tension in the string greateror is it the same in both cases? Justify your answer
The tension in the string is same in both cases because of the similar mass or weight of the bodies.
Is tension be the same in both cases?Both pulleys have the same mass and same diameter, but one is solid and the other is a hoop, then there is same tension occurs in both cases because tension depends on mass which is similar on both side. We know that tension is the opposite force of the weight.
So we can conclude that the tension in the string is same in both cases because of the similar mass or weight of the bodies.
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with full explaniation
A. The ball's (vertical) velocity [tex]v[/tex] at time [tex]t[/tex] is
[tex]v(t) = 30\dfrac{\rm m}{\rm s} - gt[/tex]
so that after 4 seconds, the ball's speed is
[tex]|v(4\,\mathrm s)| = \left|30\dfrac{\rm m}{\rm s} - \left(10\dfrac{\rm m}{\mathrm s^2}\right) (4\,\mathrm s)\right| = \boxed{10\frac{\rm m}{\rm s}}[/tex]
(The velocity is -10 m/s, so the ball is falling back down at this point.)
B. At maximum height, the ball has zero velocity, so it takes
[tex]30\dfrac{\rm m}{\rm s} - gt = 0 \implies t = \dfrac{30\frac{\rm m}{\rm s}}g = \boxed{3\,\mathrm s}[/tex]
for the ball to reach this height.
C. The height of the ball [tex]y[/tex] at time [tex]t[/tex] is
[tex]y(t) = \left(30\dfrac{\rm m}{\rm s}\right) t - \dfrac g2 t^2[/tex]
The maximum height is attained by the ball at 3 seconds after it's thrown, so
[tex]y_{\rm max} = \left(30\dfrac{\rm m}{\rm s}\right) (3\,\mathrm s) - \dfrac{10\frac{\rm m}{\mathrm s^2}}2 (3\,\mathrm s)^2 = \boxed{45\,\mathrm m}[/tex]
D. The time it takes for the ball to reach its maximum height is half the time it spends in the air. So the total airtime is [tex]\boxed{6\,\mathrm s}[/tex].
Put another way: When the ball returns to the height from which it was thrown, its final velocity has the same magnitude as its initial velocity but points in the opposite direction. This is to say, after the total time the ball is in the air, it's final velocity will be -30 m/s. Then the total airtime is
[tex]30\dfrac{\rm m}{\rm s} - gt = -30\dfrac{\rm m}{\rm s} \implies t = \dfrac{60\frac{\rm m}{\rm s}}g = \boxed{6\,\mathrm s}[/tex]
Put yet another way: Solve [tex]y(t) = 0[/tex] for [tex]t[/tex]. I don't see a need to elaborate...
Assume that the thickness of the tissue between the interior and the exterior of the body is =3 cm and that the average area through which conduction can occur is =1.5 m2 . Find heat transferred by conduction per hour if temperature difference between the inner body and the skin is ∆ = 2 0C. Coefficient of thermal for tissue without blood is = 18 × 10−2 Cal /m-hr- 0C.
The heat transferred by conduction per hour if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.
What is heat conduction?Heat conduction is the method of transfer of heat by the direct contact of the two or more solid bodies.
Given is the thickness of tissue x = 3cm =-0.03m, Area of conduction A = 1.5 m², Change in temperature of skin ΔT =2°C, and coefficient of thermal for tissue without blood is k = 18 × 10⁻² Cal /m-hr- °C
The conduction heat rate in hr is
Q =kAΔT /x
Q =18 × 10⁻² x 1.5 x 2 /0.03
Q =18 Joule / hr
Thus, the heat transferred by conduction per hour, if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.
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A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
The coefficient of friction between the road and the car's tire is determined as 0.78.
Acceleration of the carThe acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
u is the initial velocity = 97 km/h = 26.94 m/sa = (26.94)²/(2 x 47)
a = 7.72 m/s²
Coefficient of frictionμ = a/g
μ = (7.72)/9.8
μ = 0.78
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consider the three vectors, A=3i+4j, B=2i-2j+4k, and C=-i+5j-3k. Show that the sum of the three vectors can alternatively be computed by first summing A and B and then summing the resultant with C or by first summing B and C and then summing the resultant with A.
Answer:
A+B+C=7j+k
Explanation:
A=3i+4j
B=2i-2j+4k
C=i+5j-3k
A=3i+4j
+
B=2i-2j+4k
i+2j+4k
A+B=i+2j+4k
+
C=-i+5j-3k
7j+k
A+B+C=7j+k
What is the equivalent resistance of a circuit that contains two 50.00
resistors connected in series with a 12.0 V battery?
Ο A. 0.120 Ω
Ο Β. 100.0 Ω
OC. 0.480
OD. 25.00
Answer:
B
Explanation:
Resistors in series add because there is only one path the current can take.
calculations
= R1 + R2
= 50 + 50
= 100
A 1.0KW kettle contains 500g of boiling water. Calculate the time needed to evaporate all the water in the kettle. (Specific latent heat of vaporization of water=2.26MJKG^-1
The time needed to evaporate all the water in the kettle is 1.13 x 10³s.
What is Power?The power is defined as the the ratio of amount of energy produced per unit time.
P =E/t
Energy is in the form of heat gained to evaporate all water in kettle.
E = Q =mL
So, P = mL/t
Substitute mass of water m = 0.5kg, latent heat of vaporization of wate L = 2.26 x 10⁶ J/kg, power P= 10³ W, we have
10³ = (0.5 x 2.26 x 10⁶) /t
t = 1.13 x 10³s
Thus, the time needed to evaporate all the water is 1.13 x 10³s.
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A bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s. What is the angular acceleration of the tires?
The angular acceleration of a bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s will be 2. 57 rad/s²
What is angular acceleration?
Angular acceleration can be defined as the time it takes for a change in angular velocity. It is denoted as 'α' with a unit of rads/s²
It is expressed thus;
α= Δω ÷ Δt
Where α = angular acceleration
Δw = change in angular velocity = velocity ÷ radius
Δ t = change in time
How to calculate the angular accelerationUsing the formula:
α= Δω ÷ Δt
v = 5. 33m/s, r = 0. 33m and t = 6.27s
Substitute the values to get the angular velocity
Δw = v÷ r = 5. 33 ÷ 0.330 = 16. 15 m/s
Substitute the value of Δw into the equation
α= Δω ÷ Δt = 16. 15 ÷ 6. 27 = 2. 57 rad/s²
Therefore, the angular acceleration of a bike with tires radius of 0. 330m, speed of 5. 33mls in 6. 27s is 2. 57 rad/s²
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You are on a boat in the middle of the Pacific Ocean at the equator traveling in a hydrofoil
going at a constant speed of 300 /. The water is perfectly still. What is your acceleration:
a) If you’re heading due North?
b) If you’re heading due East?
c) If you’re heading straight up (something probably went wrong at this point).
You may assume the following:
The earth has a radius of 6371 km.
The earth makes one full revolution every 24 hours.
The gravitational constant at sea level is 9.81 m/s2
East and North are relative to the Earth’s axial north, not magnetic north.
a) If you’re heading due North, total acceleration will be 0.0478 m/s².
b) If you’re heading due East, acceleration will be 0.0803 m/s².
c) If you’re heading straight up, no exact value for acceleration.
What is acceleration?The time rate of change of velocity is known as acceleration.
a)If the boat is going along North from equator, there will be only centrifugal force acting along upward direction which will be balanced by the weight of boat.
Angular velocity ω= (2π / (3600 x 24)) = 7.27 x 10⁻⁵ rad/s
Acceleration due to centrifugal force a = 0.0337m/s²
Centrifugal force due to its own velocity 300 m/s is v²/r = 0.0141 m/s².
Total acceleration will be 0.0141 +0.0337 =0.0478m/s².
Thus, if you’re heading due North, total acceleration will be 0.0478 m/s².
b) If the boat goes along East, Coriolis acceleration is in the direction of centrifugal force.
Coriolis acceleration is (2 x 7.27 x 10⁻⁵ x300 ) = 0.0436m/s².
Total centrifugal acceleration due to centrifugal force = 300 + (7.27 x 10⁻⁵ x 6371 x 10³) = 0.0436 m/s².
Total acceleration is 0.0436 + 0.0367 = 0.0803m/s².
Thus, if you’re heading due East, acceleration will be 0.0803 m/s².
c) When boat is going upward, centrifugal force will change with time. but Coriolis acceleration remains same.
Thus, if you’re heading straight up, no exact value for acceleration.
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The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^24 kg. What is the gravitational force between these two objects?
Answer:
3 820 885 N
Explanation:
Gravitational equation
F = G m1 m2 / r^2
G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2
F = 6.6713 x 10^-11 * 4.41 x 10^5 * 5.97 x 10^24 / ( 6.78x 10^6)^2
= 3820885 .3 N
You decide to test both appraoches. First, you attach a parachute to a
0.07 kg egg and toss if from a window 30 m above the ground. The egg
reaches and maintains a falling speed of 0.5 m/sec. Upon reaching the
ground, the egg rapidly decelerates to 0 m/sec over the course of only
0.01 seconds. What force (expressed in N) did the egg experience upon
hitting the ground?
The force (expressed in N), the egg experience upon hitting the ground is 3.5N.
What is Net force?When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.
Given is an 0.07 kg egg and toss if from a window 30 m above the ground. The egg reaches and maintains a falling speed of 0.5 m/sec. Upon reaching the ground, the egg rapidly decelerates to 0 m/sec over the course of only 0.01 seconds.
Average force = Mass x accelerationF = m x (Vf -Vi)/t
F = 0.07 x (0 - 0.5)/0.01
F = -3.5N
Thus, the magnitude of force on egg upon hitting the ground is 3.5N.
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a pool ball is rolling along a table3 with a constant velocity. the components of its velocity vector are Vx=0.5 m/s and Vy=0.8 m/s. Calculate the distance it travels in 0.4 s
The distance traveled by the pool ball is 0.376 m.
To calculate the distance traveled by the ball, we need to first find the resultant velocity of the ball.
What is resultant velocity?This is the single velocity obtained when two or more velocities are combined.
To calculate the resultant velocity, we use the formula below.
Formula:
Vr = √(Vx²+Vy²)............. Equation 1Where:
Vr = Resultant velocity.From the question,
Given:
Vx = 0.5 m/sVy = 0.8 m/sSubstitute these values into equation 1
Vr = √(0.5²+0.8²)Vr = √(0.25+0.64)Vr = √(0.89)Vr = 0.94 m/sFinally, to calculate the total distance, we use the formula below.
Formula:
d = (Vr)t.......... Equation 2Where:
d = Distance traveled by the ballt = timeFrom the question,
Given:
Vr = 0.94 m/st = 0.4 sSubstitute into equation 2
d = 0.4×0.94d = 0.376 mHence, the distance traveled by the pool ball is 0.376 m.
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do most of the physical digestion take place in the top half of the digestive tract (mouth esophagus,and stomach ) or the bottom half (the intestine)?
1. Identify Give three examples of kinetic energy and three examples of potential energy .
Answer:
Examples of Gravitational Potential Energy
A raised weight.
Water that is behind a dam.
A car that is parked at the top of a hill.
A yoyo before it is released.
River water at the top of a waterfall.
A book on a table before it falls.
A child at the top of a slide.
Ripe fruit before it falls.
Explanation:
Any object in motion is using kinetic energy: a person walking, a thrown baseball, a crumb falling from a table, and a charged particle in an electric field are all examples of kinetic energy at work.
edward travels 150 kilometers due west and then 200 kilometers in a direction 60 degrees north of west.what is his displacement in the westerly direction?
The displacement of Edward in the westerly direction is determined as 180.27 km.
Displacement of EdwardThe displacement of Edward is calculated as follows;
R² = a² + b² - 2abcosθ
R² = 150² + 200² - 2(150 x 200) x cos60
R² = 32500
R = 180.27 km
Thus, the displacement of Edward in the westerly direction is determined as 180.27 km.
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Einstein published three major scientific papers. One of these put forward a new way of calculating the size of molecules. Another explained Brownian motion — the random dance performed by specks of dust trapped in a fluid. Einstein suggested that the tiny particles making up the fluid — its atoms or molecules — were bouncing against the specks of dust and causing the motion. These papers helped to establish the reality of atoms and molecules. Another of Einstein’s 1905 papers explained the photoelectric effect — the way that metals could emit electrons (tiny charged particles) from their surface when light was shone on them. Most scientists believed that light travelled in waves — like sound or water waves. But Einstein suggested that the photoelectric effect could be explained if light could also behave as a stream of tiny packets of energy.
(2) Einstein’s paper on the photoelectric effect helped give birth to quantum theory, and it was for this paper that Einstein received the Nobel Prize in 1922. Quantum theory led, in the 1920s and 1930s, to another revolution in physics. Physicists showed that, as well as light waves behaving like particles, particles could sometimes act as waves. This theory established “objective probability” in physics. This was the idea that completely unpredictable chance events can take place at the subatomic level. Einstein never fully accepted the prevalent interpretation of quantum theory. But, while many of these interpretations involve wild metaphysical flights of fancy, the physical results are, like those of relativity, very well established.
(3) Marrying together the two pillars of 20th century physics — relativity and quantum theory — is a central problem for physics even today. Successfully doing this may require a revolution in science similar to those begun by Newton and Einstein. There are three main interconnected driving forces for such changes in science. The first is the development of technology. Changes in technology can make new experiments possible and they also influence the problems that scientists develop an interest in. Newton was fascinated by the new machines of the 17th century. Similarly, Einstein was fascinated by electricity and magnetism. This influence also works in a negative way. The governments and multinationals that control technology are often able to dictate what is researched.
(4) The second factor driving scientific progress is the way that the dominant ideas in society change. Ideas from the broader culture can impinge upon science. Newton’s ideas were part of a revolutionary new attempt at a rational explanation of both nature and society. On the other hand, the dominant ideas in society can also limit the development of science. This is most obvious in the social sciences, where delving too deeply into how society is organized might raise difficult questions for our rulers. Less is at stake in the natural sciences. Indeed, improvements in natural sciences are vital to our rulers if they want to compete effectively with each other. But the distorted worldview of capitalism still impacts on science. Extremely narrow and specialized bodies of knowledge develop—creating problems for scientists trying to bring about the kind of sweeping revolution heralded by Einstein.
(5) Finally, science moves forward because scientists seek to develop logically consistent theories. This can push them beyond the dominant or common sense ideas of their time. Einstein’s breakthrough cannot be reduced simply to changes in technology or wider cultural and ideological shifts. Science is not simply the gathering and ordering of data about the outside world. It also requires abstraction—developing theories about the underlying laws of nature that are usually not immediately apparent. This crucial role of theory is not just a feature of the natural sciences.
(6) Einstein argued that “common sense is the prejudices acquired by age 18”. Marxist theory, which is a social science topic, challenges “common sense” political ideas. If we, according to this theory, want to change the world, we need to combine our actions with theory that digs below the surface appearance of society to understand how the system works.
What is the main issue being discussed?
The main issue that is being discussed in the manuscript is the development of science.
What factors affect development of science?Capitalism impacts on science because extremely narrow and specialized bodies of knowledge creating problems for scientists to bring revolution. The dominant ideas in the society can also limit the development of science.
So we can conclude that the main issue that is being discussed in the manuscript is the development of science.
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I need help with this question (open-ended question) 50 points (NO BOTS!!!)
Choose TWO locations below (A, B, C) and provide a two-sentence weather forecast for that location. Your forecast must include Wind direction, general wind speed (high, moderate, low) general temperatures (cold, cool, mild, warm, hot), and sky conditions(precipitation/stormy, sunny cloudy).
The weather forecast for the locations includes:
London - 21°C°, Precipitation: 60%, Humidity: 79%, Wind: 11 km/h.
New York - 26°C°, Precipitation: 50%, Humidity: 76%, Wind: 19 km/h.
What is weather?It should be noted that weather is the state of the atmosphere with respect to heat or cold, calm or storm, clearness or cloudiness.
Weather forecast is an analysis of the state of the weather in an area with an assessment of the likely developments.
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