Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
First step :
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
next :
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x = [tex]\sqrt{\frac{2*23.27}{5029} }[/tex] = 0.0962 m
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval
Answer:
The number of revolutions is 44.6.
Explanation:
We can find the revolutions of the wheel with the following equation:
[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]
Where:
[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s
t: is the time = 8 s
α: is the angular acceleration
We can find the angular acceleration with the initial and final angular velocities:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s
[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]
Hence, the number of revolutions is:
[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]
Therefore, the number of revolutions is 44.6.
I hope it helps you!
a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration
Answer:
0
Explanation:
a = dv/dt
if v is constant than the slope of the v graph will be 0, so dv/dt is 0
a= 0
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration
Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;
[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]
Therefore, the frequency of this mode of vibration is 138.87 Hz
Why are objects measured?
In order to find out how long/wide/heavy/high/dense/deep/ massive/voluminous/reflective/opaque/ tansparent/warm/cold/hard/soft/ malleable/flexible/rigid/radioactive/old/ valuable/symmetrical/flat/regular/ irregular they are.
In a way that you can easily and conveniently describe to other people.
why is it wrong to leave our light on
Answer:
you will get huge electricity bills ............
Images formed by a convex mirror are always
Answer:
Images formed by a convex mirror are always virtual
Explanation:
A virtual image is always created by a convex mirror, and it is always situated behind the mirror. The picture is vertical and situated at the focus point when the item is far away from the mirror. As the thing approaches the mirror, the image follows suit and increases until it reaches the same height as the object.
OAmalOHopeO
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answer:
Average Velocity = 3.63 m/s
Explanation:
First, we will calculate the total displacement of the quarterback, taking forward direction as positive:
Total Displacement = 15 m - 3 m + 24 m = 36 m
Now, we will calculate the total time taken for this displacement:
Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s
Therefore, the average velocity will be:
[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]
Average Velocity = 3.63 m/s
Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances
Answer:
I = 0.0857 A
Explanation:
Given that,
Power consumed by the cellphone, P = 300 mW
The voltage of the battery, V = 3.5 V
Let I is the current flowing through the cell-phone. We know that,
P = VI
Where
I is the current
So,
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]
So, the current flowing the cell-phone is 0.0857 A.
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?
Answer:
The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".
Explanation:
According to the question,
The work will be:
⇒ [tex]Work=-\frac{kQq}{R}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]
[tex]=\frac{0.3978}{\varepsilon }[/tex]
[tex]=4.49\times 10^{10} \ joules[/tex]
Thus the above is the correct answer.
We have that the workdone is mathematically given as
W=4.49*10e10 J
From the question we are told
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?WorkdoneGenerally the equation for the workdone is mathematically given as
W=-kQq/R
Therefore
0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2
Hence
W=4.49*10e10 JFor more information on Charge visit
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If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?
Answer:
the current that flows through the component is 2.42 A
Explanation:
Given;
resistance of the electrical component, r = 53 Ω
the voltage of the source, V = 128 V
The current that flows through the component is calculated using Ohm's Law as demonstrated below;
[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]
Therefore, the current that flows through the component is 2.42 A
A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?
Answer:
201.5537 mph
Explanation:
Given the following data;
Speed = 90.1 m/s
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the formula;
Speed = distance/time
To convert this value into miles per hour;
Conversion;
1 meter = 0.000621 mile
90.1 meters = 90.1 * 0.000621 = 0.05595 miles
1 metre per second = 2.237 miles per hour
90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour
90.1 m/s = 201.5537 mph
The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?
Answer:
Acosθ
Explanation:
The x-component of a vector is defined as :
Magnitude * cosine of the angle
Maginitude * cosθ
The magnitude is represented as A
Hence, horizontal, x - component of the vector is :
Acosθ
Furthermore,
The y-component is taken as the sin of the of the angle multiplied by the magnitude
Vertical, y component : Asinθ
Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.
Answer:
Atmospheric pressure at Badwater is 1.01022 atm
Explanation:
Data given:
1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa
Elevation (h) = 86m
gravity (g) = 9.8 m/s2
Density of air P = 1.225 kg/m3
Therefore pressure at bad water Pb = Pi + Pgh
Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)
Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa
hence:
Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?
Answer:
a n c
Explanation:
A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.
The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?
Answer:
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
Explanation:
The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:
[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)
Where:
[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.
[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.
[tex]\Delta P[/tex] - Pressure change, in pascals.
If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:
[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]
[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]
[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
How are elastic and inelastic collisions different?
A: Elastic collisions occur when the colliding objects move separately after the collision; after inelastic collisions, the objects are connected and move together.
B: Elastic collisions occur when the objects are going the same direction when they collide; inelastic collisions occur when the objects are going in opposite directions when they collide.
C: Momentum is conserved in elastic collisions; momentum is not conserved in inelastic collisions.
D: Elastic collisions occur between objects of the same mass; inelastic collisions occur between different masses.
Answer:
a
Explanation:
Answer:
the answer is c
'
Explanation:
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
Learn more about Coulomb's law:
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You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec
The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.
Answer:
P = 10135.6 Pa
Explanation:
For this exercise we use that the pressure varies with the height
P = P₀ + ρ g h
where h is the height from the head to the heart, which is approximately
h = 40 cm = 0.40m and P₀ is the head pressure P₀ = 6000 Pa
P = 6000 + 1055 9.8 0.40
P = 6000 + 4135.6
P = 10135.6 Pa
What are the differences among elements, compounds, and mixtures?
Answer:
Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.
••••••••••••••••
Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.
•••••••••••••••
A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements.
------------------------------
Hope it helps...
Have a great day!!!
Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.
3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive
Answer:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
Explanation:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.
After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.
Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor
Answer:
560.54 Ω
Explanation:
Applying,
V = IR'............... Equation 1
Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors
make R' the subject of the equation
R' = V/I............ Equation 2
From the question,
Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A
Substitute these values into equation 2
R' = 5/(2.23×10⁻³ )
R' = 2242.15 Ω
Since the fours resistor are connected in series and they are equal,
Therefore the values of each resistor is
R = R'/4
R = 2242.15/4
R = 560.54 Ω
Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?
Answer:
The force on the second piston is 5246.5 N .
Explanation:
Area of first piston, a = 3.002 cm^2
Area of second piston, A = 315 cm^2
Force on first piston, f = 50 N
let the force of the second piston is F.
According to the Pascal's law
[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]
The image of the object formed by the lens is real, enlarged and inverted. What is the kind of lens ?
Answer:
Converging (convex) lens.
Explanation:
A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.
Basically, there are two (2) main types of lens and these includes;
I. Diverging (concave) lens.
II. Converging (convex) lens.
A converging (convex) lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. Thus, this type of lens is usually thin at the lower and upper edges and thick across the middle.
Basically, the image of the object formed by a converging (convex) lens. lens is real, enlarged and inverted.
A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds
Answer:
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Explanation:
If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.
V(f) = V(i) + a*t
Final velocity = initial velocity + acceleration x time
Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Answer:
100 m
Explanation:
Given,
Initial velocity ( u ) = 0 m/s
Acceleration ( a ) = 2 m/s^2
Time ( t ) = 10 sec s
To find : Displacement ( s ) = ?
By 2nd equation of motion,
s = ut + at^2 / 2
= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2
= 0 + ( 2 ) ( 100 ) / 2
= 200 / 2
s = 100 m
how can scientific method solve real world problems examples
Which of the following statements is correct about the magnitude of the static friction force between an object and a surface?
a. Static friction depends on the mass of the object.
b. Static friction depends on the shape of the object.
c. Static friction depends on what the object is made of but not what the surface is made of.
d. None of the above is correct.
Answer:
Static friction depends on the mass of the object.
Explanation:
Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;
1) nature of the object and the surface(how rough or smooth the surfaces are)
2)surface area of the object and the surface
3) mass of the object
Since;
F=μmg
Where;
μ= coefficient of static friction
m= mass of the object
g= acceleration due to gravity
Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.
Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces an amount of electric energy with the hot reservoir at 500 K during Day One and then produces the same amount of electric energy with the hot reservoir at 600 K during Day Two. The thermal pollution was:
Answer: hello your question lacks some vital information below is the complete question
Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was
answer:
Total thermal pollution = 2.5 * 10^6 J
Explanation:
Low temperature reservoir = 300 K
hot reservoir temperature = 500 K
Electrical energy produced by plant ( W ) = 1 * 10^6 J
lets assume ; Q1 = energy absorbed , Q2 = energy emitted
W = Q1 - Q2 or Q2 = Q1 - W ( we will apply this as the formula for determining thermal pollution )
For day 1
T1 = 500k , T2 = 300k
applying Carnot engine formula
W / Q1 = 1 - T2/T1
∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J
thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J
for Day 2
T1 = 600k, T2 = 300k
Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J
Thermal pollution; Q2 = Q1 - W = 1 * 10^6 J
Therefore the Total thermal pollution = 1 * 10^6 + 1.5 * 10^6 = 2.5 * 10^6 J
