Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
Why would physics be used to study light emitted by a star?
O A. Stars form interesting shapes in the sky.
B. Light is very pretty.
O C. The positions of stars control our lives.
O D. Light is a form of energy.
Answer:
O D.
Explanation:
Physics has an aspect that deals with the study of energy
Answer:
D. Light is a form of energy
Explanation:
what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
[tex]F_{e}[/tex] - W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.
Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑ [tex]q_{i} / r_{i}[/tex]
where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?
Answer:
The glider’s speed after the skydiver lets go is 26 m/s
Explanation:
To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum
Mathematically;
mv = mv + mv
so 680 * 26 = (680-60)v + 60 * 26
17680 = 620v + 1560
17680-1560 = 620v
16120 = 620v
v = 16120/620
v = 26 m/s
An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
Find out more information about Force here:
https://brainly.com/question/25239010
A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.
Required:
a. How much work is done by the tension?
b. How much work is done by gravity?
The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
The given parameters;
mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 NThe work done by the tension is calculated as follows;
W = Fd
W = 15 x 0.15
W = 2.25 J
The work done by gravity is calculated as;
W = (25 x 9.8) x 0.15
W = 36.75 J
Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
Learn more here: https://brainly.com/question/19498865
Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration
Answer:
The average velocity for this trip is 0 km/hr
Explanation:
We know that average velocity = total displacement/total time.
Now, its displacement is d = final position - initial position.
Since the car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.
So, its displacement is d = 0 km - 0 km = 0 km.
Since the total time for the round trip is 2 hours, the average velocity is
total displacement/ total time = 0 km/2 hr = 0 km/hr.
So the average velocity for this trip is 0 km/hr
Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?
Answer:
0.46N
Explanation:
See attached file
A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N
The velocity and force are required.
The speed of the racket is 8.7 m/s
The required force is 471.43 N.
[tex]m_1[/tex] = Mass of racket = 1000 g
[tex]m_2[/tex] = Mass of ball = 60 g
[tex]u_1[/tex] = Initial velocity of racket = 12 m/s
[tex]u_2[/tex] = Initial velocity of ball = -15 m/s
[tex]v_1[/tex] = Final velocity of racket
[tex]v_2[/tex] = Final velocity of ball = 40 m/s
[tex]\Delta t[/tex] = Time = 7 ms
The equation of the momentum will be
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]
Force is given by
[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]
Learn more:
https://brainly.com/question/19689434?referrer=searchResults
The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]
Explanation:
Given: Speed of red light = 700 nm
= [tex]700\times10^{-9}[/tex] m
[tex]= 7\times10^{-7}[/tex] m
Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]
Speed of light = [tex]3\times10^8[/tex] m
Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]
[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]
Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]
The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].
You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d = [tex]\frac{W}{F}[/tex]
d = [tex]\frac{3}{9}[/tex]
d = 0.33 m
The distance d the block ice moved is 33 cm.
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
Zack is driving past his house. He wants to toss his physics book out the window and have it land in his driveway. If he lets go of the book exactly as he passes the end of the driveway. Should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)? Explain.
Answer:
Zack should direct his throw outward and toward the back of the car.
Explanation:
As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.
The solution is throw 3.
I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
Which statement best applies Newton’s laws of motion?The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.
When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.
The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.
Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
Learn more about skydiver on:
https://brainly.com/question/29253407
#SPJ6
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m 3 kg/m3 , length 81.2 cmcm and diameter 2.60 cmcm from a storage room to a machinist. Calculate the weight of the rod, www. Assume the free-fall acceleration is ggg = 9.80 m/s2m/s2 .
Answer:
The weight of the rod is 32.87 N
Explanation:
Density of the rod = 7800 kg/m
length of the rod = 81.2 cm = 0.812 m
diameter of rod = 2.60 cm = 0.026 m
acceleration due to gravity = 9.80 m/s^2
The rod can be assumed to be a cylinder.
The volume of the rod can be calculated as that of a cylinder, and can be gotten as
V = [tex]\frac{\pi d^{2} l}{4}[/tex]
where d is the diameter of the rod
l is the length of the rod
V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3
We know that the mass of a substance is the density times the volume i.e
mass m = ρV
where ρ is the density of the rod
V is the volume of the rod
m = 4.3 x 10^-4 x 7800 = 3.354 kg
The weight of a substance is the mass times the acceleration due to gravity
W = mg
where g is the acceleration due to gravity g = 9.80 m/s^2
The weight of the rod W = 3.354 x 9.80 = 32.87 N
21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
I got it right on my quiz!
I hope this helps!! :))
A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm. The interference pattern on the screen 4.2 m away shows the first maximum 5.5 mm from the center of the pattern. What is the wavelength of the light in nm
Answer:
Explanation:
The double slit interference phonemene is described for the case of constructive interference
d sin θ= m λ (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
The double slit interference phonemene is described for the case of constructive interference
d sin θ = m lam (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in equation 1
d y / L = m λ
λ = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 mm = 0.44 10⁻³ m
y = 5.5 mm = 5.5 10⁻³ m
L = 4.2m
m = 1
let's calculate
λ = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
λ = 5.76190 10-7 m
let's reduce to num
lam = 5.56190 10-7 m (109 nm / 1m)
lam = 556,190 nmtea
we substitute
without tea = y / L
we substitute in equation 1
d y / L = m lam
lam = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 me = 0.44 10-3 m
y = 5.5 mm = 5.5 10-3
L = 4.2m
m = 1
let's calculate
lam = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
lam = 5.76190 10⁻⁷ m
let's reduce to num
lam = 5.56190 10⁻⁷ m (109 nm / 1m)
lam = 556,190 nm
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
please brainliest!!!
Explanation:
V1/√T1 =V2/√T2
V1 = 331m/s
T1 = 0°C = 273k
V2 = ?
T2 = 35°c = 308k
331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/sThe copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
To work on your car at night, you use an extension cord to connect your work light to a power outlet near the door. How would the illumination provided by the light be affected by the length of the extension cord
Answer:
The longer the cord, the lower the illumination
Explanation:
The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.
Long wires have more electrical resistance than shorter ones.
Let us consider this formula:
Resistance =[tex]\frac{\rho L}{A}[/tex]
From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.
What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?
Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
λ = 5.2 x 10⁻⁷ m = 520 nm
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file
A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
Learn more about force here https://brainly.com/question/25239010
#SPJ2
To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface
Answer:
Explanation:
jjjjjjjjjjjjjjjj