Answer:
165.8 V/m
Explanation:
The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)
Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)
V = Q/C
V = Q ÷ 2πε₀L/㏑(b/a)
V = Q㏑(b/a)/2πε₀L
So, the potential difference per unit length V' is
V' = V/L = Q㏑(b/a)/2πε₀
Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have
V' = Q㏑(b/a)/2πε₀
V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)
V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)
V' = 0.16575 × 10³ V/m
V' = 165.75 V/m
V' ≅ 165.8 V/m
A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2850 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s
Answer:
0.1 s
Explanation:
The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log
So F - f = ma
F - μmg = ma
F/m - μg = a
So, substituting the values of the variables into the equation, we have
a = F/m - μg
a = 2850 N/300 kg - 0.45 × 9.8 m/s²
a = 9.5 m/s² - 4.41 m/s²
a = 5.09 m/s²
Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.
So, making t subject of the formula, we have
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (v - u)/a
t = (0.5 m/s - 0 m/s)/5.09 m/s²
t = 0.5 m/s ÷ 5.09 m/s²
t = 0.098 s
t ≅ 0.1 s
BRAINLY PLS HELP ME!!!
Should the Us government regulate sugar? In once sentence write down what you are preparing to argue or what stance you are going to take. This will help you to create a starting point for your idea
Answer:
yes
Explanation:
Sugar can cause health problems.
I believe that sugar should be regulated by the government since it causes heath problems. As according to this website heathline.com that up to one third of the population in America is obese. As well as that regulating sugar will greatly help lowering that number, seeing that many food products have tons of sugar in them. By lowering the sugar and regulating it can cause less obesity among the average person in the US. As stated by HeathlyFoodAmerca.org consuming too much sugar can cause increased heath problems as shown, heart disease, diabetes, and teeth decay and this is why I believe that the U.S government should regulate sugar more.
You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.
Answer:
exactly 1273 N whether or not the box moves.
Explanation:
In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction
Therefore as per the given situation, the above represent the answer
Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W
Compute the work done on the table:
W = Fd = (320 N) (32 m) = 10,240 J
Divide this by the given time duration to get the power output:
P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W
what is the frequency of a wave related to
Answer:
Frequency is the number of complete oscillations or cycles or revolutions made in one second.
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 1.8 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 28.0 m/s when they reach the bottom of the ramp. You determine that for a 75kg skier with good form, friction and air resistance will do total work of magnitude 3500 J on him during his run down the slope. What is the maximum height (h) for which the maximum safe speed will not be exceeded?
Answer:
44.6 m
Explanation:
From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.
E = E'
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J
Substituting the values of the variables into the equation, we have
U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂
mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J
9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J
(735 kgm/s²)h + 75 kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J
(735 kgm/s²)h + 121.5 kgm²/s² = 29400 kgm²/s² + 3500 J
(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J
(735 kgm/s²)h + 121.5 J = 32900 J
(735 kgm/s²)h = 32900 J - 121.5 J
(735 kgm/s²)h = 32778.5 J
h = 32778.5 J/735 kgm/s²
h = 44.6 m
So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.
g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.
Answer:
[tex]V=8.08m/s[/tex]
Explanation:
From the question we are told that:
Height[tex]h=5.00m[/tex]
Mass [tex]m=0.750kg[/tex]
Radius [tex]r=4.00cm=>0.04m[/tex]
Generally the equation for Total energy is mathematically given by
[tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]
Therefore
[tex]V=\sqrt{\frac{4gh}{3}}[/tex]
[tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]
[tex]V=8.08m/s[/tex]
2) A Ship has an area of cross - section at the water line of 2000m²
(a) By what deck does the ship sink in fresh.
water, when it loads a cargo of 4000 tonnes
(b) if the ship + Cargo has a displacement tonnage
of 12300 tonnes; by what amount will the ship
rise in the water when it sails from fresh water
into Seawater (density of Sea water - 1025kgm-²
Answer:
a. 2 m
b. 0.15 m
Explanation:
(a) By what deck does the ship sink in fresh.
water, when it loads a cargo of 4000 tonnes
We know that the upward force , U on the ship equal the weight of fresh water displaced, W = ρVg where ρ = density of fresh water = 1000 kg/m³, V = volume of water displaced and g = acceleration due to gravity.
So, U = W = ρVg
This upward force must equal the weight of the ship, W' = mg where m = mass of ship = 4000 tonnes = 4000 × 1000 kg = 4 × 10⁶ kg
So. U = W'
ρVg = mg
V = m/ρ = 4 × 10⁶ kg/10³ kg/m³ = 4 × 10³ m³
Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in water h.
So, V = Ah
Thus h = V/A = 4 × 10³ m³/2000 m³ = 2 m
So, the ship sinks to a depth of 2 m in fresh water.
(b) if the ship + Cargo has a displacement tonnage
of 12300 tonnes; by what amount will the ship
rise in the water when it sails from fresh water
into Seawater (density of Sea water - 1025kgm⁻³
We know that the upward force , U' on the ship in fresh water equals the weight of fresh water displaced, W" = ρV'g where ρ = density of fresh water = 1000 kg/m³, V' = volume of water displaced and g = acceleration due to gravity.
So, U = W" = ρV'g
This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg
So. U' = W₁
ρV'g = m₁g
V' = m₁/ρ = 1.23 × 10⁷ kg/10³ kg/m³ = 1.23 × 10⁴ m³
Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in fresh water h'.
So, V' = Ah'
Thus h = V'/A = 1.23 × 10⁴ m³/2000 m³ = 6.15 m
So, the ship sinks to a depth of 0.6 m in fresh water.
Also,
We know that the upward force , U" on the ship in sea water equals the weight of sea water displaced, W₂ = ρ'V"g where ρ' = density of sea water = 1025 kg/m³, V"= volume of water displaced and g = acceleration due to gravity.
So, U" = W₂ = ρ'V"g
This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg
So. U" = W₁
ρ'V"g = m₁g
V" = m₁/ρ' = 1.23 × 10⁷ kg/1.025 × 10³ kg/m³ = 1.2 × 10⁴ m³
Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in sea water h".
So, V" = Ah"
Thus h" = V'/A = 1.2 × 10⁴ m³/2000 m³ = 6 m
So, the ship sinks to a depth of 6 m in fresh water.
So, the rise in height from fresh water to sea water is Δh = h' - h" = 6.15 m - 6 m = 0.15 m
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff?
Answer:
144 meters
Explanation:
it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .
so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144
If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?
Answer:
72 is the premimum of the insurance.
Explanation:
Below is the given values:
The loading = 0.4
Coinsurance rate = 0.2
Number of units = 100
Total number of units = 100 * 0.4 = 40
Remaining units = 60 * 0.2 = 12
Add the 60 and 12 values = 60 + 12 = 72
Thus, 72 is the premimum of the insurance.
A cat pushes a porcelain statue off a bookshelf with a speed of 0.5 m/s and it smashed on the floor 0.85 sec later.
Answer:
167?
Explanation:
i added both
A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices
Answer:
Points away from the disk along the z-axis.
Explanation:
Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.
Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.
Two point charges exert a 6.10 N force on each other. What will the force become if the distance between them is increased by a factor of 8
Answer:
0.0953125 N
Explanation:
Applying,
F = kq'q/r²................. Equation 1
Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.
From the equation,
If both charges remain constant,
Therefore,
F = C/r²
C = Constant = product of the two charge(q' and q) and k
Fr² = F'r'²................ Equation 2
From the question,
Given: F = 6.10 N
Assume: r = x m, r' = 8x
Substitute these value into equation 2
6.1(x²) = F'(8x)²
F' = 6.1/64
F' = 0.0953125 N
Hence the new force will become 0.0953125 N
A remote ranch has a cylindrical water storage tank. It has a vertical central axis, a diameter of 24 ft, the sides are 5 ft high. The depth of this water is 4 ft. How much work (in ft-lb) would be required to pump all of the water over the upper rim?
Answer:
Explanation:
From the given information:
The diameter of the pool = 24 ft
The radius will be = 24 ft/2 = 12 ft
The volume of water V = πr²(Δx)
V = π× 12²×(Δx)
V = 144π(Δx)
Le's assume water weighs 62.5 lb/ft³;
Then:
the Force (F) will be:
= 144π(Δx) * 62.5
= 9000πΔx lb
Also, the side of cylindrical water = 5 ft while its depth = 4ft
As such, each slide of water d = 5 - x, and the region is between 0 and 4.
∴
The required work is:
[tex]W = \int^4_0 (5-x) 9000 \ \pi dx \\ \\ W = 9000 \int^4_0 (5-x) \ dx \\ \\ W = 9000 \pi \Big [5x - \dfrac{x^2}{2} \Big]^4_0 \\ \\ W = 9000 \pi \Big [5*4- \dfrac{4^2}{2} \Big] \\ \\ W = 9000 \pi \Big [20-8 \Big] \\ \\ W = 9000 \pi \Big [12 \Big] \\ \\ W = 9000 \pi (12) \\ \\ \mathbf{W = 108000 \pi \ ft.lb}[/tex]
PLEASE ANSWR 1ST AND I WILL MARK U BRAINLIEST
Two statements are given- one labeled Assertion (A) and the other labeled Reason ®. Select
the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
a. Both A and R are true, and R is correct explanation of the assertion.
b. Both A and R are true, but R is not the correct explanation of the assertion.
c. A is true, but R is false.
d. A is false, but R is true.
Assertion: An object has a negative acceleration.
Reason: The velocity of an object decreases in the same direction.
Answer:
Where is the R statement?
How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *
Answer:
you drove 50km
Explanation:
10×5 hope this helps
Answer:
50 Km
Explanation:
This is how far you have got on your journey if traveling like this.
Please Mark as Brainliest
Hope this Helps
What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs
An electric field is given by in units of N/C when is in meters. What is the potential difference from point B at (0,7) m to point A at (7,0) m
Complete Question
An electric field is given by E= (5x, 0, 0) in units of N/C when x is in meters. What is the potential difference VA - VB from point B at (0,7) m to point A at (7,0) m? V
Answer:
[tex]V_A-V_B=-122.5V[/tex]
Explanation:
From the question we are told that:
Position of point [tex]B=(0,7) m[/tex]
Position of point [tex]A=(7,0) m[/tex]
Generally the equation for pd across the points is mathematically given by
[tex]V_A-V_B=-\int_0^7(5x)d[/tex]
[tex]V_A-V_B=-[\frac{5x^2}{2}]_0^7[/tex]
[tex]V_A-V_B=-[\frac{5(7)^2}{2}][/tex]
[tex]V_A-V_B=-122.5V[/tex]
Spin cycles of washing machines remove water from clothes by producing a large radial acceleration at the rim of the cylindrical tub that holds the water and clothes. Suppose that the diameter of the tub in a typical home washing machine is 50 cm.
Required:
What is the rotation rate, in rev/min, of the tub during the spin cycle if the radial acceleration of points on the tub wall is 3g?
Answer:
The rate of rotation is 1123.6 rpm.
Explanation:
diameter of tub =50 cm
radius of tub, r = 25 cm = 0.25 m
Acceleration, a = 3 g
The centripetal acceleration is given by
[tex]a = r w^2\\\\3\times 9.8 =0.25 \times w^2\\\\w = 117.6 rad/s\\\\w = 2\pi n \\\\117.6 = 2\times 3.14 \times n\\\\n = 18.73 rps = 1123.6 rpm[/tex]
what units of measurement measures both velocity and speed
Answer:
[tex]metre \: per \: second[/tex]
Explanation:
Velocity is a derived quantity and the S.I unit is metre per second.Speed is also a derived quantity which is has the S.I unit to be metre per second.
A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)
Answer:
The force between the 10 th car and the 11 th car is 13636.4 N.
Explanation:
Force, F = 150 kN
acceleration, a = 2 m/s^2
Let the mass of each car is m. \Total numbers of cars = 11
F = n m a
150000 = 11 x m x 2
m = 6818.18 kg
The force between the 10 th and 11 th car is
T = ma = 6818.18 x 2 = 13636.4 N
A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water
Answer:
The speed of the swimmer in stil water is 0.5 m/s
Explanation:
Given;
total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s
width of the river, = 200 m
Please find the image attached for explanation.
differentiate between step up and step down transformer
Answer:
The main difference between the step-up and step-down transformer is that the step-up transformer increases the output voltage, while the step-down transformer reduces the output voltage.
Given that o.2i+bj+o.4k is a unit vector,what is the value of b?
Answer:
b = 0.89
Explanation:
The given vector is, [tex]A=0.2i+bj+0.4k[/tex]
A is a unit vector
We need to find the value of b.
For a unit vector, |A| = 1
So,
[tex]0.2^2+b^2+0.4^2=1\\\\0.04+b^2+0.16=1\\\\0.2+b^2=1\\\\b^2=1-0.2\\\\b=0.89[/tex]
So, th value of b is 0.89.
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time
Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg
assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K
Answer:
T = 2398 K
Explanation:
To calculate the emission of the light bulb we use the law is Stefan
P = σ A e T⁴
as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)
T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]
let's calculate
T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]
T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]
T = 2,398 10³ K
T = 2398 K
Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by Icm= 1/2mL^2.
Required:
Find Icmd the moment of inertia of the rod with respect to a parallel axis through one end of the rod.
Answer:
right now I see some of you have a great day
Answer: (mI^2)/3
Explanation:
The parallel axis theorem for the calculation of inertia is: I = I CM + Md^2
So, I is the apathy from an axis that is at distance d from the center of mass and LCM the apathy when the axis passes through the center of mass. Do to this, the axis passes through the end of the rod. In analysis, d=l/2
So, we have the equation:
I = mI^2/12 + m (1/2)^2 = mI^2/12 + mI^2/4 = mI^2/12 + 3mI^2/12 = mI^2/3
This relents us the terminal result: (ml^2)/3.
Problem
A charged particle is moving in the presence of uniform magnetic field. The mass of the particle
is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the
beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).
a.) How large will the x component of the velocity of the particle be in t = 2 s?
b.) Where will the particle be in t = 3.14 s?
c.) How large will the magnitude of the velocity be in t = 2.5 s?
Answer:
Answer is a I checked the work
Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Force and power. Part A How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hphp
Answer:
The number of crates is 84580.
Explanation:
mass, m = 30 kg
height, h = 0.9 mm
Power, P = 0.5 hp = 0.5 x 746 W = 373 W
time, t = 1 minute = 60 s
Let the number of crates is n.
Power is given by the rate of doing work.
[tex]P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580[/tex]
To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity.
Required:
a. A plane accelerates from rest at a constant rate of 5.00m/s^2 along a runway that is 1800m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off?
b. What is the distance dfirst traveled by the plane in the first second of its run?
c. What is the distance dfirst traveled by the plane in the first second of its run?
Answer:
(a)
67.1 s
(b) 2.5 m
(c) 2.5 m
Explanation:
initial speed, u = 0 m/s
final speed, v = 70 m/s
acceleration, a = 5 m/s2
distance, s = 1800 m
(a) Use third equation of motion
[tex]v^2= u^2 + 2 a s \\\\v^2 = 0 + 2 \times 5\times 1800\\\\v =134.2 m/s[/tex]
Let the time is t.
Use first equation of motion
v = u + at
134.2 = 0 + 5 t
t = 67.1 s
(b) Use second equation of motion
[tex]s = u t +0.5 at^2\\\\s = 0 +0.5\times 5\times 1 \\\\s = 2.5 m[/tex]
(c) Use second equation of motion
[tex]s = u t +0.5 at^2\\\\s = 0 +0.5\times 5\times 1 \\\\s = 2.5 m[/tex]
