R = R1 + R2 + R3
This is the formula for resistance in a series circuit.
During which type of change should you expect to see different chemical
symbols before and after the change?
A. Chemical
B. Phase
C. Nuclear
D. Physical
During the nuclear change we can expect to see different chemical symbols before and after the change
Chemical and nuclear reactions are quite different from one another. Atoms can share electrons with other atoms or participate in an electron transfer to increase their stability in chemical reactions. In nuclear reactions, the atom's nucleus stabilizes itself by going through some sort of alteration. Compared to the energies involved in chemical reactions, the energies produced in nuclear reactions are many orders of magnitude higher. Environmental factors like temperature or pressure do not significantly affect nuclear reactions like they do with chemical ones.
Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei
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A bus travels from Houston, Texas to Dallas, Texas, in 4.7 h with an average velocity of 76 km/h to the north. What is the bus’s displacement?
The bus's displacement from Houston to Texas with average velocity = -16.170 km/h
Evaluation :
average velocity = ( initial velocity -- final velocity )
( v-u )
displacement = [tex]\frac{v-u}{t}[/tex]
displacement = [tex]\frac{0 - 76 }{4.7}[/tex]
= - 16.170 km/h
Average velocity :
Average velocity is determined as the change in position or displacement (∆x) divided by the time period (∆t) in which the displacement results . The average velocity could be positive or negative based upon the sign of the displacement. The SI unit of average velocity is defined as meters per second (m/s or ms-1).
Displacement :The displacement is commonly the difference in the position of the two marks and is independent of the path covered when traveling between the two of the marks. The distance covered , however, is the total length of the path chosen between the two marks. Displacement defined as a vector quantity that mean to "how far out of the place an object is"; it's the object overall changes in the position.
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As you go up the y-axis, what happens to the number of sprouted bean seeds?
A. Sprouted bean seeds decrease.
B. Sprouted bean seeds increase.
C. Sprouted bean seeds remain constant.
D. None of the above
As we go up the y-axis, the number of sprouted bean seeds increase (option B).
What is a graph?Graph is a data chart intended to illustrate the relationship between a set (or sets) of numbers (quantities, measurements or indicative numbers) and a reference set.
In a graph, there are two axes as follows;
Y-axis or vertical axisX-axis or horizontal axisAccording to this question, a graph of number of sprouted bean seeds on the y-axis is plotted against temperature on the x-axis.
We can observe that as we go up the y-axis, the number of sprouted bean seeds increase.
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At this rate, how long does it take to accelerate from 70 km/h to 110 km/h ?
Answer:
[tex]{ \tt{acceleration = \frac{ \triangle \: v}{t} }} \\ \\ { \tt{a = \frac{110 - 70}{t} }} \\ \\ { \tt{t = \frac{40}{a} } \: s}[/tex]
For a body under free fall; a = g = 9.8
[tex]{ \tt{t = \frac{40}{9.8} }} \\ \\{ \tt{t = 4.08 \: s}}[/tex]
Which element has similar properties to neon
Answer:
d) Krypton
Explanation:
Krypton has similar properties to that of neon, as they are noble gases (in same group). Hence, option (d) is the answer.
Write an experiment to show motion of particles in the matter
The experiment to show the random motion of the particles is the movement of particles in a solution is the simplest representation of motion of particles.
What is the experiment to show the motion of particle in a matter?Pick up two glass beakers. Fill one beaker with hot water and the second one with cold water. Add KMnO4 crystals in very small quantities to both. We thus see that KMnO4 particles move faster in hot water than in cold water.
What is the experiment to show evidence for attraction between the particles:Pick up some iron and a piece of chalk. We learn that it's quite difficult to work with iron. This leads us to believe that the Forces can affect matter particles. among them. The pressure kept between particles and the power of the attraction in materials is at its strongest between particles the least in gases.
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A swimmer jumps from a waterfall at a velocity of 1.90 m/s to the lake below. How high is the waterfall if the swimmer lands 2.5 m away from the base of the waterfall?
At what velocity does the swimmer enter the water?
1. The waterfall is 8.54 m high
2. The velocity with which the swimmer enters the water is 14.836 m/s
How to determine the time to reach the waterInitial velocity (u) = 1.9 m/sHorizontal distance (s) = 2.5 mTime (t) = ?s = ut
2.5 = 1.9 × t
Divide both sides by 1.9
t = 2.5 / 1.9
t = 1.32 s
1. How to determine the height of the waterfallTime (t) = 1.32 sAcceleration due to gravity (g) = 9.8 m/s²Height (t) =?h = ½gt²
h = ½ × 9.8 × 1.32²
h = 8.54 m
2. How to determine the velocityThe velocity with which the swimmer enters the water can be obtained as follow:
Initial velocity (u) = 1.9 m/sAcceleration due to gravity (g) = 9.8 m/s²Time (t) = 1.32 sFinal velocity (v) = ?v = u + gt
v = 1.9 + (9.8 × 1.32)
v = 1.9 + 12.936
v = 14.836 m/s
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A body, initially at rest, explodes into two masses M, and M₂ that
move apart with speeds v, and v₂ respectively. What is the ratio ?
V₂
A.
C.
M₁
M₂
(M₁2
1|2
B.
D.
M₂
M₁
M.
(M₂₁
Using conservation of momentum we will get that the ratio of velocity of mass M₁ and mass M₂ is ratio of M₂ to M₁
i.e. v₁/v₂=M₂/M₁
What is conservation of momentum ?
conservation of momentum states that in an isolated system the initial momentum will always be equal to final momentum.
given:
initial velocity of the object= 0m/s
initial momentum of the object = 0kg.m/s
now the mass M breaks into 2 pieces such that:
M₁ is the mass of first piece
M₂ is the mass of first piece
velocity of M₁ is v₁
velocity of M₂ is -v₂ as to conserve momentum
using conservation of momentum we get,
initial momentum = final momentum
0 = M₁×v₁ + M₂×(-v₂)
M₁×v₁= M₂×v₂
v₁/v₂ = M₂/M₁
using conservation of momentum we will get that the ratio of velocity of mass M₁ and mass M₂ is ratio of M₂ to M₁
i.e. v₁/v₂=M₂/M₁
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Consider the five pairs of long, parallel wires shown. The arrows indicate the direction of the current in each wire, with small
arrows representing a current of 3 A and large arrows representing a current of 9 A. For each pair, determine in which one of the
regions the net magnetic field is zero someplace:
on the left of both wires
between the two wires
on the right of both wires
• none of the above (i.e., nowhere)
The Net magnetic field will be at the left in case I, right in Case II, In-between in Case III, and Nowhere in Case IV and V.
Case I
As the current in both wires is not equal, the net magnetic field will not be zero in the middle. It will be zero somewhere near the 3 A current carrying wire and farther from the 9A current carrying wire. Which is to the Left.
Case II
It is similar to the case I, as the current in both wires is not equal, and the net magnetic field will not be zero in the middle. It will be zero somewhere near the 3 A current carrying wire and farther from the 9A current carrying wire. Which in this case is to the Right.
Case III
As the current in both wires is the same and equal to 3 A, the net magnetic field will be zero in the middle. Which is in Between.
Case IV
As the current flowing in the wire is opposite in the direction, the magnetic field will always be in the same direction. So the magnetic field will be zero Nowhere.
Case V
It is similar to the above case IV, as the current flowing in the wire is opposite in the direction, and the magnetic field will always be in the same direction. So the magnetic field will be zero Nowhere.
Therefore, the Net magnetic field will be at left in case I, right in Case II, In-between in Case III, and Nowhere in Case IV and V.
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Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at launch? Neglect air resistance
The speed of the ball at launch is 23m/s.
After 0.5 seconds, the ball initially passed the pole.
Then it will again reach the top of the pole after 4.1 seconds.
Consequently, the ball's travel time above the pole is
Δt = 4.1 - 0.5 = 3.6sec
Now the time it will take to reach the highest point from the top of the pole is given as t = Δt/2 = 1.8sec.
Now the total time to reach the top from the bottom is given as 2.3sec.
After 0.5 seconds, the ball initially passed the pole.
Consequently, the ball's travel time above the pole is
[tex]v_{f} - v_{i} = at[/tex]
[tex]v_{i}=22.54m/s[/tex]
also the speed after 0.5 s is
[tex]v_{f} = 17.64m/s[/tex]
So, the speed of the ball at launch is 23m/s.
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Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 150 V. An electron enters the region between them at a speed of 2.6 × 105 m/s by passing through a small hole in the negative plate and continues moving toward the positive plate. Assume the electric field between the plates is uniform.
The correct answer is V_f = 1.81 × 10^14.
D = 2cm = 0.02 m
ΔV = 150 V
V_i = 2.6 × 10^5 m/s
The coulomb force,
F = qE
ma = qE
a = qE/m
Potential difference
ΔV = ED
E = ΔV/D
a = qΔV/mD
Using equation of motion,
V_f^2 = V_i^2 + 2ad
V_f = √V_i^2 + 2ad
V_f = √V_i^2 + 2d ( qΔV/mD)
here q= 1.6 × 10^-19 C
m = 9.11 × 10^-31 kg
for d= 0.1 cm = 0.001 m
V_f = √(2.6 × 10^5)^2 + 2 (0.001)(1.6 × 10^-19 C × 150V/9.11 × 10^-31 × 0.02)
V_f = √(6.76 × 10^10) + 0.002(24× 10^-18 C)/1.822× 10^-32)
V_f = √(6.76 × 10^10) + 48 × 10^-15/1.822× 10^-32
V_f = √(6.76 × 10^10) + 26.3 × 10^17
V_f = √33.06 × 10^27
V_f = 1.81 × 10^14
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and gravity is exerted on an object. DISTANCE VOLUME GRAVITY MASS both impact how much
Answer:
Explanation:
yes, it will.40
Question 10 of 10 Why are experiments often performed in laboratories? A. You can eliminate variables like air resistance. B. Scientists only want to know what happens in labs. C. It is harder to control variables in a laboratory. D. They are better at representing the real world. Need answer asap.
What precise meaning do you attach to the statement r = (24.0 ± 0.3) mm, where r is the radius of a tube.
The information provided above demonstrates that the range of the tube's radius within which we are certain is from 23.7 to 24.3.
Accurate findings are desired when any quantity is being measured. Closeness of the measurements to a particular value is referred to as accuracy (like the theoretical value). It is not insignificant that measurements can contain inaccuracies; as a result, the outcome frequently includes a margin error. The sentence above displays the radius's measured value along with its margin of error.
In other words, the aforementioned statement represents a quantity's numerical value along with its tolerance, or the only permissible (and conceivable) values to ensure that the measurement is accurate. The information provided above demonstrates that the range of the tube's radius within which we are certain is from 23.7 to 24.3.
These are all valid radius values, along with all others within this interval. By increasing the number of trials in a measurement, the value for the upper and lower bounds of the error can be decreased.
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no sunlight
mineral-rich
extremely hot
chemosynthetic bacteria
These are all habitat features of __________________ within the oceans on Earth.
Responses
A estuariesestuaries
B benthic zonesbenthic zones
C kelp forestskelp forests
D hydrothermal ventshydrothermal vents
Answer: Hydrothermal vents is your answer.
Explanation:
it's very hot, there is no light, there are bacteria, and there it's mineral-rich.
How to find the angle in a projectile motion when displacement and heigh are the same
projectile's vertical speed v yv y v, start , y, drops prior to reaching its maximum height since its acceleration is in the opposite direction. Since the object's height is rising, the initial direction of the velocity is upward
.
At the projectile's highest point, vertical velocity zeroes out. After reaching the highest point, the vertical speed increases due of the same-direction acceleration. As the object's height lowers, the vertical velocity has a downward direction.
In projectile The beginning vertical velocity determines the maximum height. Increasing the launch angle raises the maximum height because steeper launch angles have a larger vertical velocity component.
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Layla jogs with an average velocity of 2.4 m/s east. What is her displacement after 46 seconds?
Answer:
[tex]{ \tt{displacement = velocity \times time }} \\ { \tt{ = 2.4 \times 46}} \\ { \tt{ = 110.4 \: m}}[/tex]
For the vectors shown in the figure, determine
(Figure 1)
Figure
B (B=26.5)
56.0%
Ā (A = 44.0)
28.0°
C(C=31.0)
< 1 of 1 >
X
the magnitude of B-3A
Express your answer using three significant figures.
|B-3A| =
The magnitude of the given vectors for B - 3A expression is determined as 105.5 m.
What is magnitude of vectors?The magnitude of a vector is the length of the vector. The magnitude of the vector a is denoted as |a|.
The magnitude of a vector represents the absolute value of the given vector.
Magnitude of B - 3AThe magnitude of the expression B - 3A is calculated as follows;
B - 3A = 26.5 - 3(44)
B - 3A = 26.5 - 132
B - 3A = -105.5
|B - 3A| = 105.5
Thus, the magnitude of the given vectors for B - 3A expression is determined as 105.5 m.
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1. Based on the mc free-body diagram, type an expression for the horizontal acceleration of the cart only in terms of T , mh , mc , μk , and g .
2.Based on the mh free-body diagram, type an expression for the vertical acceleration of the hanging mass.
3. As shown in the first figure, the horizontal acceleration of the cart to the right must be equal to the downward vertical acceleration of the hanging mass if the string does not stretch. Given that both the cart and the hanging mass have the same acceleration, eliminate the tension force ( T ) from your answers to Q1 and Q2 to find an expression for the acceleration only in terms of mh , mc , μk , and g .
(a) The expression for the horizontal acceleration of the cart is
a = (T - μN)/Mc.
(b) The expression for the vertical acceleration of the hanging mass is
a = (T - Mhg)/Mh.
(c) The expression for the acceleration of both cart and the hanging mass is a = (Mhg - μN) / (Mc - Mh).
Horizontal acceleration of the cart
The horizontal acceleration of the cart is calculated as follows;
The horizontal acceleration of the cart is determined from the net horizontal force acting on the system.
T - μN = Mca
where;
Mc is mass of the carta is the acceleration of the carta = (T - μN)/Mc
Vertical acceleration of the hanging massT - Mhg = Mha
where;
Mh is the mass of the hanging massa = (T - Mhg)/Mh
Acceleration of both objectsIf the acceleration of the cart and the hanging mass are equal, our final equation becomes;
From first equation;
a = (T - μN)/Mc
Mca = T - μN
Mca + μN = T
From second equation;
a = (T - Mhg)/Mh
Mha = T - Mhg
Mha + Mhg = T
Solve the two equation together;
Mca + μN = Mha + Mhg
Mca - Mha = Mhg - μN
a(Mc - Mh) = Mhg - μN
a = (Mhg - μN) / (Mc - Mh)
Thus, the expression for the acceleration of both cart and the hanging mass is a = (Mhg - μN) / (Mc - Mh).
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Highway safety engineers build soft barriers so that cars hitting them will slow down at a safe rate. A person wearing a safety belt can withstand an acceleration of 300 m/s/s. How thick should barriers be to safely stop a car that hits the barriers at 76.86 km/h?
The thickness of the barrier that will stop a car travelling at 76.86 km/h is 0.76 m.
Thickness of the barriers
The thickness of the barriers can be calculated using a kinematic equation as shown below;
v² = u² + 2ad
where;
d is the thickness of the barrierv is the final velocityu is the initial velocity a is the acceleration of the carv is the final velocityThe given final velocity of the car;
v = 76.86 km/h = 21.35 m/s
Thickness of the barrier that will stop a car travelling at 76.86 km/h is calculated as follows;
d = v²/2a
d = (21.35²) / (2 x 300)
d = 0.76 m
Thus, the thickness of the barrier that will stop a car travelling at 76.86 km/h is 0.76 m.
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A 1.2kg stone is tied to string and swung in a vertical circle with a radius of 0.85 m. The string can withstand a tension of 40.0 N. At the maximum speed can the stone move at the bottom of its path without the string breaking
The maximum speed of the stone tied to the string is 5.32 m/s.
The mass of the stone is m = 1.2 kg.
The mass is tied to a string and it swung in a vertical radius of r = 0.85 cm.
The tension on the string is T = 40 N.
The centripetal force of a body moving in a circle of radius r is given as:
T = ( mv² / r )
Where T is the tension, m is the mass of the body and r is the radius of the circle.
Then,
v² = ( T × r ) / m
v = √ [ T × r / m ]
v = √ [ 40 × 0.85 / 1.2 ]
v = √ [ 34 / 1.2 ]
v = √28.3
v = 5.32 m/s
The maximum speed of the stone before the string breaks is 5.32 m/s.
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1. How much heat input is needed to
raise the temperature of a 20kg
vat made of iron and filled with a 3kg of water from 10°C to 90°C? Take the
Cp of iron and water to be 450J/kg:
land 4186.J/k9°C respectively
By using Specific heat capacity, 741.76 x [tex]10^4[/tex]heat energy is needed to raise the temperature.
The amount of heat required to increase the temperature of 1 gram of a substance by 1 degree Celsius (°C) is known as the specific heat capacity.
The importance of specific heat capacity may be summed up as the quantity of energy needed to heat or cool an item of a given mass by a certain amount.
Given:
Mass of vat, mvat = 20kg
Mass of water, mwater = 3kg
ΔT = 90 – 10 = 80℃
Specific heat capacity of iron, Ciron = 450J/kg
Specific heat capacity of water, Cwater = 4186J/kg
To find:
Heat energy required, Q = ?
Formula:
C = Q / (m ΔT)
Calculations:
Q = mCΔT
Q = Qempty + Qwater
Q = (20 x 450 x 80) + (3 x 4186 x 80)
Q = 172.464 x [tex]10^4[/tex]J
Result:
Heat energy required is 172.464 x [tex]10^4[/tex]J
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You jog at 9.5 km/h for 8.00 km; then you jump into a car and ride an additional 16.0 km. What average speed must the car have for the average speed for the entire 24.0-km trip to be 20 km/h
Average speed must the car have for the average speed for the entire 24.0 km trip 67.039 km/hr.
Equation :Average speed for entire journey (jogging + drive) = 20 km/hr
total distance traveled = 24 km
total time spent (on jog + drive) = total distance/average speed
total time spent (on jog + drive) = 24/20
total time spent (on jog + drive) = 1.2 hr
Jogging:
Speed = 9.5 km/hr
Distance = 8.0 km
Time spent jogging = Distance/Speed
= 8.0 / 9.5 = 0.842 hr
Driving car :
Speed = ?
Distance = 24 km
time spent driving = total time spent on journey - time spent jogging
time spent driving = 1.2 - 0.842 = 0.358 hr
average speed for driving = distance / time
average speed for driving = 24/0.358
average speed for driving = 67.039 km/hr
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Now think creatively about who you might have been had you been raised by other parents. Prepare three brief descriptions of how you might have developed using any three to four styles of parenting .
by using all the four type of parenting here is the descriptive answer which include 4 types of parenting by parents.
When a child queries the justification for a regulation, authoritarian parents are infamous for responding, "Because I said so," They are more concerned with obeying than they are with negotiating. Parental authority figures may substitute punishment for instruction. So instead of instructing a child on how to make better decisions
Parents who are authoritative put time and effort into avoiding behavioral issues before they arise. Additionally, they employ positive discipline techniques like reward and praise systems to reinforce good conduct.
Parents that are liberal with their children typically act more like friends than parents. They frequently encourage their kids to talk to them about their issues, but they typically don't make an attempt to discourage poor decisions or misbehavior because they are either uninvolved parents who don't understand how kids develop, or they are unaware of it.
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What two gases together make up about 99% of Earth's atmosphere? a. Carbon dioxide and nitrogen
A: Nitrogen and oxygen
B: Nitrogen and helium
C: Argon and oxygen
D: Trace gases and nitrogen
If a liter of gasoline cost $0.58, and a car gets 11 km/L, what would it cost to travel 100km?
Answer: $638,00
Explanation: Hi Tania... I believe we're in the same class and I was just doing the Science homework when I saw your question... Take this as a gift for doing the Spanish homework last time... Thx! Oh, btw that's the procedure and the response.
X=(1100*0.58)/1= 638,00
Which of the following is a form of identity fraud?
A. Chipping
B. Skimming
C. Chirping
D. Skipping
4. Invalid experiments sometimes result from scientists influencing them
unfairly. This is called:
O Bias
O Scientific Method
O Controlled Experiment
Hypothesis
Answer: Bias
Explanation:
Which sport car motion does Not get any kind of "acceleration?" When a car
A- turns a corner
B- brakes to a stop
C- starts from a traffic stop
D- moves with constant velocity
Answer:
D- moves with constant velocity
Explanation:
As we know that acceleration is tge rate of change in velocity.
Since velocity is constant, the change in velocity is zero (∆v=0)
[tex]{ \tt{acceleration = \frac{ \triangle v}{t} }} \\ \\ { \tt{acceleration = \frac{0}{t} }} \\ \\ { \tt{acceleration = 0}}[/tex]
3) Two particles are travelling along a straight line AB of length 20m.
At the same instant one particle starts from rest at A and travels
towards B with a constant acceleration of 2 ms72 and the other
particle starts from rest at B and travels towards A with a constant
acceleration of 5 ms2, find how far from A the particles collide.
At distance of 40/7 m from A the particles will collide.
Let particles collide at C
And let distance AC = x and BC = y
then, x+y = 20
As the equation used for distance required here is: s= ut +1/2 a[tex]t^{2}[/tex] and here u=0
Then, 1/2 (2)[tex]t^{2}[/tex] +1/2 (5)[tex]t^{2}[/tex] =20
7/2[tex]t^{2}[/tex] =20
[tex]t^{2}[/tex] = 40/7
Now
x= 1/2 (2)[tex]t^{2}[/tex] = 40/7 m
Therefore, at distance of 40/7 m from A the particles will collide.
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