Answer:
As for the definitions, the tendency of two or more different molecules to bond with each other is known as Adhesion, whereas the force of attraction between the same molecules is known as Cohesion.
hopefully this helps
Answer:
Adhesion is the force of attraction between molecules of different substances while cohesion is the force of attraction between molecules of same substances.
A truck contains both hydraulic brakes (reliability 0.96) and mechanical brakes (reliability 0.99). What is the probability of stopping the truck (i) while moving at a high speed, assuming both systems must work to avoid hitting a vehicle stalled in the roadway, and (ii) while moving at a low speed, assuming either system will stop the truck in time if they function properly
Answer:
i) 0.9504
ii) 0.0452
Explanation:
Given data: reliability of hydraulic brakes= 0.96
reliability of mechanical brakes = 0.99
So the probability of stopping the truck = 0.96×0.99= 0.9504
At low speed
case: A works and B does not
= 0.96×(1-0.99) = 0.0096
case2 : B works and A does not
= 0.99×(1-0.96) = 0.0396
Therefore, probality of stopping = 0.0096+0.0396 = 0.0492
The diagram shows two
charged objects, X and Y.
Based on the field lines, what are the charges of the
objects?
X: positive
Y: negative
O X: negative
Y: positive
O x: negative
Y: negative
O x: positive
Y: positive
Answer:
X: positive
Y: positive
Explanation:
The electrical field lines have a specified direction based upon the sign of the charge. The electrical field lines have a direction coming out of the charge if the sign of the charge is positive. While the electrical field lines have a direction going inside the charge if the sign of the charge of negative.
It can be clearly seen in the diagram that the electric field lines are coming out of both charges. Therefore, both the charges are positive.
Hence, the correct option is:
X: positive
Y: positive
Will award brainliest and upvote if you can help answer!
Answer:
Omqnp
Explanation:
I have done this before and got it correct
Photons with the highest frequency have the lowest _____.
A. wavelength
B. speed
C. color
D. energy
Answer:
a wavelength
Explanation:
In A. C. motor capacitor is used
p.Can dicouly [BPKIHS 2000
to reduce ripples b. to decrease A. C.
c. to increase A. C. d. to decrease D. C.
a.
34
Answer:
d. to decrease D. C.
Explanation:
d. to decrease D. C.
Answer the following question according to formal grammatical rules.
Which of the following sentences corrects the pronoun errors in this sentence?
If a person wants to help their community, they should research opportunities.
A. If a person wants to help their community, you should research opportunities.
B. If you want to help your community, a person should research opportunities.
C. If people want to help their community, they should research opportunities.
D. If a person wants to help it, they should research opportunities.
If people want to help their community, they should research opportunities corrects the pronoun errors in this sentence.
What is a Pronoun?This is defined as a word which is used instead of a noun and examples include he, she etc.
In this scenario, the word 'people' agrees with the possessive adjective 'their' and the personal pronoun 'they', which is referred to the third-person plural hence the reason why option C is the most appropriate choice.
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Answer:
D
Explanation:
D
Sodium and ice a physical or chemical reaction
Answer:
physical change uwu owo hope this helps you uwu silly goose
Two identical thermally insulated spherical tanks, A and B, are connected by a valve. Initially tank A contains 20mol of an ideal diatomic gas, tank B is evacuated, and the valve is closed.
If the valve is opened and the gas expands isothermally fromA to B, what is the change in the entropy of the gas?
Answer:
Explanation:
Volume of gas is doubled isothermally . Let the temperature of the gas be T .Work done by the gas
W = 2.303 nRT log 2 V / V
Since expansion is isothermal , Δ E = 0
Q = ΔE + W
Q = W
Q = 2.303 nRT log 2 V / V
Q / T = 2.303 n R log 2
Change in entropy = ΔS = Q / T = 2.303 n R log 2
= 2.303 x 20 x 8.3 x .3010 J T⁻¹.
= 115 J T⁻¹.
The required change in the entropy of the gas is 115 J/K.
Given data:
The number of moles of diatomic gas in tank A is, n = 20 mol.
The expansion takes place isothermally, which means the volume is doubled at the constant temperature. Therefore, work done during the isothermal expansion is,
W = 2.303 nRT log 2 V / V
Since expansion is isothermal , Δ E = 0. Then by using the first law of thermodynamics as,
Q = ΔE + W
Q = W
Q = 2.303 nRT log 2 V / V
Q / T = 2.303 n R log 2
Also, we that the expression for the change in Entropy is,
ΔS = Q / T = 2.303 n R log 2
Solving as,
ΔS = 2.303 x 20 x 8.3 x .3010
ΔS = 115 J/K
Thus, we can conclude that the required change in the entropy of the gas is 115 J/K.
Learn more about the entropy change here:
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if the reading of a mercury barometer is 75.58cm at the base of a mountain and 66.37cm at the summit,what is the height of the mountain? (density of mercury= 13600kg/m³, average density of air=1.25kg/m³
Answer:
Explanation:
What are the benefits of 3D design and 3D printing software?
The height of the mountain is 1002.048 m.
What is pressure?The perpendicular force per unit area, or the stress at a point within a confined fluid, is referred to as pressure in the physical sciences.
Atmospheric pressure is the force of the atmosphere pressing down on each square meter of Earth's surface. Pascals are used to express pressure in SI units; one pascal is equivalent to one newton per square meter.
Given parameters:
at the base of a mountain, the reading of a mercury barometer is = 75.58cm = 0.7558 m.
at the summit of a mountain, the reading of a mercury barometer is = 66.37cm = 0.6637 m.
Let the height of the mountain = h.
Then,
h × density of air × g = ( 0.7558 - 0.6637) m × density of mercury × g
⇒ h × 1.25kg/m³ = 0.0921 m × 13600kg/m³
⇒ h = 1002.048 m.
Hence, the height of the mountain is 1002.048 m.
Learn more about pressure here:
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A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded:
Time since brakes applied (sec) 0 2 4 6
Velocity (ft/s) 87 44 16 0
Give lower and upper estimates (using all of the available data) for the distance the car traveled after the brakes were applied.
Answer:
Explanation:
deceleration during first 2 s = (87 - 44) / 2 =21.5 m /s²
deceleration during next 2 s = (44 - 16) / 2 =14 m /s²
deceleration during last 2 s = (16 - 0) / 2 = 8 m /s²
If we assume the acceleration to be uniform during 3 readings
distance travelled during first 2 s
v² - u² = 2 a s
87² - 44² = 2 x 21.5 x s
s = 7569 - 1936 / 43
= 176 m
distance travelled during next 2 s
44² - 16² = 2 x 14 x s
s = 1936 - 256 / 28
= 60 m
distance travelled during last 2 s
16² - 0² = 2 x 8 x s
s = 256 / 16
= 16 m
Total distance travelled = 176 + 60 + 16 = 252 m
If we assume the acceleration as non- uniform , we first calculate average acceleration as follows .
average acceleration = (87 - 0)/ 6 = 14.5 m /s²
v² - u² = 2 a s
87² - 0² = 2 x 14.5 x s
s = 7569 / 29
= 261 m
Hence lower estimate of distance travelled = 252 m ; upper estimate of distance travelled = 261 m .
How much stronger is the electric force between two
protons than the gravitational force between them?
Answer:
Explanation:
The electric force between two protons is 10^36 times stronger than the gravitational force between them.
Answer:
Electric force is 1.23 x 10³⁶ times stronger than gravitational force.
Explanation:
First, we will calculate the gravitational force using Newton's Law:
[tex]F_G = \frac{Gm^2}{r^2}\\\\[/tex]
where,
F_G = Gravitational Force = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²
m = mass of proton = 1.67 x 10⁻²⁷ kg
r = distance between protons
Therefore,
[tex]F_G = \frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.67\ x\ 10^{-27}\ kg)^2}{r^2}\\\\F_G = \frac{1.86\ x\ 10^{-64}\ Nm^2}{r^2}\\\\[/tex]
Now, we will calculate the electrostatic force using Colomb's Law:
[tex]F_E = \frac{kq^2}{r^2}\\\\[/tex]
where,
F_E = Electrostatic Force = ?
k = Colomb's Constant = 9 x 10⁹ Nm²/C²
q = charge of proton = 1.6 x 10⁻¹⁹ C
r = distance between protons
Therefore,
[tex]F_E = \frac{(9\ x\ 10^{9}\ Nm^2/kg^2)(1.6\ x\ 10^{-19}\ C)^2}{r^2}\\\\F_E = \frac{2.3\ x\ 10^{-28}\ Nm^2}{r^2}\\\\[/tex]
Dividing both forces:
[tex]\frac{F_E}{F_G} = \frac{2.3\ x\ 10^{-28}}{1.86\ x\ 10^{-64}}[/tex]
F_E = 1.23 x 10³⁶ F_G
Therefore, electric force is 1.23 x 10³⁶ times stronger than gravitational force.
describe los ejercicios que caliente las extremidades superiores del cuerpo
Answer:
todas las cosas que la cuerda de salto hará
Explanation:
1. Mueva suavemente las muñecas y prepárese para levantar la parte superior del cuerpo.
2.Activa tus hombros sin presionarlos.
3. Calienta todo tu cuerpo elevando tu frecuencia cardíaca y cebando para la coordinación de todo el cuerpo que necesitas para el entrenamiento de la parte superior del cuerpo.
The maximum allowable resistance for an underwater cable is one hundredth of an ohm per
meter and the resistivity of copper is 1.54 x 10-80m.
a) Calculate the smallest cross sectional area of copper cable that could be used.
Answer:
A = 1.54 x 10⁻⁵ m² = 15.4 mm²
Explanation:
The resistance of a wire can be given by the following formula:
[tex]R = \frac{\rho L}{A}\\\\A = \frac{\rho L}{R} = \frac{\rho}{\frac{R}{L} }[/tex]
where,
A = smallest cross-sectional area = ?
ρ = resistivity of copper = 1.54 x 10⁻⁸ Ωm
[tex]\frac{R}{L}[/tex] = resistance per unit length of wire = 0.001 Ω/m
Therefore,
[tex]A = \frac{1.54\ x\ 10^{-8}\ \Omega m}{0.001\ \Omega/m}[/tex]
A = 1.54 x 10⁻⁵ m² = 15.4 mm²
a spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?
a. 0.0014 meters
b. 0.45 meters
c. 2.2 meters
d. 5.0 meters
show your work.
Answer:
b. 0.45 meters
Explanation:
Given the following data;
Spring constant, k = 330 N/m
Force = 150 N
To find the extension of the spring;
Mathematically, the force exerted on a spring is given by the formula;
Force = spring constant * extension
Substituting into the formula, we have;
150 = 330 * extension
Extension, e = 150/330
Extension, e = 0.45 meters
An automobile traveling at a speed of 28.3 m/s applies its brakes and comes to a stop in 4.9 s. If the automobile has a mass of 1.1 x 103 kg, what is the average horizontal force exerted on it during braking? Assume the road is level.
Find Magnitude and direction
WILL GIVE BRAINLIEST
Answer:
i say it would be 32.5 m if the brakes comes and stops. the road is 3.7 x 105
Explanation:
A 1.55 kg frictionless block is attached to an ideal spring with force constant 340 N/m . Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 14.0 m/s.
A.) Find the amplitude of the motion.
B.) Find the maximum acceleration of the block.
C.) Find the maximum force the spring exerts on the block.
Answer:
Explanation:
Let amplitude be A . At displacement A , velocity will be zero , so
1/2 k A² = 1/2 m v²
340 A² = 1.55 x 14²
A = .9452 m
= 94.52 cm
B ) angular frequency of oscillation ω = √ ( k / m )
= √ ( 340 / 1.55 )
ω = 14.81 radian /s
maximum acceleration = ω²A
= 14.81² x 0.9452
= 207.32 m/s²
C) Maximum force = mass x maximum acceleration
= 1.55 x 207.32
= 321.36 N .
When oxygen in water interacts with iron it results in O A rust forming O B) a chemical change C) oxidation O D) all of the above
Alexis is in her Toyota Camry and trying to make a turn off an expressway at 19.0 m/s. The turning radius of the level curve is 35.0 m. Her car has a mass of 1240 kg. Determine the acceleration, net force and minimum value of the coefficient of friction which is required to keep the car on the road.
Answer:
The solution is given below:
Explanation:
The computation is shown below:
The acceleration is
= v^2 ÷ r
= 19^2 ÷ 35 m
= 10.3 m/s^2
According to the newton second law
F = mac
= 1240 kg (10.3 m/s^2)
= 12772 N
And,
frictional force (f) = \mu N
f = \mu (mg) = mac
\mu = ac ÷ g
= 10.3 m/s^2 ÷ 9.80 m/s^2
= 1.0510
In these ways it can be determined
The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-N applied force. The uniform horizontal rod has a mass of 15 kg and is welded to the block at B. Friction in the guide is negligible.
Required:
Compute the bending moment M exerted by the weld on the rod at B. The bending moment is positive if counterclockwise, negative if clockwise.
Answer:
The bending moment is 459.16 N.m
Explanation:
From the given information;
Let's assume that the angle is 66°
Then, the free body diagram is draw and attached in the file below.
Now, the calculation of the acceleration from the first part of the free body diagram is:
[tex]\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a \\ \\ a = \dfrac{347.906}{50} \\ \\ a = 6.96 \ m/s^2[/tex]
Bending moment M:
From the second part of the diagram:
[tex]\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435 \\ \\ M = 238.435 + 220.725 \\ \\ \mathbf{M = 459.16 \ N.m}[/tex]
A house is advertised as having 1460 square feet under roof. What is the area of this house in square meters?
Answer: 131.92 m
Explanation: The problem give above is a conversion problem. To convert any measurement from one unit to another, we need a relation for the original units to that of the desired unit. This relation is called the conversion factor. It is used by multiplying or dividing the factor depending on what is the unit of the original measurement. For this case, we need to convert from ft^2 to m^2. The conversion factor would be 1 foot is equal to 0.3048 m. So, we would be getting the square of the factor and then multiply to the original value.
1420 ft^2 ( 0.3048 m / 1 ft)^2 = 131.92 m
please help meeeee :((((((((((
Answer:
hello me
Explanation:
help me thisanswer
When an object is put into a liquid, it experiences a buoyant force that is equal to the weight of the liquid the object displaces. The force on the wire is given as the block is slowly lowered into the liquid (position is given in "centimeters", which you have to change to "meters") and force is given in "newtons"). Choose a mass of the block between 0.125 kg and 0.375 kg and a density of the liquid between 500 kg/m^3 and 1000 kg/m^3. The object is in static equilibrium when the clock stops.
Required:
a. What is the weight of the block and the tension, F, in the string when the block is in the liquid?
b. What is the volume of the block in the liquid—either the submerged part of the block if the block is partially submerged when you paused it or the entire block if it is completely submerged (the dimension of the block that is into the screen is 5.00 cm = 0.0500m)?
c. What is the volume of the water that is displaced by the block (the dimension of both water containers into the screen is 10.00 cm = 0.100m)?
Answer:
A) F = 1.8375 N, B) V = 1.25 10⁻⁴ m³, C) V= 1.25 10⁻⁴ m³
Explanation:
This exercise should select some values such as the weight of the block m=0.250 kg and the density of the liquid ρ = 500 kg / m³ (water); with these values answer the question, for this we use the static equilibrium relations.
Σ F = 0
F + B - W = 0
the expression for thrust is
B = ρ g V_liquid
A) the weight of the block is
W = m g
W = 0.250 9.8
W = 2.45 N
Let's look for the maximum push
V_body = V_liquid
B = 500 9.8 0.05³
B = 0.6125 N
we substitute
F = W - B
F = 2.45 - 0.6125
F = 1.8375 N
B) As the body is totally submerged, the volume of the liquid and the body are equal
V = l³
V = 0.05³
V = 1.25 10⁻⁴ m³
C) the volume of water displaced is equal to the volume of the body
V_liquid = 1.25 10⁻⁴ m³
We know the mass and luminosity of the star. We also know the mass of the spacecraft and the area of its stellar sail. We must find the net force exerted on the spacecraft, as well as the direction of its acceleration. We are asked to find both the net force exerted on the spacecraft and the direction of the acceleration. Which principle is best used to find the direction of the acceleration once the net force has been found
Answer:
Newton's principle of force, the variation of the momentum and the acceleration have the same direction
Explanation:
To find the force exerted by the star on the sail of the ship, the expression of the momentum of the radiation is used, which in the case of a total reflection
p = [tex]\frac{2U}{c}[/tex]
where p is the momentum and U the incident energy and c the speed of light.
The strength I could get from
F = [tex]\frac{dp}{dt}[/tex]
the direction of the acceleration is the direction of the momentum, therefore the principle to find the acceleration of is Newton's principle of force, the variation of the momentum and the acceleration have the same direction.
A heat engine operating between a pair of hot and cold reservoirs with respective temperatures of 500 K and 200 K will have what maximum efficiency?
A) 60%
B) 50%
C) 40%
D) 30%
Answer:
The maximum efficiency of the heat engine is 60%.
Explanation:
Given that,
The temperature of hot reservoir = 500 K
The temperature of cold reservoir = 200 K
We need to find the maximum efficiency. The formula for the efficiency of heat engine is given by :
[tex]\eta=1-\dfrac{T_c}{T_H}[/tex]
Put all the values,
[tex]\eta=1-\dfrac{200}{500}\\\\\eta=0.6[/tex]
or
[tex]\eta=60\%[/tex]
So, the maximum efficiency of the heat engine is 60%.
Starting from rest, a car takes 2.4 s to travel the first 15 m. Assuming a constant acceleration, how long will it take the car to travel the next 15 m?
Answer:
The time it will take the car to travel the next 15 m is 1 s.
Explanation:
Given;
initial velocity of the car, u = 0
time of motion, t = 2.4 s
initial distance traveled, d = 15 m
determine the constant acceleration of the car;
d = ut + ¹/₂at²
15 = 0 + (0.5 x 2.4²)a
15 = 2.88a
a = 15 / 2.88
a = 5.21 m/s²
The velocity of the car at the end of the first 15 m
v - u + at
v = 0 + 5.21 x 2.4
v = 12.5 m/s
The time to cover the next 15 m;
d₂ = vt + ¹/₂at²
15 = 12.5t + (0.5 x 5.21)t²
15 = 12.5t + 2.61t²
0 = 2.61t² + 12.5t - 15
Use the formula method to solve the above quadratic equation.
a = 2.61
b = 12.5
c = -15
[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-12.5 \ \ +/- \ \ \sqrt{12.5^2 - (4\times 2.61 \times -15)} }{2\times 2.61}\\\\t = 0.9937 \ s \ \ \ or \ \ \ -5.78 \ s\\\\time \ can \ only \ be \ positive;\\\\t \approx \ 1\ s[/tex]
Therefore, the time it will take the car to travel the next 15 m is 1 s.
54. Which of the following behaviors is most closely
associated with the foot-in-the-door phenomenon?
(A) Beth continues to participate in class because
she is positively reinforced.
(B) Adam is sleeping while the rest of his
classmates are working on their group
project
(C) Sutan asks his father for $5, and when he
agrees, Sutan asks him for $15 more.
(D) James feels pressure to go to the movies with
his friends even though he prefers to go
bowling.
(E) Diana feels guilty because she did not help
her family clear the table after dinner.
Answer:
C
Explanation:
Foot-in-the-door technique is a compliance tactic that aims at getting a person to agree to a large request by having them agree to a modest request first
The behavior that should be closely associated is option c. Sutan asks his father for $5, and when he agrees, Sutan asks him for $15 more.
What is the foot-in-the-door phenomenon?It is a tactic of compliance that focuses on agreeing on the person by having a request.
It shows the connection that lies between the person who is asking for a request and the person who is being asked.
So based on this, option c is correct.
learn more about behavior here: https://brainly.com/question/15661050
The chart shows the masses of selected particles.
Particle
Mass
2350
137 9 u
6
92
138
Ba
K:
on
94.9u
1.0 u
In the equation u + Ön – 133Ba + K+ + 36n+E , the energy E is equivalent to a mass of
1.
0.2 u
2.
2.0 u
3.
2.2 u
4.
0.0 u
Answer: 0.2 u
Explanation:
The atomic mass of nuclide is written as a superscript. On the left (reactants) aide of the nuclear equation shown, the total atomic mass is 235.0 + 1.0 u = 236 u. On the right (products) side reaction, the total atomic mass is 137.9u + 94.9u + 3 x 1.0u = 235.8 u. Therefore, 0.2 mass units must have been converted into energy.
A 1.6-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an unstretched length of 6 in. The collar is released from being held at A, and it passes ponit B with a speed of 16.0 ft/s. Determine the spring constant. The circular rod is kept from moving.
Answer:
k = 652 lb/ft
Explanation:
Given :
Weight of the collar = 1.6 lb
The upstretched length of the spring = 6 in
Speed = 16 ft/s
PA = 8 + 10
= 18 inch
Let the initial elongation be [tex]$\Delta x_i$[/tex]
∴ [tex]$\Delta x_i$[/tex] = 18 - 6
= 12 inch = 1 foot
[tex]$PB = \sqrt{13^2+5^2}$[/tex]
= 13.925 inch
Final elongation in the spring
[tex]$\Delta x_B = 7.928 $[/tex] inch = 0.66 feet
Applying the conservation of the mechanical energy between A and B is
[tex]$K.E_A+P.E_{g,A}+P.E_{sp,A}= K.E_B+P.E_{g,B}+P.E_{sp,B} $[/tex]
[tex]$0+mg_r+\frac{1}{2}k(\Delta x_i)^2=\frac{1}{2}mv_B^2+0+\frac{1}{2}k(\Delta x_B)^2$[/tex]
[tex]$\frac{1}{2}k[(1)^2-(0.66)^2]=\frac{1.6}{2}\times (16)^2-1.6 \times 32 \times \frac{5}{12}$[/tex]
[tex]0.281 \ k =204.8-21.33[/tex]
k = 652 lb/ft
which is a non-renewable resource used to heat homes a A. solar energy
B.natural gas
C. WINDS
D.woods
Answer:
B. natural gas
Answer:
B natural gas
Explanation:
natural gas is used to make electricity
1 pts
Harry spent his summer playing guitar on the corner trying to earn some extra money. He would keep his guitar
case open for people to drop extra change into. Sometimes people would put money in the case and sometimes
people would be rude. Putting money in the hat is a to Harry's behavior.
Tir
Atte
1M
Negative reinforcement
Positive reinforcement
O Negative punishment
Positive punishment
Question 2
1 pts
O
SIC
acer
Answer:
Positive reinforcement