Answer:
7.91 m/s
Explanation:
Speed: This can be defined as the ratio of this to time. Speed can also be the magnitude of a velocity or speed is velocity without direction.
The S.I unit of speed is m/s.
From the question,
K.E = 1/2(mv²)................ Equation 1
Where K.E = Kinetic Energy, m = mass of the runner, v = velocity of the runner.
make v the subject of the equation
v = √(2K.E/m).................Equation 2
Given: K.E = 1.09×10³ J, m = 34.8 kg.
Substitute into equation 2
v = √(2× 1.09×10³/34.8)
v = √(62.64)
v = 7.91 m/s.
Hence the speed of the runner = 7.91 m/s
A metal ring 4.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.200 T/s.
(a) What is the magnitude of the electric field induced in the ring?
(b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
Explanation:
a) d[phi]/dt = (dB/dt)*Acos(0) = (-0.20)*(pi(2.25*10^-2)^2) = -3.98*10^-4 Wb
E = (1/2r*pi)*(d[phi]/dt) = -2.8*10^-3 N/C
b) Clockwise because The induced magnetic field will be in the direction to oppose the change. Since the magnetic flux from the magnets is decreasing, the induced magnetic field will be in the same direction as the magnet's field.
Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in opposite directions.
(d) Calculate the change in kinetic energy of the system, in joules, in this case.
Answer:
a) 6.57 m/s
b) 53.75 J
c) 6.37 m/s
d) -98.297 J
Explanation:
mass of player = [tex]m_{p}[/tex] = 117.5 kg
speed of player = [tex]v_{p}[/tex] = 6.5 m/s
mass of ball = [tex]m_{b}[/tex] = 0.43 kg
velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s
Recall that momentum of a body = mass x velocity = mv
initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s
initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s
initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] = 2482.187 J
a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.
for this first case that they travel in the same direction, their momenta carry the same sign
[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
where v is the final velocity of the player.
inserting calculated momenta of ball and player from above, we have
763.75 + 11.395 = (117.5 + 0.43)v
775.145 = 117.93v
v = 775.145/117.93 = 6.57 m/s
b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J
change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained
c) if they travel in opposite direction, equation becomes
[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
763.75 - 11.395 = (117.5 + 0.43)v
752.355 = 117.93v
v = 752.355/117.93 = 6.37 m/s
d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex] = 2383.89 J
change in kinetic energy = 2383.89 - 2482.187 = -98.297 J
that is 98.297 J lost
Damon purchased a pair of sunglasses that were advertised as being polarized. Describe how Damon could test the sunglasses to verify they are polarized.
Answer:
To verify that they're polarized, he could hold the two lenses perpendicular (90 degrees) to each other, one lens in front of the other, and point it at a light source. If no light passes through then the lenses are polarized
The test of Polarization of pair of sunglasses is , hold the two lenses perpendicular to each other, one lens in front of the other, and point it towards a light source. If no light passes through then the lenses are polarized.
When a beam of light is reflected from a smooth surface, such as water or ice, it becomes polarized.Polarized light irritates the eyes and makes it hard to see clearly.For example, when fishing on a sunny day, you wouldn't see through the water. You would only see a reflection of the sun hitting the water.
Polarized lenses will neutralize the reflection of the water, and you will be able to into the water.To verify that pair of sunglasses are polarized, he could hold the two lenses perpendicular to each other, one lens in front of the other, and point it towards a light source. If no light passes through then the lenses are polarized.
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A stellar object is emitting radiation at 3.55 mm. If a detector is capturing 3.2×108 photons per second at this wavelength, what is the total energy of the photons detected in 1.0 hour?
Answer:
E = 6.45 x 10⁻¹¹ J
Explanation:
First we need to find total number of photons detected in 1 hour. Therefore,
No. of Photons = n = (3.2 x 10⁸ photons/s)(1 h)(3600 s/1 h)
n = 11.52 x 10¹¹ photons
Now, the energy of these photons can be given by the formula:
E = nhc/λ
where,
E = Total Energy of the Photons = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of radiation = 3.55 mm = 3.55 x 10⁻³ m
Therefore,
E = (11.52 x 10¹¹)(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(3.55 x 10⁻³ m)
E = 6.45 x 10⁻¹¹ J
Suppose that 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm. (a) How much work is needed to stretch the spring from 41 cm to 45 cm? (Round your answer to two decimal places.) J (b) How far beyond its natural length will a force of 15 N keep the spring stretched? (Round your answer one decimal place.)
Answer:
Explanation:
Work done on a spring is expressed as [tex]W = 1/2 ke^{2}[/tex]
k is the elastic constant
e is the extension of the material
If 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm, then;
Work done = 4J and the extension e = 47 cm - 36 cm; e = 11 cm
11cm = 0.11m
Substituting the given values into the equation above to get the elastic constant;
[tex]W = 1/2 ke^{2}\\4 = 1/2k(0.11)^{2} \\8 = 0.0121k\\k = 8/0.0121\\k = 661.16N/m[/tex]
a) In order to determine the amount of work needed work is needed to stretch the spring from 41 cm to 45 cm, wre will use the same formula as above.
[tex]W = 1/2ke^{2} \\e = 0.45 - 0.41\\e = 0.04 m\\ k = 661.16N/m[/tex]
[tex]W = 1/2 * 661.16 * 0.04^{2} \\W = 330.58*0.0016\\W = 0.53J (to\ 2d.p)[/tex]
b) According to hooke's law, F = ke where F is the applied force
We are to get the extension when a force of 15N is applied to the original length of the material.
e = F/k
e = 15/661.16
e = 0.02 m (to 1 d.p)
This means that the natural length of the spring will be stretched by 0.02 m when a force of 15N is applied to it.
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Given data :
Landing speed of Jet = 200 km/h
Distance = 425 m
Total mass of aircraft = 140 Mg with mass center at G
Determine the reaction N under the nose of wheel B First step : calculate the value of the Jet accelerationJet acceleration = 1 / (2 *425) * (200² - 60² ) * 1 / (3.6)²
= 3.3 m/s²
Next step : determine the reaction N under the nose of WheelReaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140). ----- ( 1 )
∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )
Hence Reaction N = 257 KN
We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
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A horizontal uniform meter stick is supported at the 50.0 cm mark. It has a mass of 0.52 kg, hanging from it at the 20.0 cm mark and a mass of 0.31 kg mass hanging from the 60.0 cm mark. Determine the position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance. Group of answer choices
Answer: 70.5 cm
Explanation:
The position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance will be at the side of 0.31kg.
You will use the moment techniques.
That is,
Sum of the clockwise moment = sum of anticlockwise moments
Please find the attached file for the remaining explanation and solution.
A box of mass 115 kg sits on an inclined surface with an angle of 59. What is the component of the weight of the box along the surface?
Answer:
966 N
Explanation:
For computing the component of the weight we first need to compute the object weight which is shown below:
Weight = (115 kg)(9.8 m/s²)
= 1,127 N
Now the weight of the box along the surface with an angle of 59 is
= (1,127 N) (sin 59 degree)
= 966 N
Hence, the weight component of the box along the surface is 966 N
A tungsten filament used in a flashlight bulb operates at 0.20 A and 3.0 V. If its resistance at 20°C is 1.5Ω, what is the temperature of the filament when the flashlight is on?
Answer:
The temperature of the filament when the flashlight is on is 2020 °C.
Explanation:
The resistivity varies linearly with temperature:
[tex] R = R_{0}[1 + \alpha*(T-T_{0})] [/tex] (1)
Where:
T: is the temperature of the filament when the flashlight is on=?
T₀: is the initial temperature = 20 °C
α: is the temperature coefficient of resistance = 0.0045 °C⁻¹
R₀: is the resistance at T₀ = 1.5 Ω
When V = 3.0 V, R is:
[tex]R = \frac{V}{I} = \frac{3.0 V}{0.20 A} = 15 \Omega[/tex]
By solving equation (1) for T we have:
[tex]T = \frac{R-R_{0}}{\alpha*R_{0}} + T_{0} = \frac{15-1.5}{0.0045*1.5} + 20 = 2020 ^{\circ} C[/tex]
Therefore, the temperature of the filament when the flashlight is on is 2020 °C.
I hope it helps you!
A long solenoid (1500 turns/m) carries a current of 20 mA and has an inside diameter of 4.0 cm. A long wire carries a current of 2.0 A along the axis of the solenoid. What is the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire
Answer:
The magnitude of the magnetic field is 55μT
Explanation:
Given;
number of turns of the solenoid per length, n = N/L = 1500 turns/m
current in the solenoid, I = 20 mA = 20 x 10⁻³ A
diameter of the solenoid, d = 4 cm = 0.04 m
The magnetic field at a point that is inside the solenoid;
B₁ = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Given;
current in the wire, I = 2 A
distance of magnetic field from the wire, r = 1 cm = 0.01 m
The magnetic field at 1.0 cm from the wire;
[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]
The magnitude of the magnetic field;
[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]
Therefore, the magnitude of the magnetic field is 55μT
The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
Given the following parameters from the question
Number of turns of the solenoid per length, n = N/L = 1500 turns/m current in the solenoid, I = 20 mA = 20 x 10⁻³ A Diameter of the solenoid, d = 4 cm = 0.04 mThe magnetic field at a point that is inside the solenoid is expressed according to the formula;
B₁ = μ₀nIWhere;
μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Next is to get the magnetic field strength in the second wire.
Current in the wire, I = 2 A Distance of magnetic field from the wire, r = 1 cm = 0.01 mThe magnetic field at 1.0 cm from the wireSubstitute into the formula:
[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]
Get the resultant magnetic field:
[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]
Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
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Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diameter and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.
Answer:
work done in pumping the entire fuel is 1399761 J
Explanation:
weight per volume of the gasoline = 6600 N/m^3
diameter of the tank = 3 m
length of the tank = 6 m
The height of the tractor tank above the top of the tank = 5 m
The total volume of the fuel is gotten below
we know that the tank is cylindrical.
we assume that the fuel completely fills the tank.
therefore, the volume of a cylinder =
where r = radius = diameter ÷ 2 = 3/2 = 1.5 m
volume of the cylinder = 3.142 x x 6 = 42.417 m^3
we then proceed to find the total weight of the fuel in Newton
total weight = (weight per volume) x volume
total weight = 6600 x 42.417 = 279952.2 N
therefore,
the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)
work done in pumping = 279952.2 x 5 = 1399761 J
Which type of reaction is shown in this energy diagram?
Answer:
Option C
Explanation:
The graph shows endothermic reaction because the reactants are lower in energy and the products are higher is energy. Endothermic reactions absorb energy having products with higher energy.
Answer:
C
Explanation:
In an endothermic reaction, the energy-time graph shows reactants are at a lower energy level than the products.
A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?
Answer:
Explanation:
a = F / m
where a is acceleration , F is thrust and m is mass
taking log and differentiating
da / a = dF / F - dm / m
(da / a)x 100 = (dF / F)x100 - (dm / m) x100
percentage increase in a = percentage increase in F - percentage increase in m
= percentage increase in acceleration a = 39 - 13 = 26 %
required increase = 26 %.
At 30.0 m below the surface of the sea (density = 1 025 kg/m3), where the temperature is 5.00°C, a diver exhales an air bubble having a volume of 0.95 cm3. If the surface temperature of the sea is 20.0°C, what is the volume of the bubble just before it breaks the surface?
Answer:
The volume is [tex]V_a = 1.510 *10^{-5} m^3[/tex]
Explanation:
From the question we are told that
The depth below the see is [tex]d_1 = 30.0 \ m[/tex]
The density of the sea is [tex]\rho_s = 1025 \ kg /m^3[/tex]
The temperature at this level is [tex]T_d = 5.00 ^oC = 278 \ K[/tex]
The volume of the air bubble at this depth is [tex]V_d = 0.95 \ cm^3 = 0.95 *0^{-6}\ m[/tex]
The temperature at the surface is [tex]T_a = 20^oC =293\ K[/tex]
Generally the pressure at the given depth is mathematically evaluated as
[tex]P_d = P_o + \rho_s * g * d[/tex]
Where [tex]P_o[/tex] is the atmospheric pressure with a constant value
[tex]P_o = 1.013 *10^{5} \ Pa[/tex]
substituting values
[tex]P_d = 1.013 * 10^{5} * + (1025 * 9.8 * 30 )[/tex]
[tex]P_d = 4.02650 * 10^{5} \ Pa[/tex]
According to the combined gas law
[tex]\frac{P_a * V_a }{T_a } = \frac{P_d * V_d }{T_d }[/tex]
=> [tex]V_a = \frac{4.026650 *10^{5} * 0.95 *10^{-6} * 293 }{278 * 1.013*10^{5} }[/tex]
=> [tex]V_a = 1.510 *10^{-5} m^3[/tex]
A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.
Answer:
Explanation:
Total mass m = 18 kg .
Spring are parallel to each other so total spring constant
= 4 x 24 = 96 N/cm = 9600 N/m
Time period of vibration
[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]
Putting the given values
[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]
= .27 s .
A circular loop of wire of radius 10 cm carries a current of 6.0 A. What is the magnitude of the magnetic field at the center of the loop
Answer:
3.77x10^-5T
Explanation:
Magnetic field at center of the loop is given as
B=uo*I/2r =(4pi*10-7)*6/2*0.1
B=3.77*10-5Tor 37.7 uTi
Cables supporting a suspension bridge have a linear mass density of 3700 kg/m; the tension is 1.7 x 10 8 N. What would be the transverse wave speed in such a cable?
Answer:
The speed is [tex]v = 214.35 \ m/s[/tex]
Explanation:
From the question we are told that
The linear mass density is [tex]\mu = 3700 \ kg/m[/tex]
The tension is [tex]T = 1.7*10^8 \ N[/tex]
The transverse wave speed is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
substituting values
[tex]v = \sqrt{\frac{1.7 *10^{8}}{3700} }[/tex]
[tex]v = 214.35 \ m/s[/tex]
How much heat is required to convert 500g of liquid water at 28°C into steam at 150°C? Take the specific heat capacity of water to be 4183J/kg°C and the latent heat of vaporization to be 2.26×10⁶J/kg
Answer:
1,327,063Joules
Explanation:
Heat energy is the energy needed to convert the state of a body from one phase to another.
According to the question, we want to calculate the total heat required to convert water into vapour (steam).
Note that before water can vapourize, it has to reach the boiling point first which is at 100°C. Heat energy needed to convert the water to 100°C is expressed as H1 = mcΔθ
m is the mass of the object in kg =0.5kg
c is the spcific heat capacity of water = 4183J/kg°C
Δθ is the change in temperature = 100-28 = 72°C
H1 = 0.5*4183*72
H1 = 150,588Joules
Energy required to convert the water to team H2 =mLsteam
Lsteam is the latent heat of vaporization = 2.26×10⁶J/kg
H2 = 0.5*2.26×10⁶
H2 = 1130000Joules
Heat energy needed to convert the water to 150°C is expressed as H3 = mcΔθ
m is the mass of the object in kg =0.5kg
c is the spcific heat capacity of steam= 1859J/kg°C
Δθ is the change in temperature = 150-100 = 50°C
H3 = 0.5*1859*50
H1 = 46,475Joules
Total Heat requires = H1+H2+H3 = 150,588Joules+1130000Joules+ 46,475Joules = 1,327,063Joules
Suppose you have a lens system that is to be used primarily for 775-nm light. What is the second thinnest coating of fluorite (calcium fluoride) that would be non-reflective for this wavelength?
Answer:
406 nm
Explanation:
We are given;
Wavelength; λ = 775 nm
Refractive index of Calcium fluoride with wavelength of 775 nm as seen in the graph attached is approximately 1.4308.
n = 1.4308
Formula for the thickness of the film that would destruct the light is;
t = (m + 0.5)(λ/2n)
Where m is the order of the thickness.
The first smallest thickness is at m = 0 while the second smallest thickness is at m = 1.
Thus;
t = (1 + 0.5)(775/(2 × 1.4308))
t ≈ 406 nm
A 900 kg roller coaster car starts from rest at point A. rolls down the track, goes
around a loop (points B and C) and then flies off the inclined part of the track (point D),
Figure 2.
The dimensions are: H =80 m.
r= 15m, h=10m and theta =9.30°
Calculate the
(a) gravitational potential energy at point A.
(b) velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J.
c) distance of the car land (in the horizontal direction) from point D if given the
velocity at point D is 37.06 m/s
I
Answer:
gravitational potential energy at point A.
A) The gravitational potential energy at point A is; 705600 J
B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s
A) Formula for gravitational potential energy is;
PE = mgh
At point A;
mass; m = 900 kg
height; h = 80 m
Thus;
PE = 900 × 9.8 × 80
PE = 705600 J
B) Kinetic energy of the roller coaster at point C is given as;
KE = PE - W
We are given Workdone; W = 264870 J
Thus;
KE = 705600 - 264870
KE = 440730 J
Thus, velocity at point C is gotten from the formula of kinetic energy;
KE = ½mv²
v = √(2KE/m)
v = √(2 × 440730/900)
v = 31.295 m/s
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A copper telephone wire has essentially no sag between poles 36.0 m apart on a winter day when the temperature is −20.0°C. How much longer is the wire on a summer day when the temperature is 34.0°C?
Answer:
The extension is [tex]\Delta L = 0.033 \ m[/tex]
Explanation:
From the question we are told that
The length of the wire on a winter day is [tex]L_w = 36.0 \ m[/tex]
The temperature on the winter day is [tex]T_w = -20.0 ^o C[/tex]
The temperature on a summer day is [tex]T_s = 34.0 ^0 C[/tex]
The the extension of the wire on a summer day is mathematically represented as
[tex]\Delta L = \alpha L_w [T_s - T_w][/tex]
Where
[tex]\alpha[/tex] is the coefficient of linear expansion of copper with a values [tex]\alpha = 17 *10^{-6}[/tex]
substituting value
[tex]\Delta L = 17 *10^{-6} * 36.0 [34 - [-20]][/tex]
[tex]\Delta L = 0.033 \ m[/tex]
How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other
Answer:
cutting the magnet in two parts each part has a North and South pole,
Explanation:
In magnetism the magnetic mono-poles are not found, this means that we do not have magnetic charges alone, therefore when cutting the magnet in two parts each part has a North and South pole, the magnetic lines go from the North pole to the South pole, see attached.
The density of the lines is approximately the intensity of the magnetic field.
A body of mass 2.5kg is raised 4.0m above the ground.Calculate the potential energy if g=10m/s squared
Answer:
100 joulesExplanation:
[tex]mass = 2.5kg\\height = 4.0\\Acceleration \: due \:to\:gravity = 10m/s^2\\\\P.E = mgh\\P.E = 2.5kg\times10\times4\\\\Potential \: Energy = 100 joules[/tex]
The potential energy if g = 10m/s² is 98 J
What is Potential Energy ?Potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.
Formula for potential energy :
P.E. = mgh
m = mass (kg)g = gravity (m/s²)h = height/distance (m)Given :
The book is held from the ground of a distance = 4.0 m,
so
h = 4.0 m
we know that the book weighs,
2.5 kg
so
m = 2.5 kg.
Now we just put it in the formula ;
PE = (2.5kg) × (9.8 m/s²) × ( 4.0 m)
P E = 98 J
Therefore, The potential energy if g = 10m/s² is 98 J
Learn more about Potential Energy here ;
https://brainly.com/question/24284560
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2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the
upthrust of the liquid on the cube.
Answer:
vpg = 0.064 N
Explanation:
Upthrust = Volume of fluid displaced
upthrust liquid on the cube g=10ms−2
vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N
vpg = 0.064 N
hope it helps.
A "590-W" electric heater is designed to operate from 120-V lines.
A)What is its operating resistance?
b)What current does it draw?
c)If the line voltage drops to 110 V, what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.)
d)The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in the previous part?
a. It will be smaller. The resistance will be smaller so the current drawn will increase, decreasing the power.
b. It will be smaller. The resistance will be smaller so the current drawn will decrease, decreasing the power.
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
d. It will be larger. The resistance will be smaller so the current drawn will decrease, increasing the power.
Answer:
a) 24.4 Ω
b) 4.92 A
c) 495.9 W
d)
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
Explanation:
b)
The formula for power is:
P = IV
where,
P = Power of heater = 590 W
V = Voltage it takes = 120 V
I = Current Drawn = ?
Therefore,
590 W = (I)(120 V)
I = 590 W/120 V
I = 4.92 A
a)
From Ohm's Law:
V = IR
R = V/I
Therefore,
R = 120 V/4.92 A
R = 24.4 Ω
c)
For constant resistance and 110 V the power becomes:
P = V²/R
Therefore,
P = (110 V)²/24.4 Ω
P = 495.9 W
d)
If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
WILL MARK BRAINLIEST!!An igneous rock has large red, black, and green crystals. How else can this rock be accurately described?
O fine texture
O cooled quickly
O intrusive origin
O created by lava
Answer:
D
Explanation:
Which of the following statements is not true?
1) The average power supplied to an inductor in an AC circuit is proportional to the angular frequency of the power source.
2) By stepping up AC voltage with a transformer, we can transport electricity across large distances with minimal power loss.
3) Voltage and current are in phase across a resistor connected to an AC power source.
4) In AC circuits, RMS stands for Root Mean Square.
Answer:
Explanation:
1 ) Average power supplied to an inductor is zero because the phase difference of potential and current is π / 2 .
So it is a wrong statement .
2 ) Step up transformer increases the voltage . At high voltage , lesser current is required to transport electrical energy . When current is reduced , the loss of energy due to heating effect is reduced .
3 ) voltage and current are in phase in resistance in ac .
3 ) RMS stands for Root Mean Square .
16. When a plastic rod is rubbed with fur, the rod will become negatively charged. Which of the following best explains how this happens? A. Electrons are transferred from the plastic rod to the fur. B. Electrons are transferred from the fur to the plastic rod. C. Negative charges are created on the plastic rod. D. Positive charges are removed from the plastic rod.
Answer:
B. Electrons are transferred from the fur to the plastic rod.
Explanation:
Triboelectricity or friction charging refers to the ability of materials to gain or lose electrons as a result of rubbing them against something. This phenomenon has been observed in the case of rubbing plastic rod against fur, or glass rod against silk.
In the context of rubbing plastic rod against fur, what happens is that the fur which has an excess of charges loses electrons to the plastic rod. This makes the plastic rod to become positively charged, and the fur, negatively charged.
What is the wavelength λλlambda of the wave described in the problem introduction? Express the wavelength in terms of the other given variables and constants
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda= \frac{2 \pi }{k}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]
The magnetic field is [tex]\= B = B_0 sin (kx -wt) \r k[/tex]
From the above equation
and k is the wave number which is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]\lambda= \frac{2 \pi }{k}[/tex]
Where [tex]\lambda[/tex] is the wavelength
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the satellite). The mass of the satellite is 200 kg.
About how much energy (in kJ) was converted to heat?
Answer:
The energy coverted to heat is 200 kilojoules.
Explanation:
GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:
[tex]m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v[/tex]
Where:
[tex]m_{M}[/tex], [tex]m_{S}[/tex] - Masses of the small meteorite and the communication satellite, measured in kilograms.
[tex]v_{M}[/tex], [tex]v_{S}[/tex] - Speeds of the small meteorite and the communication satellite, measured in meters per second.
[tex]v[/tex] - Final speed of the satellite-meteorite system, measured in meters per second.
The final speed of the satellite-meteorite system is cleared:
[tex]v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}[/tex]
If [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex] and [tex]v_{S} = 0\,\frac{m}{s}[/tex], the final speed is now calculated:
[tex]v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}[/tex]
[tex]v = 0.1\,\frac{m}{s}[/tex]
Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:
[tex]K_{S} + K_{M} - K - Q_{disp} = 0[/tex]
[tex]Q_{disp} = K_{S}+K_{M}-K[/tex]
Where:
[tex]K_{S}[/tex], [tex]K_{M}[/tex] - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.
[tex]K[/tex] - Kinetic energy of the satellite-meteorite system, measured in joules.
[tex]Q_{disp}[/tex] - Dissipated heat, measured in joules.
The previous expression is expanded by using the definition for the translational kinetic energy:
[tex]Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}][/tex]
Given that [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex], [tex]v_{S} = 0\,\frac{m}{s}[/tex] and [tex]v = 0.1\,\frac{m}{s}[/tex], the dissipated heat is:
[tex]Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right][/tex][tex]Q_{disp} = 200000\,J[/tex]
[tex]Q_{disp} = 200\,kJ[/tex]
The energy coverted to heat is 200 kilojoules.