Answer:
Truck [tex]\dfrac{g}{10}[/tex]
Road [tex]-\dfrac{g}{10}[/tex]
Explanation:
[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]
[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]
Frictional force is given by
[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]
Net acceleration is given by
[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]
The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.
Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10-m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope (5-m diameter) on Palomar Mountain in California
Answer:
Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Explanation:
If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.
Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
What is surface tension
Answer:
Surface tension is, the surface where the water meets the air, water molecules cling even more tightly to each other.
Which of the following is not a unit of speed *
a) m/s
b) km/s
c) mph [miles per hour]
d) light year
Answer:
I think it's light year but there shouldn't be also km/s but km/h
A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has a diameter of 20 cm. What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s? Hint: To find total linear acceleration (in m/s2)you need to have tangential and radial acceleration . Tangential acceleration can be calculated using angular acceleration at=rα To find redial acceleration you need to calculate final linear speed of that point and use ar=vf2r . And vf=rωf.
Answer:
The total linear acceleration is approximately 0.246 meters per square second.
Explanation:
The total linear acceleration ([tex]a[/tex]) consist in two components, radial ([tex]a_{r}[/tex]) and tangential ([tex]a_{t}[/tex]), in meters per square second:
[tex]a_{r} = \omega^{2}\cdot r[/tex] (1)
[tex]a_{t} = \alpha \cdot r[/tex] (2)
Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:
[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex] (3)
Where:
[tex]r[/tex] - Radius of the wheel, in meters.
[tex]\omega[/tex] - Angular speed, in radians per second.
[tex]\alpha[/tex] - Angular acceleration, in radians per square second.
Given that wheel accelerates uniformly, we use the following kinematic equation:
[tex]\omega = \omega_{o}+ \alpha\cdot t[/tex] (4)
Where:
[tex]\omega_{o}[/tex] - Initial angular speed, in radians per second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]r = 0.1\,m[/tex], [tex]\alpha = 2\,\frac{rad}{s^{2}}[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex] and [tex]t = 0.60\,s[/tex], then the total linear acceleration is:
[tex]\omega = \omega_{o}+ \alpha\cdot t[/tex]
[tex]\omega = 1.2\,\frac{rad}{s}[/tex]
[tex]a_{r} = \omega^{2}\cdot r[/tex]
[tex]a_{r} = 0.144\,\frac{m}{s^{2}}[/tex]
[tex]a_{t} = \alpha \cdot r[/tex]
[tex]a_{t} = 0.2\,\frac{m}{s^{2}}[/tex]
[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex]
[tex]a \approx 0.246\,\frac{m}{s^{2}}[/tex]
The total linear acceleration is approximately 0.246 meters per square second.
In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 4.07 kg. They then hang the object on a pivot located 0.155 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 247 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis
Answer:
I = 0.65 kgm²
Explanation:
Since the mass is an inertial pendulum, we use the formula for the period, T of an inertial pendulum.
T = 2π√(I/mgh) where I = moment of inertia of object about pivot point, m = mass of object5 = 4.07 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.155 m.
Given that the team measures 113 cycles of oscillation in 247 s, the period, T = time of oscillations/total number of oscillations = 247 s/113 oscillations = 2.186 s/oscillation
So, T = 2.186 s
We now find I by making it subject of the formula in the equation for T.
So,
T = 2π√(I/mgh)
dividing both sides by 2π, we have
T/2π = √(I/mgh)
squaring both sides, we have
(T/2π)² = [√(I/mgh)]²
T²/4π² = I/mgh
multiplying both sides by mgh, we have
T²mgh/4π² = I
I = T²mgh/4π²
substituting the values of the variables into the equation, we have
I = T²mgh/4π²
I = (2.186 s)² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 4.778 s² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 29.539 kgm²/4π²
I = 0.748 kgm²
Now I = I' + mh² (parallel axis theorem) where I' = moment of inertia of object about its center of mass, m = mass of object = 4.07 kg and h = distance of center of mass object from pivot point.
So, I' = I - mh²
Substituting the values of the variables into the equation, we have
I' = I - mh²
I' = 0.748 kgm² - 4.07 kg × (0.155 m)²
I' = 0.748 kgm² - 4.07 kg × 0.02403 m²
I' = 0.748 kgm² - 0.098 kgm²
I = 0.65 kgm²
which of the following best defines spring constant ?
a. the amount of force needed to extend or compression of a spring for every 1 kilogram of the spring.
b. the amount of force needed every 1 meter of stretch or compression of the spring.
c. the amount of energy needed to extend or compress a spring for every 1 kilogram of mass of the spring.
d. the amount of energy needed for every 1 meter of stretch or compression of the spring.
Answer:
your answer gonna be The letter C is the correct answer
The instantaneous speed of a cat running 400 meters in 20 seconds is 20 meters per second.
OTrue
O False
Answer:
true
Explanation:
400/20=20
Answer:
True
Explanation:
400/20=20mps
plz mark me as brainliest.
10 POINTS!! SPACE QUESTION!!
Answer: A (first option)
Explanation: In general, the surface temperatures decreases with increasing distance from the sun. Any warm object in space loses heat via infrared radiation; also visible radiation if it is hot enough. The hotter is the object, the faster it loses heat. Consequently any object warmed by the Sun stays at equilibrium temperature - it warms up until the radiative heat loss equals to the heat received, and cannot get any warmer beyond that.
Planet X has a moon similar to Earth’s moon.
Which path would this moon’s orbit take?
All the questions are in the photos above. Thanks guys!
Answer:
right
Explanation:
What are the two forces acting on a balanced hot air balloon.
of
O a. Gravity and Weight
O b.
Air resistance and upthrust
O c. Applied force and upthrust
d.
Gravity and air resistance
Answer:
I want to say option C) Applied force and upthrust.
Explanation:
Sorry if my answer is wrong. You can look in your book or study up on the website.
Consider the model of the energy transformation of this system. What can you say about N and M in the model?
A) M < N
B) N = M
C) N < M
D) N + M =
Model of the energy transformation of this system we can say about N and M in the model is M < N.
What is energy?Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
The model shows the energy transformation in electrical appliance Fan the is lass of energy in form of heat so M < N.
Model of the energy transformation of this system we can say about N and M in the model is M < N.
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Consider a block of mass m at rest on an inclined plane of angle theta. An acrobat of mass m_A is standing on the top corner of the block so that their weight presses vertically (perpendicular to the flat ground under the incline) downward. If the coefficient of static friction between the block and the incline is mu_s find an expression relating m_A, m, theta, and mu_s just before the block begins slipping. If the condition for slipping does not involve some of these parameters, leave them out.
Enter your expression as best you can and I will check it. Do not forget to upload your work as well.
Answer:
μ = tan θ
Explanation:
For this exercise let's use the translational equilibrium condition.
Let's set a datum with the x axis parallel to the plane and the y axis perpendicular to the plane.
Let's break down the weight of the block
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
The acrobat is vertically so his weight decomposition is
sin θ = = wₐₓ / wₐ
cos θ = wₐ_y / wₐ
wₐₓ = wₐ sin θ
wₐ_y = wₐ cos θ
let's write the equilibrium equations
Y axis
N- W_y - wₐ_y = 0
N = W cos θ + wₐ cos θ
X axis
Wₓ + wₐ_x - fr = 0
fr = W sin θ + wₐ sin θ
the friction force has the formula
fr = μ N
fr = μ (W cos θ + wₐ cos θ)
we substitute
μ (Mg cos θ + mg cos θ) = Mgsin θ + mg sin θ
μ = [tex]\frac{(M +m) \ sin \ \theta }{(M +m) \ cos \ \theta }[/tex]
μ = tan θ
this is the minimum value of the coefficient of static friction for which the system is in equilibrium.
has a man he has married many women but has never been married before who is he
Answer:
Explanation:
The answer is a priest or a moulana
A plano concave lens is one that has a surface that curves inward, making the lens thinner at its center than at its edges. Its effect on light is :_______.
a.) to make it appear as if incident parallel light is coming from a point close to the lens on the opposide side from the light source
b.) the diverge in incident beam of light
c.) to converge light to a focus
d.) to make it appear as if incident parallel light is coming from a point close to the lens on the same side as the light source
Answer:
b.) the diverge in incident beam of light
Explanation:
A lens refers to any refracting surface. The manner in which a lens is curved tells whether it is a concave(diverging) lens or a convex (converging) lens.
A convex lens is thicker at the center than at the edges. It makes light to converge at a point.
A concave lens is is thinner at the center than at the edges. It makes light to appear to diverge from a point, hence the answer above.
⦁ An electron is moving through a 10 T magnetic field at a speed of 3.5 x107m/s perpendicular to the direction of the field. What is the force that applied on the charge?
Answer:
F = 5.6 10⁻¹¹ N
Explanation:
The magnetic force is given by
F = q v x B
The bold indicate vectors, the scalar form of this expression is
F = q v B sin θ
in this case they indicate that the speed and the magnetic field are perpendicular, so the angle is 90º and the sin 90 = 1
F = q v B
the magnitude of the force is
F = 1.6 10⁻¹⁹ 3.5 10⁷ 10
F = 5.6 10⁻¹¹ N
If objects are traveling in opposite directions, what do you know about the signs of their momenta?
Explanation:
the impulse and momentum change on each object are equal in magnitude and opposite in directions. Thus the total momentum is reserved
They have no overall momentum at all. They are travelling in opposing directions yet having the same mass and velocity. Their momentum vectors add up to exactly zero when added together.
What If objects are travelling in opposite directions?When two objects collide, opposite-direction forces of equal magnitude are exerted to each item. When there are such pressures, it usually happens that one item speeds up and gains momentum, while the other object slows down (lose momentum).
Think about a situation where two similar objects are going in opposite directions at the same speed. It's noteworthy to note that despite both items moving, the momentum of the system as a whole is zero because the oppositely oriented vectors cancel out.
Therefore, Every object experiences an equal but opposite change in impulse and momentum. Thus, the entire momentum is held back.
Learn more about objects here:
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You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons to ensure that on the order of 1% of these copper nuclei absorb a neutron. After activation, you turn off the neutron flux and then use a highly efficient detector to monitor the gamma radiation that comes out of the sample. Assume half of the 66Cu nuclei emit a 1.04-MeV gamma ray in their decay. (The other half of the activated nuclei decay directly to the ground state of 66Ni.) (Enter your answer using one of the following formats: 1.2e-3 for 0.0012 and 1.20e 2 for 120.)
Required:
a. If after 10 min (two half-lives) you have detected 10000 MeV of photon energy at 1.04 MeV, approximately how many 65Cu atoms are in the sample?
b. Assume the sample contains natural copper. Refer to the isotopic abundances listed in your text (Chemical and Nuclear Information for Selected Isotopes) and estimate the total mass of copper in the sample.
Answer:
a) number of copper atoms 65 (⁶⁵Cu) is 7.692 10⁶ atoms
b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg
Explanation:
a) For this exercise let's start by using the radioactive decay ratio
N = N₀ [tex]e^{- \lambda t}[/tex]o e - lambda t
The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are
N = ½ (½ N₀) = ¼ N₀
N₀ = 4 N
in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV
# _atoms = 1 atom (photon) (E / Eo)
# _atoms = 1 10000 / 1.04
# _atoms = 9615,4 atoms
N₀ = 4 #_atoms
N₀ = 4 9615,4
N₀= 38461.6 atoms
in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms
N_activated = 2 # _atoms
N_activated = 2 38461.6
N_activated = 76923.2
also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment
N_activated = 0.01 N_total
N_total = N_activated / 0.01
N_total = 76923.2 / 100
N_total = 7.692 10⁶ atoms
so the number of copper atoms 65 (⁶⁵Cu) is 7.692 10⁶
b) the natural abundance of copper is
⁶³Cu 69.17%
⁶⁵Cu 30.83%
Let's use a direct proportion rule. If there are 7.692 10⁶ ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%
# _63Cu = 69.17% (7.692 10⁶ / 30.83%)
# _63Cu = 17.258 10⁶ atom ⁶³Cu
the total amount of comatose is
#_total Cu = #_ 65Cu + # _63Cu
#_total Cu = (7.692 + 17.258) 10⁶
#_total Cu = 24.95 10⁶
the atomic mass of copper is m_Cu = 63.546 u
m_total = #_totalCu m_Cu
m_total = 24.95 10⁶ 63,546 u
m_total = 1.585 10⁹ u
let's reduce to kg
m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)
m_total Cu = 2.632 10⁻¹⁸ kg
Imagine you are on a space mission and you are 6 AU's from the Sun and you use a light sensor to measure the brightness of the Sun. The amount of sunlight received per square centimeter would be different by what factor compared to the same measurement on Earth at AU
Answer:
36 times less.
Explanation:
The distance from you to the sun is 6AU's, and from the sun to the earth is 1 AU.
Therefore,
At Earth sunlight received per unit cm² is:
[tex]I_{earth} = \dfrac{I_o}{4 \pi \times (1)^2}[/tex]
[tex]I_{earth} = \dfrac{I_o}{4 \pi}[/tex]
[tex]I_{me} =\dfrac{I}{4 \pi (6)^2}[/tex]
[tex]I_{me} = \dfrac{I_o}{36(4 \pi)}[/tex]
Thus, [tex]I_{earth} = 36 \times I_{me}[/tex]
Thus, the right answer is 36 times less.
Student 1 and student 2 talk about the speed of the truck
Answer:
Explanation:
any more explanation?
Two long, straight wires are separated by a distance of 32.2 cm. One wire carries a current of 2.75 A, the other carries a current of 4.33 A. (a) Find the force per meter exerted on the 2.75-A wire. (b) Is the force per meter exerted on the 4.33-A wire greater than, less than, or the same as the force per meter exerted on the 2.75-A wire
Answer:
a)[tex]\frac{F_1}{L}=1.95*10^-^5N[/tex]
b)[tex]\frac{F_2}{L}=1.95*10^-^5N[/tex]
Explanation:
From the question we are told that:
Distance between wires [tex]d=32.2[/tex]
Wire 1 current [tex]I_1=2.75[/tex]
Wire 2 current [tex]I_2=4.33[/tex]
a)
Generally the equation for Force on [tex]l_1[/tex] due to [tex]I_2[/tex] is mathematically given by
[tex]F_1=I_1B_2L[/tex]
Where
B_2=Magnetic field current by [tex]I_2[/tex]
[tex]B_2=\frac{\mu *i_2}{2\pi d}[/tex]
Therefore
[tex]F_1=I_1B_2L[/tex]
[tex]F_1=I_1(\frac{\mu *i_2*l_1}{2\pi d})L[/tex]
[tex]\frac{F_1}{L} =\frac{4*\pi*10^{-7}*2.75*4.33*100 }{2*\pi*12.2 }[/tex]
[tex]\frac{F_1}{L}=1.95*10^-^5N[/tex]
b)
Generally the equation for Force on [tex]I_2[/tex] due to [tex]I_1[/tex] is mathematically given by
[tex]F_2=I_2B_1L[/tex]
Where
B_1=Magnetic field current by [tex]I_2[/tex]
[tex]B_1=\frac{\mu *I_1}{2\pi d}[/tex]
Therefore
[tex]\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})[/tex]
[tex]\frac{F_2}{L}=1.95*10^-^5N[/tex]
Student Exploration: Energy Conversion in a SystemNCVPS Chemistry Fall 2014Vocabulary: energy, gravitational potential energy, heat energy, kinetic energy, law of conservation of energy, specific heat capacityPrior Knowledge Questions (Do these BEFORE using the Gizmo.)A battery contains stored energy in the form of chemical energy.1. What are some examples of devices that are powered by batteries? ____________________________________________________________________________________________2. What different forms of energy are dmonstrated by these devices? ___________________ _________________________________________________________________________Gizmo Warm-upEnergy constantly changes from one form to another, but in a closed system, the total amount of energy always remains the same. This concept is known formally as the law of conservation of energy.The Energy Conversion in a System Gizmo™ allows you to observe the law of conservation of energy in action. In the Gizmo, a suspended cylinder has gravitational potential energy. When the cylinder is released, the
Answer:
a) Battery-operated devices have: small led lights, flashlights, wireless keyboards and mice, watches, electronic weights
b) ed lights and flashlights transform into light energy and thermal energy
c) Em₀ = U = m gh, Em_f = K = ½ m v²
Explanation:
In this exercise ask to complete the sentences
a) Battery-operated devices have: small led lights, flashlights, wireless keyboards and mice, watches, electronic weights
b) These devices transform the chemical energy stored in the batteries into other forms of energy.
Led lights and flashlights transform into light energy and thermal energy
Keyboards transform into electromagnetic energy that is emitted
clocks transform to mechanical energy from the movement of the needles
Electronic weights transforms the chemical energy of the baria into gravitational potential energy that prevents the movement of the plate and this translates into the reading of the body weight
c) The total energy of the cylinder mechanical energy when sustained is
Em₀ = U = m gh
it is transformed as it descends into kinetic energy, at any point
Emₙ = K + U = 1/2 m v² + m g y
at the lowest point of the trajectory all energy is transformed
Em_f = K = ½ m v²
ILL GIVE BRAINLIST PLS In which circuit does charge reverse direction many times per second?
A. A DC circuit
B. A combined circuit
C. A parallel circuit D. An AC circuit
Answer: D. An AC circuit
Explanation:
I took it on a test and it was correct ; )
What is a negative effect of increased carbon dioxide within the carbon cycle?
Answer:
Carbon dioxide controls the amount of water vapor in the atmosphere and thus the size of the greenhouse effect. Rising carbon dioxide concentrations are already causing the planet to heat up
Explanation:
Hope it helps! Correct me if I am wrong
Im sure about my answer
PLZ help 10 points!!! space question!
Answer:
B. They are smaller and made of rocky material
Explanation:
i think it's right??
Chemotherapy is used to treat
cervical cancer.
endometriosis.
undescended testicles.
benign prostatic hypertrophy.
Answer:
Cervical cancer
Explanation:
Chemotherapy is a common treatment or most cancers
Answer:
A
Explanation:
A skater spins with an angular speed of 5.9 rad/s with her arms outstretched. She lowers her arms, decreasing her moment of inertia by a factor of 1.7. Ignoring friction on the skates, determine the ratio of her final kinetic energy to her initial kinetic energy.
Answer:
the ratio of her final kinetic energy to her initial kinetic energy is 1.7.
Explanation:
Given;
initial angular speed, ω₁ = 5.9 rad/s
let her initial moment of inertia = I₁
her final moment of inertia [tex]I_2 = \frac{I_1}{1.7}[/tex]
Apply the principle of conservation of angular momentum to determine the final angular speed of the girl;
[tex]\omega_1I_1 = \omega_f I_2\\\\\omega_f = \frac{\omega _1 I_1}{I_2} \\\\\omega_f = \frac{5.9 \times I_1}{I_1/1.7} \\\\\omega = 5.9 \times 1.7 \\\\\omega_f = 10.03 \ rad/s[/tex]
The initial rotational kinetic energy is given as;
[tex]K.E_I = \frac{1}{2}I_1 \omega_I ^2[/tex]
The final rotational kinetic energy is given as;
[tex]K.E_f = \frac{1}{2}I_2 \omega_f ^2[/tex]
The ratio of her final kinetic energy to her initial kinetic energy is given as;
[tex]\frac{K.E_f}{K.E_I}= \frac{\frac{1}{2}I_2 \omega_f^2 }{\frac{1}{2} I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_2 \omega_f^2}{ I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_1/1.7 \times \omega_f^2}{ I_1 \times \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ \omega_f^2}{ 1.7 \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ (10.03)^2}{ 1.7(5.9)^2} = \frac{17}{10} = 1.7[/tex]
Therefore, the ratio of her final kinetic energy to her initial kinetic energy is 1.7.
A geologist notices that a river is eroding its valley at a constant rate. Knowing the height of the valley walls, how could the geologist figure out when the river started carving the valley?
A.
Count growth rings of trees growing on the valley floor.
B.
Divide the height of the valley walls by the rate of erosion.
C.
Fill up the river valley with rocks, and time how long it takes the rocks to wash out.
D.
Sit and observe the river for a few hours until the valley walls double in height.
Answer:
B.
Divide the height of the valley walls by the rate of erosion.
Explanation:
There is a relationship between the rate of erosion and the hieght at which it is eroded according to Newton's law of motion. In the case of the scenario above, the best way to determine the time the river started carving the valley would be the division of the height of the valley walls by the rate of erosion.
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest. It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.48 kg and 2.77 kg, and the length of the wire is 1.11 m. Find the velocity of the ball just after the collision.
Answer: Velocity of the ball just after the collision is -1.414 m/s.
Explanation:
As energy is conserved in a reaction so here, energy before collision will be equal to the energy after collision.
[tex]E_{before} = mgh = E_{after} = \frac{1}{2}mv_{o}^{2}[/tex]
where,
m = mass
g = gravitational energy = [tex]9.8 m/s^{2}[/tex]
h = height or length
[tex]v_{o}[/tex] = initial velocity
Also here, height is the length of wire. Let the height be denoted by 'L'. Therefore,
[tex]\frac{1}{2}mv_{o}^{2} = mgL\\v_{o}^{2} = 2gL\\v_{o} = \sqrt{2gL}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.11 m}\\= 4.66 m/s[/tex]
Formula used to calculate velocity after the collision is as follows.
[tex]v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}][/tex]
where,
[tex]v_{f ball}[/tex] = final velocity of ball after collision
[tex]m_{ball}[/tex] = masses of ball
[tex]m_{block}[/tex] = masses of block
Substitute the values into above formula as follows.
[tex]v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}]\\= 4.66 m/s [\frac{1.48 kg - 2.77 kg}{1.48 kg + 2.77 kg}]\\= 4.66 m/s \times (-0.303)\\= -1.414 m/s[/tex]
Thus, we can conclude that velocity of the ball just after the collision is -1.414 m/s.
Can someone take there time and answer this :)
Answer: I think B.)
Explanation: