Answer:
1000.
Explanation:
Thousand more steps could be taken when the number of cells was increased from 3 to 5 because in the data, the number of cells i.e. 5 was achieved on 1000 steps. According to the data, the number of cells i.e. 1 was achieved on 800 steps, the number of cells i.e. 5 was achieved on 1000 steps, the number of cells i.e. 2 was achieved on 600 steps, the number of cells i.e. 3 was achieved on 400 steps and the number of cells i.e. 4 was achieved on 200 steps. So we can say that 1000 steps are needed for the number of cells increased from 3 to 5.
How energy is obtained due to flow of charges?
A student is investigating the affect of different salts on melting points. Four patches of ice of equal
size are roped off and a
different type of salt is poured on each, one receives table salt (NaCl), one
receives Calcium Chloride (CaCl2), one receives Potassium Carbonate (KCO3) and the fourth
receives inert sand instead. Each patch receivęs an equal amount of salt or sand. The student
measures the volume of ice remaining and subtracts it from the original volume of ice to see how
much melted away. What is a control variable in this experiment?
A. The size of the ice patches.
B. The type of salt applied to the ice.
C. The amount of ice that melted.
D. None of these.
Answer:
A. The size of the ice patches
Explanation:
In an experiment, the control variable also known as the CONSTANT is the variable that must be kept uniform or the same for all groups throughout the experiment in order not to influence the outcome of the experiment.
According to the experiment described in this question, the effect of different salts on melting points is investigated by a student. Sodium chloride (NaCl), Calcium Chloride (CaCl2), Potassium Carbonate (KCO3) and inert sand are the four types of salt used. The volume of the ice used and melted was finally measured. This means that the SIZE OF THE ICE PATCHES USED is the control variable of the experiment as the same size was used for all groups throughout.
Answer:
the size of the ice patches
Explanation:
What is the magnitude of the electric field at a point 0.0055 m from a 0.0025
C charge?
kg
Use E = and k=9.00 x 10 N.m²/C2.
O A. 7.4 x 1011 N
O B. 2.0 x 1010 N
O C. 4.1 x 10°N
OD. 7.9 x 1012 N
Answer:
Explanation:
The equation for the electric field is
[tex]E=\frac{kQ}{r^2}[/tex] so filling in:
[tex]E=\frac{9.00*10^9(.0025)}{(.0055)^2}[/tex] which in the end gives you
E = 7.4 × 10¹¹, choice A
An unstrained horizontal spring has a length of 0.40 m and a spring constant of 340 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
(a) Both the charges are positive or negative.
(b) Teh value of each charge is 1.53 x 10^-5 C.
Explanation:
Spring constant, K = 340 N/m
Natural length, L = 0.4 m
stretch, y = 0.033 m
(a) Let the charge on each sphere is q and they repel each other so the nature of charge of either sphere may be both positive or both negative.
(b) The electrostatic force is balanced by the spring force.
[tex]\frac{kq^2}{(L + y)^2}=Ky\\\\\\\frac{9\times 10^9 q^2}{(0.4 +0.033)^2} = 340\times0.033\\\\q= 1.53\times 10^{-5} C[/tex]
what is the relationship between Hectare and cubic meter
What is the enthalpy change, AH, for this reaction? Show your work to receive full credit. URGENT PICTURE INCLUDED !!!
Answer:
150 kJ
Explanation:
Applying,
ΔH = Energy level of Product(E') - Energy level of reactant(E)
Where ΔH = enthalpy change, E' and E = energy level of the product and the reactant respectively
ΔH = E'-E............. Equation 1
From the diagram,
Given: E' = 200 kJ, E = 50 kJ
Substitute these values into equation 1
ΔH = 200-50
ΔH = 150 kJ
Hence the enthalpy change for the reaction is 150 kJ
Complete the sentence below using one of the following
words: equilibrium, flux, motion.
If supported at its center of gravity, an object will remain in ______ any position.
Answer:
equilibrium
Explanation:
acceleration will remain constant
A 1.25 kg block is attached to a spring with spring constant 17.0 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 49.0 cm/s . What are You may want to review (Pages 400 - 401) . Part A The amplitude of the subsequent oscillations
Answer:
The amplitude of the subsequent oscillations is 13.3 cm
Explanation:
Given;
mass of the block, m = 1.25 kg
spring constant, k = 17 N/m
speed of the block, v = 49 cm/s = 0.49 m/s
To determine the amplitude of the oscillation.
Apply the principle of conservation of energy;
maximum kinetic energy of the stone when hit = maximum potential energy of spring when displaced
[tex]K.E_{max} = U_{max}\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\where;\\\\A \ is \ the \ maximum \ displacement = amplitude \\\\mv^2 = kA^2\\\\A^2 = \frac{mv^2}{k} \\\\A = \sqrt{\frac{mv^2}{k}} \\\\A = \sqrt{\frac{1.25\ \times \ 0.49^2}{17}} \\\\A = 0.133 \ m\\\\A = 13.3 \ cm[/tex]
Therefore, the amplitude of the subsequent oscillations is 13.3 cm
How many more neutrons are in a I SOTOPE of copper-14 than in standard carbon atom
Answer:
2 more neutrons
Explanation:
To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:
For carbon–14:
Mass number = 14
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
14 = 6 + Neutron
Collect like terms
14 – 6 = Neutron
8 = Neutron
Neutron number = 8
For carbon–12:
Mass number = 12
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
12 = 6 + Neutron
Collect like terms
12 – 6 = Neutron
6 = Neutron
Neutron number = 6
SUMMARY:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Finally, we shall determine the difference in the neutron number. This can be obtained as follow:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Difference =?
Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)
Difference = 8 – 6
Difference = 2
Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)
The 75.0 kg hero of a movie is pulled upward at a constant velocity by a rope. What is the tension on the rope?
Answer:
750 N
Explanation:
the tension on the rope is the weight of the hero
If an object undergoes a change in momentum of 10 kg m/s in 3 s ,then the force acting on it is
Answer:
Force = 3.333 Newton
Explanation:
Given the following data;
Change in momentum = 10 Kgm/s
Time = 3 seconds
To find the force acting on it;
In Physics, the change in momentum of a physical object is equal to the impulse experienced by the physical object.
Mathematically, it is given by the formula;
Force * time = mass * change in velocity
Impulse = force * time
Substituting into the formula, we have;
10 = force * 3
Force = 10/3
Force = 3.333 Newton
After landing the aeroplane's momentum becomes zero. Explain how
the law of conservation holds here.
Answer:
The law of momentum conservation can be stated as follows. For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
a body is moving along a circular path 'r'. what will be the distance and displacement of the body when it completed half a revolution?
After half of a revolution ...
==> Distance = π•r
==> Displacement = 2•r
This question is divided into two parts. This is part (a) of the question. A proton accelerates from rest in a uniform electric field of 580 N/C. At some later time, its speed is 1.00 x 106 m/s. (a) Find the magnitude of the acceleration of the proton. (Mass of the proton is 1.67 x 10-27 kg and charge is 1.60 x 10-19 C) (in the following options 10^10 m/s^2 is 1010 m/s2)
Answer:
The acceleration of proton is 5.56 x 10^10 m/s^2 .
Explanation:
initial velocity, u = 0
Electric field, E = 580 N/C
final speed, v = 10^6 m/s
(a) Let the acceleration is a.
According to the Newton's second law
F = m a = q E
where, q is the charge of proton and m is the mass.
[tex]a= \frac{q E}{m}\\\\a = \frac{1.6\times10^{-19}\times 580}{1.67\times 10^{-27}}\\\\a= 5.56\times 10^{10} m/s^2[/tex]
A uniform metre ruler scale balanced at 40 cm mark, when weight 25 gf and 10gf are suspended at 10cm mark and 75 cm mark respectively.Calculate the weight of the metre scale.
Answer:
40 gf.
Explanation:
Please see attached photo for diagram.
In the attached photo, W is the weight of metre rule.
The weight of the metre rule can be obtained as follow:
Clockwise moment = (W×10) + (10×35)
Clockwise moment = 10W + 350
Anticlock wise moment = 25 × 30
Anticlock wise moment = 750
Clockwise moment = Anticlock wise moment
10W + 350 = 750
Collect like terms
10W = 750 – 370
10W = 400
Divide both side by 10
W = 400/ 10
W = 40 gf
Thus, the weight of the metre rule is 40 gf
Answer:
40 gf
Explanation:
The balance point of the uniform meter rule with the suspended weights = 40 cm = The pivot point
The location where the 25 gf weight is suspended = 10 cm
The location where the 10 gf weight is suspended = 75 cm
Let W represent the weight of the meter rule.
We have that the location of the application of the weight of the meter rule is at the center, 50 cm mark, point
Given that the meter rule is balanced, and taking moment about the pivot point, we have;
The moment om the left hand side, LHS, of the pivot point = The moment on the right hand side, RHS, of the pivot point
The moment on the LHS = 25 gf × (40 cm - 10 cm) = 750 gf·cm
The moment on the RHS = W × (50 cm - 40 cm) + 10 gf × (75 cm - 40 cm)
The moment on the RHS = W·(10 cm) + 350 gf·cm
∴ 750 gf·cm = W·(10 cm) + 350 gf·cm
W·(10 cm) = 750 gf·cm - 350 gf·cm = 400 gf·cm
W = (400 gf·cm)/(10 cm) = 40 gf
The weight of the meter scale (rule), W = 40 gf.
what are MA and VR of a lever?
Explanation:
Mechanical advantage (MA) = Load/Effort. Velocity ratio (VR) = distance effort moves/ distance load moves in the same time
what force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s
Answer:
[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times \frac{20}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]
A satellite of mass 5460 kg orbits the Earth and has a period of 6520 s
A)Determine the radius of its circular orbit.
B)Determine the magnitude of the Earth's gravitational force on the satellite.
C)Determine the altitude of the satellite.
Answer:
what if I do and b then someone else c I don't have enough time pls
please help me with my question due tomorrow morning,
Answer:
D)7 1/2 or 15/2
Explanation:
Let's calculate Combined resistance of the parallel first
1/Rt= 1/2+1/6=4/6
Rt=6/4 which is also equal with 3/2
Now let's add it with the series one
Rt= 6+3/2
=15/2 And when we put that un a mixed fraction 7 1/2
How do space probes make it past the asteroid belt without crashing into asteroids?
Answer:
The thing is space is really vast like really big so even though the asteroid belt looks really cramped it isn't. There's a lot of space between asteriods and using simple navigation and maneuvering, space probes can easily make it through without the threat of crashing.
Explanation:
20 swings takes 5 seconds in the pendulum. calculate the periodic time of the swing
Answer:
0.25s
Explanation:
5/20 = 0.25s
This might be correct
A mass that weighs 8 lb stretches a spring 24 in. The system is acted on by an external force of 4 sin 4t lb. If the mass is pulled down 6 in. and then released, determine the position of the mass at any time. Determine the first four times at which the velocity of the mass is zero
Answer:
[tex]t = \frac{\pi}{8}, \frac{\pi}{4}, \frac{3\pi}{8}, \frac{3 \pi}{4}[/tex]
Explanation:
The equation of force is
F = 4 sin 4 t
Compare with the standard equation
f = A sin wt
where, w is the angular frequency and A is the amplitude.
Now
w = 4 rad/s
Let the time period is T.
the relation for the time period is
[tex]T = \frac{2\pi}{w}\\\\T = \frac{2 \pi}{4}\\\\T = \frac{\pi}{2}[/tex]
the time period is defined as the time taken by the body to complete one oscillation.
So, the velocity is zero at the extreme points where the object is at time, T/4 and its odd T/2, 3T/4, 3T/2, etc.
So, the velocity is zero at time
[tex]t = \frac{\pi}{8}, \frac{\pi}{4}, \frac{3\pi}{8}, \frac{3 \pi}{4}[/tex]
A boy travels 12km east wards to a point B and then 5km southwards to another point C. Calculate the difference between the magnitude of the displacement of the boy and the distance travelled by him
The difference b/w the displacement and total distance traveled is 4km.
Explanation.
▪ total distance - displacement
= 17 km - 13 km
= 4 km...answer
for the equation BaCI2 + Na2SO4 > BaSO4 + 2NaCI
A. reactants: 1 ;products: 1
B. reactants: 1 ;products: 2
C. reactants: 2 ;products: 1
D. reactants: 2 ;products: 2
A student attaches a rope to a box and pulls the box up a ramp as shown below. The ramp has a rough surface. When
drawing the free body diagram for the box, the friction force should be directed:
O up and to the right
down and to the left
up and to the left
to the left
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Which would be used to measure the distance between the earth and a planet ,meter ruler or a measuring tape? Why?
Answer:
parallax
Due to foreshortening, nearby objects show a larger parallax than farther objects when observed from different positions, so parallax can be used to determine distances. To measure large distances, such as the distance of a planet or a star from Earth, astronomers use the principle of parallax.
URGENT
A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes.
a. What was his average speed?
b. What was his displacement?
PLEASE SHOW ALL WORK
Answer:
(a) 3.93 m/s
(b) 861.66 m
Explanation:
A = 4.5 m/s [27° S of W] for 3.0 minutes
B = 3.5 m/s [35° S of E] for 4.1 minutes
Distance A = 4.5 x 3 x 60 = 810 m
Distance B = 3.5 x 4.1 x 60 = 861 m
(a) The average speed is defined as the ratio of the total distance to the total time.
Total distance, d = 810 + 861 = 1671 m
total time, t = 3 + 4.1 = 7.1 minutes = 7.1 x 60 = 426 seconds
The average speed is
[tex]v=\frac{1671}{426}=3.93 m/s[/tex]
(b)
[tex]\overrightarrow{A} = 810(- cos 27 \widehat{i} - sin 27 \widehat{j})=- 721.7 \widehat{i} - 367.7 \widehat{j}\\\\\overrightarrow{B} = 861( cos 35 \widehat{i} - sin 35 \widehat{j})= 705.3 \widehat{i} - 493.8 \widehat{j}\\\\\overrightarrow{C} = (- 721.7 + 705.3) \widehat{i} - (367.7 + 493.8) \widehat{j} \\\\\overrightarrow{C}= - 16.4 \widehat{i} - 861.5 \widehat{j}[/tex]
The magnitude is
[tex]C =\sqrt{16.4^2+861.5^2} = 861.66 m[/tex]
a car is travelling at 36 km per hour if its velocity increases to 72 km per hour in 5 seconds then find the acceleration of car in SI unit
Answer:
36 km /h means 10 m/s. Increase in speed is 10m/s in 5 s . Acceleration is ( 10/5 ) = 2 m/s^ 2.
a= 2m/s²
Explanation:
U=36km/h
V=72km/h
T=5s
Conversion of Km to m and H to s
1km = 1000m
36km=36×1000 = 36000m
1H = 3600s
For U, 36000/3600
=10m/s
For V,
72km= 72×1000 =72000
72000/3600
20m/s
a=(V-U)/T
a=(20-10)/5
a= 10/5
a= 2m/s²
A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 22 m/s and the other ball, of mass 1.3 kg, is moving downward at 11 m/s. How high do the combined two balls of putty rise above the collision point
Answer:
The height balls rise above the collision point, is approximately 7.37 meters
Explanation:
The given parameters just before the collision are;
The mass, m₁ and velocity, v₁ of the ball moving upward are;
m₁ = 3.0 kg, v₁ = 22 m/s
The mass, m₂ and velocity, v₂ of the ball moving downward are;
m₂ = 1.3 kg, v₂ = -11 m/s (downward motion)
The type of collision = Inelastic collision
We note that the momentum is conserved for inelastic collision
Let, [tex]v_f[/tex], represent the final velocity of the balls after collision, we have;
∴ Total initial momentum = Total final momentum
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·[tex]v_f[/tex]
Therefore, we get;
m₁·v₁ + m₂·v₂ = 3.0 kg × 22 m/s + 1.3 kg × (-11) m/s = 51.7 kg·m/s
(m₁ + m₂)·[tex]v_f[/tex] = (3.0 kg + 1.3 kg) ×
∴ 51.7 kg·m/s = 4.3 kg × [tex]v_f[/tex]
[tex]v_f[/tex] = (51.7 kg·m/s)/4.3 kg ≈ 12.023 m/s
The final velocity, [tex]v_f[/tex] ≈ 12.023 m/s
The maximum height, h, the combined balls will rise from the point of collision, moving upward at a velocity of [tex]v_f[/tex] ≈ 12.023 m/s, is given from the kinetic equation of motion, v² = u² - 2·g·h, as found follows
At maximum height, we have;
[tex]h_{max} = \dfrac{v_f^2}{2 \cdot g }[/tex]
Therefore;
[tex]h_{max} \approx \dfrac{12.023^2}{2 \times 9.81 } \approx 7.37[/tex]
The height the combined two balls of putty rise above the collision point, [tex]h_{max}[/tex] ≈ 7.37 m.
Q011) The Doppler effect a. occurs when the frequency of sound waves received is lower if the wave source is moving toward you than if it's moving away. b. occurs when the pitch of a sound gets lower if the source is receding. c. is the basic explanation for the blue shift of light in our Universe. d. can be applied only to sound waves.
Answer:
Option (c) is correct.
Explanation:
The apparent change in the frequency of light due to the relative motion between the source and the observer is called Doppler's effect.
When the source is moving towards the observer which is at rest, the apparent frequency increases and if the observer is moving away the frequency of sound decreases.
It occurs for both light and sound.
So, to explain the blue shift of light in the universe is due to the Doppler's effect of light.
