Answer:
T = 556K
Explanation:
For this exercise let's use Stefan's law
P = σ A e T⁴
where the stafan-Boltzmann constant is 5,670 10⁻⁸ W / m² K⁴.
In this case the area of a circular cylinder is
A = π r²
the radius is
r = d / 2 = 0.150 m
A = π (0,150)²
A = 7.0685 10⁻² m²
let's clear
T⁴ = P /σ A e
T⁴ = 100 / (5.670 10⁻⁸ 7.0685 10⁻² 0.26)
T⁴ = 9.5966 10¹⁰ = 959.66 10⁸
T = 5.56 102 K
T = 556K
The temperature of the filament is 366.3 K
Using the equation for radiated power, P
P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴, ε = emissivity = 0.26, A = area of cylindrical tungsten filament = πDh (since it radiates through its sides)where D = diameter of tungsten filament = 0.40 mm = 0.4 × 10⁻³ m and h = length of tungsten filament = 30.0 cm = 0.3 m and T = temperature of tungsten filament.
Since we require the temperature of the filament, making T subject of the formula, we have
T = ⁴√(P/σεA)
T = ⁴√(P/σεπDh)
Since P = 100 W, substituting the values of the variables into the equation, we have
T = ⁴√(P/σεπDh)
T = ⁴√(100 W/5.67 × 10⁻⁸ W/m²K⁴× 0.26 × π× 0.4 × 10⁻³ m × 0.3 m)
T = ⁴√(100 W/0.5558 W/K⁴ × 10⁻⁸)
T = ⁴√179.93 × 10⁸ K⁴)
T = 3.663 × 10² K
T = 366.3 K
So, the temperature of the filament is 366.3 K
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please answer this question
Answer:
Pic not clear.........
HELPPP MEEE PLEADEEEE How could a cow have more momentum than a tiger ? Explain
Answer:
If a cow has a greater mass than the tiger, it has greater momentum, but only if they are moving at the same speed, or the tiger is going slower (i think)
Explanation:
the quantity of motion of a moving body, measured as a product of its mass and velocity.
Have you ever written a bio-data or an application letter? Share your experience in the
space below. (e.g. How did you find the experience? What was hard? What was easy?)
Leave the space blank if you have never written any of these.
Answer:
I found the experience tasking
Explanation:
I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.
I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn
The experience of writing a bio-data or an application letter was quite a tasking goal, where proper structure is required.
The given problem is based on the fundamentals of bio-data. Biodata is a document that is used to display the biographical data about the work experience in any organization.
As per my experience it was neither hard nor easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way. I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn.Thus, it is concluded that writing a bio-data or an application letter was quite a tasking goal, where proper structure is required.
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if 49 jules of work is done when a 7 newton wagon of cheeseburgers is pulled, how far does it move?
A person picking apples stand on a ladder 3.0 m above the ground. He throws them into
a basket 2.0 m away. How fast must the person throw the apple in order for it to land in
the basket?
Answer:
The speed the apple must be thrown in order for it to land in the basket is 2.554 m/s.
Explanation:
Given;
height above the ground, h = 3.0 m
horizontal distance, X = 2.0 m
The time to drop from the given height;
h = ¹/₂gt²
[tex]t = \sqrt{\frac{2h}{g} }\\\\t = \sqrt{\frac{2*3}{9.8} }\\\\t = 0.783 \ s[/tex]
The horizontal speed traveled by the apple is given by;
vₓ = X / t
vₓ = 2 / 0.783
vₓ = 2.554 m/s
Therefore, the speed the apple must be thrown in order for it to land in the basket is 2.554 m/s.
Your eyes see different wavelengths of light as different ________.
amplitude
color
detectors
spectrum
Answer:
Its color
Explanation: I got it right
A ferry crossed the Potomac River four times in one day. The ferry traveled the same distance for each trip, but it took different amounts of time. The time for each trip is shown in the table below. Time for Ferry to Cross on Different Trips Trip Number Time Trip 1 10 minutes Trip 2 11 minutes Trip 3 9 minutes Trip 4 12 minutes On which trip did the ferry travel the fastest? A. Trip 3 B. Trip 2 C. Trip 4 D. Trip 1
Explanation:
trip 3 ( 9 minutes)
If the ferry is moving at a faster speed, it will travel the same distance in a shorter time. The ferry crossed the river the fastest on Trip 3 because it took the shortest amount of time on that trip.
n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Use the measured length to predict the theoretical pendulum period with a range of error (use the error propagation method you learned in Lab 1). ii. Compute the percentage difference (as defined in Lab 1) between the measured value and the predicted value .
Answer:
The answer is "[tex](1.265 \pm 0.010) \ s \ and \ 0.709 \%[/tex]"
Explanation:
In point i:
[tex]T_{theo}= 2\pi \sqrt{\frac{l}{g}}[/tex]
[tex]=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s[/tex]
If error in the theoretical time period :
[tex]\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}[/tex]
[tex]= 0.010 \ s[/tex]
[tex]T_{theo} = (1.265 \pm 0.010) \ s[/tex]
In point ii:
[tex]\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100[/tex]
[tex]= \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%[/tex]A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.
For the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.
Given the data in the question;
Speed of the car; [tex]v_1 = 15m/s[/tex]
Speed for the car to double its kinetic energy; [tex]v_2 = \ ?[/tex]
Using the expression for kinetic energy:
[tex]K_E = \frac{1}{2}mv^2[/tex]
Where m is the mass and v is the velocity
Now, Initial Kinetic Energy will be;
[tex]K_E_1 = \frac{1}{2}mv_1^2 \\\\K_E_1 = \frac{1}{2}*m*(15)^2\\\\K_E_1 = \frac{1}{2}*m*225\\\\K_E_1 = 112.5*m[/tex]
For the kinetic energy to become double
[tex]K_E_2 = 2 * K_E_1\\\\\frac{1}{2}mv_2^2 = 2 * ( 112.5 * m)\\\\\frac{1}{2}mv_2^2 = 2m( 112.5)\\\\v_2^2 = 4( 112.5)\\\\v_2^2 = 450\\\\v_2 = \sqrt{450}\\\\v_2 = 21.21m/s[/tex]
Therefore, for the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.
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A force of 29.4 N is required to pull a 100 kg ice block at constant speed along an ice surface. Find the coefficient of kinetic friction for ice on ice.
Answer:
The value is [tex]\mu_k = 0.03[/tex]
Explanation:
From the question we are told that
The force applied on the ice is [tex]F = 29.4 N[/tex]
The mass of the ice block is [tex]m = 100 \ kg[/tex]
Generally for the ice block to move at constant speed , the force applied on it must be equal to the kinetic frictional force which is mathematically represented as
[tex]F_F = m* g * \mu_k[/tex]
=> [tex]F = F_F = m* g * \mu_k[/tex]
=> [tex]29.4 = 100 * 9.8 * \mu_k[/tex]
=> [tex]\mu_k = 0.03[/tex]
what are fundamental quantities
Answer: length, luminous intensity,mass, time, temperature, electric current, amount of a substance.
Explanation:
2 A car rounding a bend travels in an arc of a circle.
a What provides the centripetal force?
b Is a larger or a smaller centripetal force required if
(i) the car travels faster,
(ii) the bend is less curved,
(iii) the car has more passengers?
Answer:
a) the centripetal force causing the cat to turn in a circular path is due to friction between the car's tyres and the surface of the road.
b)
1) a larger centripetal force is required if the car travels faster
2) a larger centripetal force is needed when the bend is less curved.
3) when a car has more passengers, the mass of the car increases, therefore, a larger force is required to move the car.
Explanation:
The friction between the car's tire and the surface provides centripetal force and if the car travels faster, the bend is curved, or the car has more passengers it requires a larger centripetal force.
What is Centripetal force?When an object is moving in a curve path then a force is acting radically in the object this force is called centripetal force, this force helps to maintain the object in the curve e.g. if a car is moving in a well, so the centripetal force is responsible, so the car is not falling. The SI unit used to represent centripetal force is Newton.
A. The friction between the car's tire and the curve surface is responsible for providing the centripetal force, and it also depends on the car's speed.
B. If a car is traveling faster than a larger centripetal force is required to maintain the curve path of the car, means if the friction between the tire and surface decreases the centripetal force also decreases.
When the bend is less curved, the radius of the curve increase and thus require more centripetal force.
When people in the car increase then the mass of the car also increases so due to this the gravitation force is now more on the car so more centripetal force I required to maintain the curve path of the car.
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A 0.5 kg ball is thrown horizontally towards a wall with a speed of 10 m/s. The initial velocity is chosen to be the positive x -direction for this question. The ball horizontally rebounds back from the wall with a speed of 10 m/s in the negative x -direction. What is momentum of the ball before it hits the wall, p? What is momentum of the ball after it hits the wall, p,? What is the change in momentum of the ball, delta p? (Give both magnitude and direction for each answer.) Is momentum conserved for the ball?
Explanation:
Step one:
given dara
mass of ball m=0.kg
initial velocity u= 10 m/s
final velocity v=10m/s
Required
momentum before and after impact
the expression for momentum P
P=mv
before impact
P=mv
P1=0.5*10
P1=5(kg⋅m/s)
after impact
P=mu
P2=0.5*10
P2=5(kg⋅m/s)
change in momentum is =P1-P2= 5-5=0
yes, momentum is conserved.
A particle with charge q and mass m moving with speed v in the x direction enters a magnetic field of strength B pointing in the y direction. The magnitude of the acceleration of the particle as it travels one semicircle is
Answer:
The acceleration of the particle as it travels one semicircle is
[tex]a= \frac{\pi R}{t^2}[/tex]
Explanation:'
Kindly see attached a sketch of a semi-circle
Step one:
given data
velocity =v
let the time taken be t
The path PQM is the distance covered
so distance [tex]d= \pi R[/tex]
we know that time= distance/velocity
t= πR/v
step two:
velocity =distance/time
[tex]velocity=\frac{\pi R}{t}[/tex]
also, we know that acceleration is velocity/time
[tex]a= \frac{\pi R}{\frac{t}{t} }[/tex]
[tex]a= \frac{\pi R}{t}*\frac{1}{t}[/tex]
[tex]a= \frac{\pi R}{t^2}[/tex]
To remove a stain using a solvent the stain has to become dissolved in the solvent
True
False
Answer:
True
Explanation:
have a good day:)
Answer: This statement is True
While Barb was playing soccer, she was kicked in the anterior thigh by an opposing teammate. Due to this injury, what muscle function may she find difficult to perform?
Answer:
knee extension is the muscle function that will be difficult to perform.
Explanation:
Barb was kicked in the anterior thigh. Now, the thigh muscles performs a combined operation of moving the knee and leg and they reside in the following compartments.
- Anterior compartment which is composed of knee joint extension and thigh flexion.
- Lateral Compartment which is composed of the tensor fasciae latae, which is a tiny muscle that abducts and centrally will make the thigh to rotate.
-Medial compartment which involves thigh addiction which is rotating of the thigh around the hips.
- Posterior compartment which involves knee joint flexion and high extension.
Thus, from the different compartments listed above, we can see that the muscles that extend the knee and flex the thigh all lie in the anterior compartment of the upper leg.
Thus, we can conclude that knee extension is the muscle function that will be difficult to perform.
What kind of energy do electric and magnetic fields contain?
A
thermal energy
B
electrical energy
C
kinetic energy
D
pot
Answer:
Electromagnetic waves
Electromagnetic waves bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not.
So the answer is B electrical energy
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A metallic spherical shell has a charge density of 1 mC/m2 on its surface. The shell has radius of 7.5 cm. Determine the electric field magnitude (in kV/m) 10 meters away from the center of the shell.
Answer:
Explanation:
charge on the shell = 4πR² x charge density
= 4 x 3.14 x 7.5² x 10⁻⁴ x 10⁻³ C
= 706.5 x 10⁻⁷ C
electric field = k Q / d² , d is distance of point from centre
= 9 x 10⁹ x 706.5 x 10⁻⁷ / 10²
= 6358.5 N /C
= 6.358 kV / m
At the focal point of a converging lens, the object cannot be seen. O A. True B. False
Answer: True
Explanation: Guessed on A p E x and it was correct
Answer:
True
Explanation:
a p e x
How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s²
Answer:
108 N
Explanation:
Use Newton's second law.
F = ma
F = (90 kg) (1.2 m/s²)
F = 108 N
5)
Using only the information available in the periodic table, consider the elements calcium and chlorine. From their
location on the periodic table, identify the oxidation state and number of valence electrons for calcium and chlorine.
Which statement most accurately describes the compound formed by calcium and chlorine?
C
A)
B)
Calcium, a nonmetal with an oxidation number of +2 will form a covalent
bond with chlorine, a halogen (nonmetal) with an oxidation number of -1
called calcium chloride (CaCl2).
Calcium, an alkaline earth metal with an oxidation number of +2 will form
covalent bond with chlorine, also a metal with an oxidation number of -1
called calcium dichloride (CaCla)
Calcium, an alkaline earth metal with an oxidation number of +2 will form
an ionic bond with chlorine, a halogen in group VILA with an oxidation
number of -1 called calcium chloride (CaCl2).
Calcium, an alkaline earth metal with an oxidation number of 2 will share
electrons to form an lonic bond with chlorine, a nonmetal with an
xidation number of -1 called calcium dichloride (CaCl).
D)
Answer:C,(Calcium,an alkaline earth metal with an oxidation number of +2 will form an ionic bond with chlorine,a halogen in group VllA with an oxidation number of -1 called calcium chloride (CaCl2)
Explanation:
on USAtestprep !!
Calcium, an alkaline earth metal with an oxidation number of +2 will form
an ionic bond with chlorine, a halogen in group VIIA with an oxidation
number of -1 called calcium chloride (CaCl₂). This is correct statement.
What is oxidation number?Simply said, the number assigned to each element in a chemical combination is the definition of an oxidation number. The total number of electrons that an atom in a molecule can share, lose, or gain while forming a chemical bond with an atom of a different element is known as the oxidation number.
Also known as oxidation state, oxidation number is a numerical value. But depending on whether we take into account the atoms' electronegativity or not, these phrases might occasionally have a different meaning. In coordination chemistry, the term "oxidation number" is often used.
According to Periodic table: calcium is a alkaline earth metal with an oxidation number of +2 whereas chlorine is a halogen in group VIIA with an oxidation number of -1. When they reacts chemically, they form an ionic compound named calcium chloride having chemical formula CaCl₂.
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If the resistance in the coil is 2.0 , what is the magnitude of the induced current in the coil while the field is changing
This question is incomplete, the complete question is;
A coil consists of 200 turns of wire. Each turn is a square of side 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to 0.50 T in 0.80 s,
a) what is the magnitude of the induced emf in the coil while the field is changing?
b) if the resistance of the coil is 2.0, what is the magnitude of the induced current in the coil while the field is changing?
Answer:
a) the magnitude of the induced emf in the coil is 4.05 V
b) induced current in the coil I is 2.025 A
Explanation:
Given that;
side of turn a = 18 cm = 0.18 m
no. of turns N = 200
dB = 0.50 T
time t = 0.80 sec
(a)
what is the magnitude of the induced emf in the coil while the field is changing?
we know that the magnetic flux is equal to the product of the magnetic field in a loop and the area of the loop so;
∅ = NBA
expression for the electromotive force is expressed as;
∈ = d∅/dt
Now replace NBA for ∅ in the above equation.
∈ = d(NBA) / dt
= NA(dB/dt)
The expression for the area of each square turn is expressed as follows
A = a²
a is the side of the turn
so we substitute the value of a
A = (0.18) ²
A= 0.0324 m²
As earlier derived
formula for the electromotive force is as follows:
∈ = NA(dB/dt)
so we substitute all our values
∈ = (200)(0.0324m²) (0.50T/0.80s)
∈ = 6.48 × 0.625
∈ = 4.05 V
Therefore the magnitude of the induced emf in the coil is 4.05 V
(b)
if the resistance of the coil is 2.0, what is the magnitude of the induced current in the coil while the field is changing?
we know that the current induced in the circuit is equal to the ratio between the electromotive forces to the resistance of the ring so;
I = ∈ / R
given that; resistance of the coil = 2.0
so we substitute
I = 4.05 / 2.0
I = 2.025 A
Therefore induced current in the coil I is 2.025 A
How much energy is required to move a 1000 kg object from the Earth's surface to an altitude twice the Earth's radius?
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
Since the object must be moved away to a distance greater than the radius of the Earth, then change in gravitational potential energy must be based on Newton's Law of Gravitation.
By the Work-Energy Theorem, the work ([tex]W[/tex]), in joules, done on the object is equal to the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]W = U_{g}[/tex] (1)
[tex]W = -G\cdot m\cdot M\cdot \left(\frac{1}{r_{f}}-\frac{1}{r_{o}} \right)[/tex] (1b)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.[tex]m[/tex] - Mass of the object, in kilograms.[tex]M[/tex] - Mass of the Earth, in kilograms.[tex]r_{o}[/tex] - Initial distance, in meters.[tex]r_{f}[/tex] - Final distance, in meters.If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 1000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]r_{o} = 6.371\times 10^{6}\,m[/tex] and [tex]r_{f} = 19.113\times 10^{6}\,m[/tex], then the energy required to move the object from the Earth's surface is:
[tex]W = -\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1000\,kg)\cdot (5.972\times 10^{24}\,kg)\cdot \left[\frac{1}{19.113\times 10^{6}\,m} - \frac{1}{6.371\times 10^{6}\,m} \right][/tex][tex]W = 4.171\times 10^{10}\,J[/tex]
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
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An object accelerates 2.0 m/s2 when a force of 25.0 Newtons is applied to it. What is the mass of the object?
Answer:
12.5 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{25}{2} \\ [/tex]
We have the final answer as
12.5 kgHope this helps you
Please select the word from the list that best fits the definition The skeletons of tiny ocean animals grow together to form ______, structures that are found only in warm, clear ocean water.
Answer:
coral reef
Explanation:
During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +48.5" m"/s at an angle of 42.0°, how long does it take the stone to hit the ground? For those settings, what is the maximum range? How high will the stones go? Show all your work
Answer:
a) t = 6.62 s
b) x = 238.6 m
c) H = 53.7 m
Explanation:
a) We can find the time of flight as follows:
[tex] y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex]y_{f}[/tex] is the final height = 0
[tex]y_{0}[/tex] is the initial height = 0
[tex]v_{0_{y}}[/tex] is the initial vertical velocity of the stone
t: is the time
g: is the gravity = 9.81 m/s²
[tex] v_{0}sin(42)t - \frac{1}{2}gt^{2} = 0 [/tex]
[tex] 48.5 m/s*sin(42)*t - \frac{1}{2}9.81 m/s^{2}*t^{2} = 0 [/tex]
By solving the above quadratic equation we have:
t = 6.62 s
b) The maximum range is:
[tex] x = v_{0_{x}}t = 48.5 m/s*cos(42)*6.62 s = 238.6 m [/tex]
c) The maximum height (H) can be found knowing that at this height the final vertical velocity of the stone is zero:
[tex] v_{f_{y}}^{2} = v_{0_{y}}^{2} - 2gH [/tex]
[tex] H = \frac{v_{0_{y}}^{2} - v_{f_{y}}^{2}}{2g} = \frac{(48.5 m/s*sin(42))^{2} - 0}{2*9.81 m/s^{2}} = 53.7 m [/tex]
I hope it helps you!
find the mass of an object with a density of 1.5 g/cm^3 and had a volume of 8cm^3
Answer:
12 gExplanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volume
From the question we have
mass = 1.5 × 8
We have the final answer as
12 gHope this helps you
This is a graph of a car speeding up and then reaching a constant speed of 105 m/s . After it finishes speeding up, how far does the car travel during the last 6 seconds of the graph, from 4 to 10 s?
60 m
Explanation:Concept Used:
We know that the area under a velocity-time graph represents the Displacement of the body
Displacement in the Last 6 seconds:
To find the Displacement in the last 6 seconds, we will find the area under the graph between x = 4 and x = 10
We can see that the shape formed is a rectangle also shown in the given graph. So, the area of the rectangle is the Displacement of the car in the last 6 seconds
Area of the Rectangle:
From the graph, we know that the rectangle is 10 (m/s) tall and 6 (s) wide
Area of Rectangle= length*Breadth
replacing the values
Area = 10 (m/s) * 6 (s)
Area = 60 m
Hence, the car travelled 60 m in the last 6 seconds of the graph
A racecar accelerates uniformly from 18.5 mil to 46.1 m/s in 2.47 seconds.
What is the acceleration of the car?*
Answer:
The acceleration of the racecar is [tex]\mathbf{11.17~m/s^2}[/tex]
Explanation:
Uniformly Accelerated Motion
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Following the definition above, the acceleration is defined as:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.
The racecar goes from vo=18.5 m/s to vf=46.1 m/s in t=2.47 seconds, thus the acceleration is:
[tex]\displaystyle a=\frac{46.1-18.5}{2.47}[/tex]
[tex]\displaystyle a=\frac{27.6}{2.47}[/tex]
[tex]a = 11.17~m/s^2[/tex]
The acceleration of the racecar is [tex]\mathbf{11.17~m/s^2}[/tex]
A boy throws a tell ball straight up . disregarding any effects of air resistance, the forces acting 9n the ball untill it returns to the ground is are