The amount of energy lost to air friction, given the data is 37.71 J
How to obtain the initial energyInitial velocity (u) = 15.1 m/sMass (m) = 450 g = 450 / 1000 = 0.45 KgInitial Energy (E₁) = ?
E₁ = ½mu²
E₁ = ½ × 0.45 × 15.1²
E₁ = 51.3 J
How to obtain the final energyFinal velocity (u) = 19.89 m/sMass (m) = 450 g = 450 / 1000 = 0.45 KgFinal Energy (E₂) = ?
E₂ = ½mv²
E₂ = ½ × 0.45 × 19.89²
E₂ = 89.01 J
How to determine the energy lostInitial Energy (E₁) = 51.3 JFinal Energy (E₂) = 89.01 JEnergy lost =?Energy lost = E₂ - E₁
Energy lost = 89.01 - 51.3
Energy lost = 37.71 J
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A proton has a positive electric charge of q = 1.6 × 10–19 coulombs. what is the electric potential at a point 5.3 × 10–11 m from the proton? 4.4 × 10–18 volts 2.7 × 101 volts 3.0 × 10–9 volts 5.1 × 101 volts
B. The electric potential at the given point from the proton is 2.7 x 10¹ volts
Electric potential of the proton
The electric potential of the proton is calculated as follows;
V = kq/r
where;
k is Coulomb's constantq is charger is distanceV = (9 x 10⁹ x 1.6 x 10⁻¹⁹)/(5.3 x 10⁻¹¹)
V = 27.2 volts
V = 2.7 x 10¹ volts
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Mithun wants to test if a convex lens produces an inverted image of an object placed at a distance away from the lens. Which of the following should he use directly as an object for the test?
a. A
b. B
c. C
d. D
Answer:
(D) any of these objects can be used as long as all points on the object lie farther away from the lens than the focal length of the lens.
Points lying between the focal point and the lens will be erect and will not form an object visible on a screen.
Vanessa made a list comparing the differences between the outer coverings in plants and animals, but she made a mistake. Read the table and then choose the statement that explains why the list is incorrect. Structure Description of Function Spines Protects the plant from predators and helps increase water loss Skin Protects the animal from changes in weather and its environment The skin is a reproductive organ. The skin offers little protection against the weather. The spines attract predators. The spines help decrease water loss.
The skin is NOT a reproductive organ and the spines attract predators are the incorrect statements.
Why the list is incorrect?The list is incorrect because skin is not a reproductive organ and the spines do not attract predators towards each other. Skin protects the body from the outer environment whereas the spines repel the predators from the plant body.
So we can conclude that the skin is NOT a reproductive organ and the spines attract predators are the incorrect statements.
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Which of these must be the same on both
sides of a balanced chemical equation?
A the number of atoms
B
the number
C the number
D the number of solids, liquids, and gases
of compounds
of molecules
Answer:
the total number of atoms of each element present is the same on both sides of the equation. (if I remember correctly)
A ball is rolling across the floor at a constant velocity. What is the value of the change in its kinetic energy, ΔEk?
Positive
Zero
Negative
B. When the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.
What is change in kinetic energy?
The change in kinetic energy of an object is the dereference between the final kinetic energy and the initial kinetic energy.
ΔK.E = K.Ef - K.Ei
ΔK.E = 0.5m(vf² - vi²)
where;
K.Ef is the final kinetic energyK.Ei is the initial kinetic energyvf is final velocityvi is initial velocityAt constant velocity, the initial velocity and final velocity are equal.
ΔK.E = 0.5m(0) = 0
Thus, when the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.
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Hi! Whoever can answer this question the best I will give the brainliest answer!!
Describe the structure of a battery and how it works.
Thank you!! ✨
What is the mass of mars
the mass of mars is
6.39 × 10^23 kg
when 82.5 calories of heat are given to a metallic rod of mass 150g its temperature raises from 20 degree celcius to 25 degree Celsius. what is the specific heat of the metal.
A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.
What is specific heat?It is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).
A metallic rod of mass 150 g (m) absorbs 82.5 cal of heat (Q) and its temperature raises from 20 °C to 25 °C. We can calculate the specific heat (c) of the metal using the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 82.5 cal / 150 g × (25 °C - 20 °C) = 0.11 cal/g.°C
A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.
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PLS HELP:
Find the magnitude of an electric field located 7.00 cm from a source charge of - 11 nC. (1nC = 1 x 10-9 C). Will the field lines point toward or away from the source charge? Show your work and include an electric field diagram assuming the source is a point charge
Answer:
E=kQ/r*2
=(9. 10*9)(11.10*-9)/(0,07)*2
=20204 N. C*-1
The field lines will point towards the source charge because it is negatively charged.
A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The magnitude of the force on the left-hand pole is mathematically given as
f'=0.167N
[tex]Q=45 \textdegree[/tex]
What is the magnitude of the force on the left-hand pole.?Generally, the equation for Force is mathematically given as
[tex]F=\frac{mg}{sin\theta}[/tex]
Therefore
[tex]F=\frac{17.1*10^{-3}*9.8}{sin45}[/tex]
F=0.237N
Considering horizontal
f'-fcos=0
Therefore
f'=0.237*cos45
f'=0.167N
In conclusion, the slope
[tex]tan\theta=lv/ln\\\\tan\theta=30/30\\\\\theta=tan^{-1}\\\\[/tex]
[tex]Q=45 \textdegree[/tex]
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A slinky spring 3m long rests on a horizontal bench with one end fixed. When the free end is suddenly pushed forward, the compression pulse travels to the fixed end and back in 2 secs. What is the speed of the longitudinal waves along the spring? What is the wavelength of the longitudinal waves produced when the free end is moved to and fro with a frequency of 2.5Hz? A slinky spring 3m long rests on a horizontal bench with one end fixed . When the free end is suddenly pushed forward , the compression pulse travels to the fixed end and back in 2 secs . What is the speed of the longitudinal waves along the spring ? What is the wavelength of the longitudinal waves produced when the free end is moved to and fro with a frequency of 2.5Hz ?
The speed of the longitudinal waves along the spring is 1.5 m/s.
The wavelength of the longitudinal waves produced when the free end is moved to and fro is 0.6 m.
What is wavelength?The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.
Wavelength λ = c/f
where c is the speed of light in vacuum = 3*10⁸ m/s
A slinky spring 3m long rests on a horizontal bench with one end fixed. When the free end is suddenly pushed forward, the compression pulse travels to the fixed end and back in 2 secs.
λ = v/f
v =fλ =λ /T where T is the time period = 2sec
Put the values we get
v = 3/2 =1.5 m/s
Thus, the speed of wave is 1.5 m/s.
Using the same expression for frequency 2.5Hz, wavelength is
λ = v/f
λ = 1.5 /2.5 = 0.6 m
Thus, the wavelength is 0.6m.
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For the circuit shown in the figure, the switch S is initially open and the capacitor is uncharged. The switch is then closed at time t = 0. How many seconds after closing the switch will the energy stored in the capacitor be equal to 50.2 mJ?
Hi there!
To start, we can use the equation for the charge of a capacitor versus time.
[tex]q(t) = C\epsilon (1 - e^{-\frac{t}{\tau}})[/tex]
[tex]q(t)[/tex] = Charge on capacitor vs. time
[tex]\epsilon[/tex] = EMF of battery (40 V)
**Note: 'Cε' is equal to the MAXIMUM CHARGE on a capacitor.
Where the time constant is given as:
[tex]\tau = RC[/tex]
[tex]\tau[/tex] = Time Constant (s)
[tex]R[/tex] = Resistance (0.5 MΩ)
[tex]C[/tex] = Capacitance (90 μF)
Let's first convert the given units to the necessary 'J', 'Ω' and 'F' to make things easier:
[tex]J = 50.2 mJ[/tex]
[tex]50.2 mJ * \frac{0.001J}{1mJ} = 0.0502 J[/tex]
[tex]R = 0.5 M\Omega\\\\0.5M\Ohm * \frac{1000000\Omega}{1M\Ohm} = 500000\Omega[/tex]
[tex]F = 90 \mu F[/tex]
[tex]90 \mu F * \frac{0.000001 F}{1\mu F} = 0.00009 F[/tex]
Now, let's calculate the time constant.
[tex]\tau = RC = (500000)(0.00009) = 45 s[/tex]
Time to use the equation for energy stored in a capacitor:
[tex]E_C = \frac{1}{2}\frac{Q^2}{C}[/tex]
Our 'Q' varies with time, so:
[tex]E(t)= \frac{1}{2}\frac{( C\epsilon (1 - e^{-\frac{t}{\tau}}))^2}{C}\\\\E(t) = \frac{1}{2}\frac{( C^2\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)}{C}\\\\E(t) = \frac{1}{2}( C\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)[/tex]
Let's now plug values into our equation.
[tex]0.0502 = \frac{1}{2}(0.00009)(40^2) (1 - e^{-\frac{t}{45}})^2[/tex]
Simplifying:
[tex]0.697 = (1 - e^{-\frac{t}{45}})^2[/tex]
Take the square root of both sides.
[tex]0.835 = 1 - e^{-\frac{t}{45}}\\\\ e^{-\frac{t}{45}} = 0.165[/tex]
Take the natural log of both sides.
[tex]-\frac{t}{45} = ln(0.165) \\\\\frac{t}{45} = 1.802\\\\t = \boxed{81.08 s}[/tex]
What is the maximum height that the water would reach in the internal network of a building, without the use of a pump, if the pressure at the sidewalk level is 1.3 kgf/cm2? (density of water 1000 kg/m3)
The maximum height to be reached by the water at the given pressure is 13 m.
Maximum height reached by the water
The maximum height to be reached by the water is calculated as follows;
P = ρgh
where;
P is the pressure of the waterρ is the density of the waterg is acceleration due to gravityh is the maximum height1 kgf/cm² = 98066.5 Pa = 98066.5 N/m²
1.3 kgf/cm² = 127486 N/m²
h = P/ρg
h = (127486)/(1000 x 9.8)
h = 13 m
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a bus strats from from rest. if the accleration is 2m/s,find the distance travelled and the velocity after 2 seconds
Answer:
Velocity: 4 m/s
Distance: 4 m
Explanation:
initial velocity: u = 0 m/s
Acceleration: a = 2 m/s²
Time taken: t = 2 seconds
Distance traveled: d = (what we need to find)
Final velocity: v = (what we need to find)
---------------------------------------
v = u + at
v = 0 + 2 × 2
v = 4 m/s
---------------------------------------
d = 0 × 2 + 1/2 × 2 × (2)²
d = 1/2 × 2 × 4
d = 4 m
A solid, homogeneous sphere with of mass of M = 2.35 kg and a radius of R = 12.9 cm is resting at the top of an incline as shown in the figure.
The height of the incline is h = 1.29 m, and the angle of the incline is θ = 20.1°. The sphere is rolled over the edge very slowly. Then it rolls down to the bottom of the incline without slipping.
What is the total kinetic energy (translational plus rotational kinetic energy) of the sphere when it reaches the bottom of the incline?
The total kinetic energy of the sphere when it reaches the bottom of the incline is 44.57 J.
Total kinetic energy of the sphereThe total kinetic energy of the sphere = translational + rotational kinetic energy.
Translational kinetic energyK.E = ¹/₂mv²
K.E(bottom) = P.E(top) = mgh
K.E = (2.35)(9.8)(1.29)
K.E = 29.71 J
Rotational kinetic energyK.E(rot) = ¹/₂Iω²
K.E(rot) = ¹/₂I(v²/R²)
where;
I is rotational kinetic energyI = ¹/₂MR²
K.E(rot) = ¹/₂( ¹/₂MR²)(v²/R²)
K.E(rot) = ¹/₄Mv²
K.E(rot) = ¹/₄M(2gh)
K.E(rot) = ¹/₂Mgh
K.E(rot) = ¹/₂(29.71) = 14.855 J
Total kinetic energy = 29.71 + 14.855 = 44.57 J
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A cup of tea at room temperature of 24°C is heated until it has twice the internal energy. Calculate the final temperature of the tea.
The final temperature of the tea cup is 100°C.
What is internal energy?The Internal energy is the energy of a substance due to to the constant random motion of its particles.
The symbol for internal energy of a substance is U and it is measured in Joules.
ΔU = q + W
q is the heat, q = mcΔTW is the mechanical work.In conclusion, the final temperature of the tea cup at room temperature of 24 °C which is heated until it has twice the internal energy is 100°C.
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The bars in the drawing above are magnets. According to chapter 21 which of the following picture(a) is true? Explain. Then explain why the other 2 pictures are incorrect. Make sure to explain all 3 sets.
Magnets are materials which are able to attract others to themselves by producing a magnetic field.
Based on the properties of magnetic lines of forces;
Picture 1 is correct because the magnetic lines of forces originate from the North pole and are directed towards the south pole.Picture 2 is incorrect because unlike poles attractPicture 3 is incorrect because like poles repel.In conclusion, the law of magnetism states that like poles repel whereas unlike poles attract.
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Latency refers to the 27 seconds of time it takes for someone to refocus on driving after they finish texting or talking on the phone. true false
The term "latency" describes the 27 seconds that pass after someone has finished texting or conversing on the phone before they can concentrate again on driving. The given statement is true.
What is latency?A measure of delay is latency. The amount of time it takes for some data to travel across a network is known as latency.
The amount of time it takes for information to travel from its source to its destination and back is often quantified as a round trip delay.
The 27 seconds that pass after someone has done texting or talking on the phone before they can focus on driving again are referred to as "latency." The assertion is accurate.
Hence, the given statement is true.
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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.
The tension in the cable is mathematically given as
T =266.659 N
What is the tension in the cable.?Generally, the equation for the angle of the boom with horizontal is mathematically given as
A = tan-1(5 /10)
[tex]A= 26.57 \textdegree[/tex]
The angle of cable with horizontal
B = tan-1(4 / 10)
B= 21.80 degrees
Hence
175.5 * cos26.57 + 94.7 *cos26.57 * 1/2 = T (sin(26.57+21.80)) * 1
T =266.659 N
In conclusion, the tension in the cable.
T =266.659 N
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A space probe sent to Mars has a weight of 10,000 N on Earth (g = 10). What is its mass?
Answer:
[tex]1,\!000\; {\rm kg}[/tex].
Explanation:
The "[tex]g[/tex]" given in the question likely refers to the strength of the gravitational field. The standard unit of [tex]g[/tex] is [tex]{\rm m \cdot s^{-2}}[/tex], same as the standard unit of acceleration. However, since [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}[/tex], the standard unit of [tex]g[/tex] is equivalent to [tex]{\rm N \cdot kg^{-1}}[/tex].
For instance, the gravitational strength on the surface of the earth is approximately [tex]g = 10\; {\rm N \cdot kg^{-1}}[/tex]. However, the question did not specify the unit of this value.
If the mass of an object is [tex]m[/tex], the weight of that object would be [tex](\text{weight}) = m\, g[/tex] when the gravitational field strength around that object is [tex]g[/tex]. It is given that for the object in this question, [tex]m\, g = (\text{weight}) = 10,\!000\; {\rm N}[/tex] when [tex]g = 10\; {\rm N \cdot kg^{-1}}[/tex]. Therefore, the mass [tex]m\![/tex] of this object would be:
[tex]\begin{aligned}m &= \frac{(\text{weight})}{g} \\ &= \frac{10,\!000\; {\rm N}}{10\; {\rm N \cdot kg^{-1}}} \\ &= 1,\! 000\; {\rm kg}\end{aligned}[/tex].
For a vector with origin at coordinate (0, 0) (horizontal, vertical) and end at (6, 8) northeast of the origin,
what is the horizontal component of the vector?
The horizontal component of the vector is determined as 6i.
Horizontal component of the vector
The horizontal component of the vector is calculated as follows;
V = (0 i, 0 j) + (6i, 8j)
V = (6i, 8j)
where;
6i is the horizontal component of the vector8j is the vertical component of the vectorThus, the horizontal component of the vector is determined as 6i.
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A roller coaster has a mass of 450 kgIt sits at the top of a hill with height 49 m. If it drops from this hill, how fast is it going when it reaches the bottom? Assume there is no air resistance or friction.)
The speed of the roller coater at the bottom of the hill is 31 m/s.
Speed of the roller coater at the bottom of the hill
Apply the principle of conservation of mechanical energy as follows;
K.E(bottom) = P.E(top)
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
v is the speed of the coater at bottom hillh is the height of the hillg is acceleration due to gravityv = √(2 x 9.8 x 49)
v = 31 m/s
Thus, the speed of the roller coater at the bottom of the hill is 31 m/s.
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Two loudspeakers, 3.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.15 m . Assume the speed of sound is 340 m/s .
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.15 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
For the given question the answer to the first part is that frequency of sound will be 404.76 Hz. and answer to the second part will be that the initial frequency for which that place will have the highest level of sound intensity
Given, the speed of sound 340 m/s
a) L1 + L2 = 3.5 m
and Ld = | L1 - L2 | = ( 1.75 + 0.21 ) - (1.75 - 0.21 ) = 1.96 - 1.54 = 0.42 m
Ld = λ/2
λ = 2Ld = 2×0.42 = 0.84 m
and finally,
f = v/λ
f = 340/0.84
f = 404.76 hertz
Frequency came out to be 404.76 hertz in this case
b) For the first frequency
0.42 = λ
f = v/λ
f = 340 / 0.42
f = 809.52 Hertz
Frequency came out to be 809.52 Hertz in this case.
To conclude with we can say that the Frequency of the sound in case on came out to be 404.76 hertz which is approximately 405 Hz after applying all the concepts and calculations, in second case first frequency for which that location will be a maximum of sound intensity came out to be 809.52 Hertz after applying all the concepts and calculations.
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What is standard one kilogram ?
Answer:
As of the 2019 redefinition of the SI base units, the kilogram is defined in terms of the second and the metre, both based on fundamental physical constants.
It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures. It was originally defined as the mass of one liter (10-3 cubic meter) of pure water. At the Earth's surface, a mass of 1 kg weighs approximately 2.20 pounds ( lb ).
Explanation:
Lindsay is planning a flight from St. Catherines to Hamilton, which lies due west of St. Catharines.
Her aircraft flies at a speed of 200 km/h in still air. A wind of 50.0 km/h is blowing from [W 60.0° N].
In what direction must she aim the aeroplane to fly directly to Hamilton? (KU, TI)
Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.
Using Sine rule to solve this question
Sine rule => SinA/a = SinB/b = SinC/c = constant
The magnitude of wind is 50 with an angle of 60 degrees.
The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.
One side is 50 km/hr at an angle of 60 degrees.
sin 60°/200 = sin θ / 50
50 × sin 60° = 200 × sin θ
√3/2 = 4 × sin θ
√3/8 = sin θ
sin θ = 0.2165
θ = sin⁻¹(0.2165)
θ = 12.5°
So Lindsay have to fly the plane in the direction of [W 12.5° S].
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Three ropes A, B and C are tied together in one single knot K. (See figure.)
If the tension in rope A is 65.3 N, then what is the tension in rope B?
The tension in the rope B is determined as 10.9 N.
Vertical angle of cable B
tanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
Angle between B and Cθ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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D
Question 2
What are the coefficients needed to balance this chemical equation?
___O₂ → CO₂ +___ H₂O
C₂H5OH +_
O 6, 2, 6, 4
O 6, 6, 2,4
O 1, 3, 2, 3
1, 2, 2, 1
2 pts
Hello and Good Morning/Afternoon:
Original Question: C₂H₅OH + __O₂ → __CO₂ + __ H₂O
To balance this equation:
⇒ must ensure that there is an equal number of elements on both sides of the equation at all times
Let's start balancing:
On the left side of the equation, there are 2 carbon molecule⇒ but only so far one on the right side
C₂H₅OH + __O₂ → 2CO₂ + __ H₂O
On the left side of the equation, there are 6 hydrogen molecules⇒ but only so far two on the right side
C₂H₅OH + __O₂ → 2CO₂ + 3H₂O
On the right side of the equation, there are 7 oxygen molecules⇒ but only so far three on the left side
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
Let's check and make sure we got the answer:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
2 Carbon ⇔ 2 Carbon
6 Hydrogen ⇔ 6 Hydrogen
7 Oxygen ⇔ 7 oxygen
Thefore the coefficients in order are:
⇒ 1, 3, 2, 3
Answer: 1,3,2,3
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Two forces F1=20N and F2=30N are acting on an object as shown in figure: F1=20N F2=30N (i) Find the net force acting on the object? (ii) State the direction of the net force acting on the object? (iii) If the body still does not move under the application of these forces, what can be the possible reason for this? Identify the name of this extra force and its direction. Two forces F1 = 20N and F2 = 30N are acting on an object as shown in figure : F1 = 20N F2 = 30N ( i ) Find the net force acting on the object ? ( ii ) State the direction of the net force acting on the object ? ( iii ) If the body still does not move under the application of these forces , what can be the possible reason for this ? Identify the name of this extra force and its direction .
The net force is the 10N force and its direction is towards F2 force.
What is resultant force?The resultant force or the net force is the actual force that acts on a body. We are told that the two forces act in opposite directions hence the net force is F2 - F1 = 30 N - 20 N = 10 N.
ii) The direction of this net force is going to be towards the 30N force.
iii) If the body does not move under the application of these forces then the F1 force must be reinforced by another force which balances it against the F2 force.
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A car moves at a speed of 45 km/h, when traveling 1500 meters, how long did it take to go that distance?
. A wheel of diameter 0.2 m starts from rest and accelerated with constant angular acceleration to an angular velocity of 900 rev min¹ in 5 s. (a) Find the position at the end of 1 s of a point originally at the top of the wheel. (b) Compute and show in a diagram the magnitude and the direction of the acceleration at the end of 1 s.
(a)The position at the end of 1 s of a point originally at the top of the wheel will be 9.425 radians.
(b)The magnitude and the direction of the acceleration at the end of 1 s will be 1.885 m/sec² clockwise.
What is angular acceleration?Angular acceleration is defined as the pace of change of angular velocity concerning time.
Given data;
Diameter of the wheel,d = 0.2 m
Initial velocity,ω₀ = 0
Angular acceleration,α = c
n = 900 rev min¹
ω = 94.2478 rad /sec
Time of rolling,t = 5 s
From Newton's first equation of motion;
ω = ω₀ + αt
94.2478 rad /sec = 0 + α(5 sec)
α = 18.85 rad/sec²
The position at the end of 1 s of a point is found as;
Θ = ω₀t+1/2(αt²)
Θ = 0+1/2[(18.85 rad/sec²) × (1 sec)²]
Θ = 9.425 radian
The magnitude and the direction of the acceleration at the end of 1 s are found as;
a = αr
a= 18.85 rad/sec² × 0.1 m
a = 1.885 m/sec²
Hence the position at the end of 1 s and magnitude and the direction of the acceleration at the end of 1 s will be 18.85 rad/sec²and the magnitude and the direction of the acceleration at the end of 1 s will be 1.885 m/sec² clockwise.
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