[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be:______
Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s
An electric bulb rated 100 W, 100 V has to be
operated aross 141.4 V, 50 Hz A.C. supply. The
capacitance of the capacitor which has to be
connected in series with bulb so that bulb will
glow with full intensity is [NCERT Pg. 251]
Answer:
The capacitance of the capacitor is 31.84 μF.
Explanation:
Given;
power rating of the bulb, P = 100 W
voltage rating of the bulb, Vr = 100 V
operating voltage of the bulb, V= 141.4 V
frequency of the AC = 50 Hz
P = IV = 100 W
V = 100 V
I =
Ic = 1 A
The voltage across the capacitor is given by;
[tex]V_c = \sqrt{V^2 - V_R^2} \\\\V_c = \sqrt{141.4^2 - 100^2} \\\\V_c =99.97 \ V[/tex]
[tex]V_c = I_cX_c\\\\V_c = I_C* \frac{1}{2\pi fC}\\\\ 99.97 = 1 * \frac{1}{2\pi *50 *C}\\\\ C=\frac{1}{2\pi *50*99.97}\\\\ C = 31.84*10^{-6} \ F\\\\C = 31.84 \ \mu F[/tex]
Therefore, the capacitance of the capacitor is 31.84 μF.
What are the standard international (si) units of distance
Answer:
meter
Explanation:
Answer: The International System of Units is a system of measurement based on 7 base units
Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.
A helicopter accelerates at a rate of 7 m/s2 for 5 seconds. Calculate its velocity if it's final velocity is 60 m/s.
Answer:
7 m/s
Explanation:
Answer: 25 m/s
Explanation:
Use formula acceleration = velocity final - velocity initial divided by time
A = 7 m/s^2
Vf = 60 m/s
Vi =
t = 5 sec
7 = 60 - x/5
Multiply both sides by 5
35 = 60 - x
Subtract 60 from both sides
-25 = -x
x = 25
Vi = 25 m/s
A 925 N crate is being pulled across a level floor
by a force F of 325 N at an angle of 25 above the
horizontal. The coefficient of kinetic friction is
0.25. Find the magnitude of the acceleration of
the crate.
The magnitude of the acceleration of the crate is 1.033 m/s².
The given parameters;
weight of the crate, W = 925 Nhorizontal force applied to the carte, F = 325 Nangle of inclination of the force, θ = 25⁰the coefficient of friction, μ = 0.25The mass of the crate is calculated as;
W = mg
[tex]m = \frac{W}{g} \\\\m = \frac{925}{9.8} \\\\m = 94.388 \ kg[/tex]
The normal force on the crate is calculated as;
Fₙ = 925 - 325 x sin(25)
Fₙ = 925 - 137.35
Fₙ = 787.65 N
The frictional force on the object is calculated as follows;
[tex]F_k = \mu F_n\\\\F_k = 0.25 \times 787.65\\\\F_k = 196.91 \ N[/tex]
The magnitude of the crates acceleration is calculated from the net horizontal force on the crate;
[tex]\Sigma F_x = 0\\\\Fcos (\theta) - F_k = ma\\\\325\times cos(25) \ - \ 196.91 = ma\\\\97.54 = ma[/tex]
[tex]a = \frac{97.54}{94.388} \\\\a = 1.033 \ m/s^2[/tex]
Thus, the magnitude of the acceleration of the crate is 1.033 m/s².
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A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted:
Complete Question
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,
A :
a force repels the glass out of the capacitor.
B :
a force attracts the glass into the capacitor.
C :
no force acts on the glass.
D :
a net charge appears on the glass.
E :
the glass makes the plates repel each other.
Answer:
The correct option is B
Explanation:
Generally when the glass dielectric is slowly inserted between the plated,
The positive plate of the capacitor will induce a negative charge on the glass while the negative plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass
B: a force attracts the glass into the capacitor.
A parallel-plate capacitor, with the air dielectric, is charged from a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates.
How is a dielectric slab inserted in a capacitor?
The dielectric slab is slowly inserted between the plates to a parallel plate capacitor while the capacitor is connected with a battery. As it is being inserted: 57. A parallel-plate capacitor, with air dielectric, is charged by the battery, after which the battery is the disconnected. A slab of glass dielectric is then slowly inserted between the plates. A parallel-plate capacitor, with the air dielectric, is charged from a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates.
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#SPJ2
What is the term for the cumulative body of observations on which scientific explanations are based?
peer review
pseudoscience
logical reasoning
empirical evidence
PLS ANSWER WILL MARK BRANLIEST!!!!!!!!!!!!
Describe the life cycle of a star before it collapses into a black hole.
Describe the life cycle of a star before it becomes a black dwarf.
What is the likely outcome of our sun? *
The sun will supernova and become a black hole.
The sun will swell, encompassing the inner planets and collapses into a dwarf star.
The sun will become a pulsar.
How Do You Know?
P.S. the how do you know is only for the last question
1) describe the life cycle of a star before it collapses into a black hole.
1) describe the life cycle of a star before it collapses into a black hole.ans: A star's life cycle is determined by its mass. The larger its mass, the shorter its life cycle. A star's mass is determined by the amount of matter that is available in its nebula, the giant cloud of gas and dust from which it was born. Over time, the hydrogen gas in the nebula is pulled together by gravity and it begins to spin. As the gas spins faster, it heats up and becomes as a protostar. Eventually the temperature reaches 15,000,000 degrees and nuclear fusion occurs in the cloud's core. The cloud begins to glow brightly, contracts a little, and becomes stable. It is now a main sequence star and will remain in this stage, shining for millions to billions of years to come. This is the stage our Sun is at right now.
2) describe the life cycle of a star before it becomes a dwarf.
ans: The life cycle of a low mass star (left oval) and a high mass star (right oval). ... As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf.
3) what is the likely outcome of our sun?
ans: All stars die, and eventually — in about 5 billion years — our sun will, too. Once its supply of hydrogen is exhausted, the final, dramatic stages of its life will unfold, as our host star expands to become a red giant and then tears its body to pieces to condense into a white dwarf.
A student must design an experiment to determine the relationship between the mass of an object and the resulting acceleration when the object is under the influence of a net force. Which of the following experiments should the student conduct in order to determine the relationship between all three quantities?
Answer choices:
A) Drop objects of different masses from a known height above the ground for multiple trials such that they reach their respective terminal speeds. Use a stopwatch to measure the time it takes each object to reach the ground, and record the mass of each object by using a mass scale.
B) Slide objects of different masses across the same rough surface so that each object travels at a constant speed while under the influence of the force of kinetic friction. Then measure the force required to keep each object at a constant speed by using a force sensor, and record the mass of each object by using a mass scale. Perform this experiment multiple times with objects of different masses.
C) Place an object on a rough surface so that the object is at rest. Use a force sensor to exert a force on the object until just after the object overcomes the force of static friction. Record this force. Repeat the experiment for objects of different masses.
D) Slide an object of known mass across a rough surface, using a constant applied force that can be measured by a force sensor. Place a motion detector behind the object so that its speed can be measured as it slides across the surface. Repeat the experiment for different applied forces.
Answer:
D) Slide an object of known mass across a rough surface, using a constant applied force that can be measured by a force sensor. Place a motion detector behind the object so that its speed can be measured as it slides across the surface. Repeat the experiment for different applied forces.
Explanation:
"The motion detector will provide information about the object’s speed as a function of time as it slides as a result of the applied force. The information about the object’s speed as a function of time can be used to determine the acceleration of the object. The force sensor measures the applied force exerted on the object, and the mass of the object is known. Therefore, this experiment can be used to determine how an object’s mass is related to the net force exerted on the object and the acceleration of the object."
It cannot be A because we need an acceleration will be determined by gravity.
It cannot be B because the term constant speed means that there is no net force, which is required by the initial question.
It cannot be C because the experiment is good for determining the coefficient of friction but not for determining how the mass relates to the acceleration.
It must be D because the object is moving and we have a motion detector, we can graph the acceleration vs time graph. So D allows you to have a lot of the different acceleration values which helps with determining the relationship between acceleration and the mass.
What are the four basics for knowledge called "innate knowledge"?
Answer: The four basics of innate knowledge are as follows:
Explanation:
1. This knowledge is learned from birth and it is inborn knowledge. It is allows the organism to act naturally. For example, a dog is not taught to pant, but it pants to reduce heat from the body.
2. It is inherent.
3. It is essential for survival.
4. It arises from intellectual knowledge rather than being learned via experiences.
Are any of the forces acting on the freezer
unbalanced?
Answer: put them in the microwave
Explanation:
Jumping on a trampoline cause you to fly up in the air. What type of newton’s law is it ?
Answer:
The Third law
Explanation:
For every action there is an equal and opposite reaction.
Answer:
First Law
Explanation:
An object at rest (not moving) will stay at rest unless an unbalanced force acts on it.
An object in motion will stay in motion (in a straight line and at a constant speed) unless an unbalanced force acts on it.
You jump down on a trampoline and fly up in the air as a result.
2. A boy walks 3 miles north then turns around and walks on the same path
south for 2.1 miles. What is the boy's displacement?
Answer:
.9
Explanation:
3-2.1=.9
A small compass is held horizontally, the center of its needle has a distance of 0.270 m directly north
of a long wire that is perpendicular to the Earth's surface. When there is no current in the wire, the
compass needle points due north, which is the direction of the horizontal component of the Earth's
magnetic field at that location. This component is parallel to the Earth's surface. When the current in
the wire is 26.3 A, the needle points 22.9∘ east of north.
(a) Does the current in the wire flow toward or away from the Earth's surface? ( 2 marks)
(b) What is the magnitude of the horizontal component of the Earth's magnetic field at the location of
the compass? (3 marks)
Answer:
Explanation:
The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .
a )
Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .
b )
magnetic field created at the magnetic needle B = 10⁻⁷ x 2I / d where I is current and d is distance .
B = 10⁻⁷ x 2 x 26.3 / .27
= 194.81 x 10⁻⁷ T
angle of deflection of solenoid = 22.9°
Tan 22.9 = B /H
.422 = 194.81 x 10⁻⁷ / H
H = 461.63 x 10⁻⁷ T
= .46 x 10⁻⁴ T .
A) The current in the wire flows towards the Earth's surface
B) The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
A) The compass needle held horizontally points in a North-south direction of the earth and also deflects eastwards when current is allowed to flow through it. The deflection of the needle indicates the presence/generation of a magnetic field on the earth surface. which is facilitated by the flow of the current in the wire towards the Earth's surface
B) Determine The magnitude of the horizontal component of the Earth's magnetic field
B ( magnetic field ) = 10⁻⁷ * 2I / d ---- ( 1 )
where : l = 26.3 A, d = 0.27 m
Back to equation ( 1 )
B = 10⁻⁷ * 2 * 26.3 / 0.27
= 194.81 * 10⁻⁷ T
Final step : Calculate the magnitude of horizontal component ( H )
Tan ∅ = B / H ---- ( 2 )
where : ∅ ( angle of deflection ) = 22.9°
∴ H = B / Tan ( 22.9° )
= ( 194.81 * 10⁻⁷ ) / 0.422
= 0.46 x 10⁻⁴ T
Hence we can conclude that The current in the wire flows towards the Earth's surface and The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
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Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Answer:
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
When a stretched rubber band snaps back it has increase
thermal energy.
potential energy.
kinetic energy.
elastic energy.
Answer:
Kinetic
Explanation:
Because it has a force pushing on it
Anything that takes up space and has mass is
a. Volume
b. Density
C. Matter
d. Viscous
Answer:
materia
Explanation:
Answer:
the answer to your question is matter
A box falls out of an airplane that is traveling horizontally at 100m/s. The plane is at an altitude of 300m.
Where does the box land relative to where it was dropped from?
Answer:
782.461 m
Explanation:
[tex]a_{x}=0[/tex] [tex]v_{xo}=100[/tex] [tex]a_{y}=-g[/tex] [tex]v_{yo}=0[/tex]
X-direction | Y-direction
[tex]x=x_{o}+v_{xo}t[/tex] | [tex]y=y_{o}+v_{yo}t+\frac{1}{2} a_{y} t^2[/tex]
[tex]x=100(7.82461)[/tex] | [tex]0=300-\frac{1}{2}(9.8)t^2[/tex]
x= 782.461 m | [tex]-300=-4.9t^2[/tex]
| [tex]\sqrt{\frac{-300}{-4.9} } =\sqrt{t^2}[/tex]
| [tex]7.82461=t[/tex]
The box is falling due the attraction of gravitational force (its weight) and it lands at approximately 782.1 meters further from where it was dropped
The reason why the above value of the location is correct is given as follows:
The known parameter of the box are;
Horizontal velocity of the box, vₓ = 100 m/s
The altitude of the plane, h = 300 m
Required:
The location where the box lands
Solution:
The time it takes the box to land is [tex]t =\sqrt{\dfrac{2 \cdot h}{g} }[/tex]
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]t =\sqrt{\dfrac{2 \times 300}{9.81} } \approx 7.821[/tex]
The time it takes the box to reach the ground, t ≈ 7.82 seconds
Horizontal distance covered by the by the box during free fall, d, = The distance the box lands relative to where it was dropped from and is given as follows;
d = vₓ × t
∴ d ≈ 100 m/s × 7.821 s = 782.1 m
Therefore;
The distance the box lands relative to where it was dropped from d ≈ 782.1 m
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Which of the following is a homogeneous mixture?
a.
salad
b.
vegetable soup
c.
salt water
d.
oil & water
Explanation: (Two liquids can have a very close boiling point) So it will be salt water. I think :C
6) The magnitude of the force the Sun exerts on Uranus is 1.41 x 1021 newtons. Explain how it is possible for the Sun to exert agreater force on Uranus than Neptune exerts on Uranus.
Answer and Explanation:
TL: DR The Sun is much more massive than Neptune — more than enough to make up for the somewhat smaller distance between the two planets at the closest approach.
[The surprise in this answer (to me, a non-astronomer), is that the gap between the orbits of Neptune and Uranus is large — half the distance from Uranus to the Sun.]
The ratio of gravitational attraction of the Sun on Uranus versus Neptune on Uranus is directly proportional to the ratio of the Sun’s mass to Neptune’s and inversely proportional to the ratio of the square of the distances (let’s use the closest approach of the two planets to one another to calculate a maximum attraction).
Numbers:
Sun’s mass: 2 x 10^30 kg
Neptune’s mass: 1 x 10^26 kg
Distance of Sun to Uranus: 3 x 10^9 km
Closest approach of Uranus and Neptune: 1.5 x 10^9 km
Without doing any arithmetic, we see that even at their closest approach, Uranus and Neptune are separated by about one-half of the Uranus to Sun distance. Squaring that ratio, we see that if the Sun and Neptune had the same mass, the attraction between the Sun and Uranus would only be about 1/4 of that between the Sun and Neptune; however, the Sun has 20000 times the mass of Neptune, so the attraction between Uranus and the Sun is about 5000 times stronger than the maximum attraction between Uranus and Neptune.
The explanation of the possibility of why sun exerts a greater force on Uranus than Neptune exerts on Uranus is; because the force was calculated to be greater.
The formula for calculating the Force of Gravity between two masses is:
F = G*m₁*m₂/r²
Where;
F = force of gravity
G = gravitational constant = 6.674 × 10⁻¹¹ N•m²/kg²
m₁ = mass of the larger object
m₂ = mass of the smaller object
r = the distance between the centers of the two masses
Now, from online values, we have the following;
mass of Neptune; m₁ = 102.413 × 10²⁴ kg
mass of Uranus; m₂ = 86.813 × 10²⁴ kg
average distance between the centers of Neptune and Uranus; r = 1.62745 × 10¹² m
Thus, force exerted by Neptune on Uranus is;
F = (6.674 × 10⁻¹¹ × 102.413 × 10²⁴ × 86.813 × 10²⁴)/(1.62745 × 10¹²)²
F = 2.240 × 10¹⁷ N
We are told that the force the Sun exerts on Uranus is 1.41 the force the Sun exerts on Uranus is 1.41 × 10²¹ N.
That is greater than the force Neptune exerts on Uranus.
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A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid cylindrical hub of mass 0.0055 kg and diameter 0.011 m . Part A Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest. Express your answer using three significant figures and include the appropriate units.
Answer: IM 95%sure that the answer is B jus took the test got the answer right
Explanation:
Answer:
sorry I forgot I wish I could help
A researcher is interested in exploring the effects of exposure to ultraviolet light on people’s emotional state. He divides his participants into three groups. The first group receives 40 minutes of ultraviolet light 3 times a week, the second group receives 20 minutes of ultraviolet light 3 times a week, and the third group receives no ultraviolet light. Tell me which group(s) were in the control group, and which were in the experimental group. (You can simply delete the incorrect answer for each group. So, for example, if you think the answer to group one is "control," you can simply delete the word "experimental" after group one)
Group One: Experimental Control
Group Two: Experimental Control
Group Three: Experimental Control
Answer:
Explanation:
In an experimental research, the control group is the group that serves as the neutral group that is not given any form of treatment and serves as the group in which the experimental groups are firstly compared to. Thus, the control group in the question described is the Third group.
While experimental groups are the groups that receive treatments required to make an inference from the experiment. From this description, it can be deduced that the First and the Second group are the experimental groups.
a car is moving eastward and speeding up. the momentum of the car is
cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?
Answer:
Displacement is 3 km North
Explanation:
Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?
Answer:
We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Explanation:
LIFTING:
When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.
PUSHING:
When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.
Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
A trebuchet launches a pumpkin at an angle of 63 degrees at an initial velocity of 51 m/ s from the ground. What is the range of the pumpkin?
Answer:
214.72m
Explanation:
The range (R) of a projectile can be calculated using the formula:
R = u²sin2θ/g
Where R = Horizontal range
u = initial velocity
θ = angle of initial velocity
g = acceleration due to gravity (9.8m/s²)
According to the provided information in this question, R= ?, u= 51m/s, θ = 63°, g = 9.8m/s².
Hence, R = u²sin2θ/g
R = 51² × sin 2×63/ 9.8
R = 2601 × sin 126/9.8
R = 2601 × 0.809/9.8
R = 2104.25/9.8
R = 214.72m
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by
Answer:
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
Explanation:
Let [tex]\vec A = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})[/tex] and [tex]\vec B = 4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex], both measured in meters. The resultant vector [tex]\vec R[/tex] is calculated by sum of components. That is:
[tex]\vec R = \vec A+\vec B[/tex] (Eq. 1)
[tex]\vec R = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})+4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex]
[tex]\vec R = (6\cdot \cos 30^{\circ}-4\cdot \sin 30^{\circ})\,\hat{i}+(6\cdot \sin 30^{\circ}-4\cdot \cos 30^{\circ})\,\hat{j}[/tex]
[tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex]
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Write a general expression for the force that the spring exerts on the mass, in term of k and x. Choose the initial position of the front of the spring to be xi
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]F_{spring} = - k * x_f[/tex]
Explanation:
Generally the force exerted on the mass by the spring is mathematically represented as
[tex]F_{spring} = - k * \Delta x[/tex]
Here the negative sign shows that the force is acting in the negative x axis
k is the spring constant
and [tex]\Delta x[/tex] is the distance through which the spring moves the mass which is mathematically evaluated as
[tex]\Delta x = x_f -x_i[/tex]
=> [tex]F_{spring} = - k * [ x_f -x_i][/tex]
Given that [tex]x_i = 0 \ m[/tex]
[tex]F_{spring} = - k * x_f[/tex]
Increase in potential energy of spring