Answer:
Statement 1: While the ball was rolling down the hill the forces acting on it were unbalanced.
Explanation:
ball is dropped from a height of 45 m on a floor. If at each collision with the floor the ball loses the nineteen percent of kinetic energy then the speed of the ball just after striking the floor second time, is (g = 10 m s–2)
The speed of the ball just after striking the floor a second time, is 30.0 m/s.
Initial height (h) = 45 m
Acceleration due to gravity (g) = 10 m/s²
Energy loss per collision (k) = 19% = 0.19
At each collision with the floor, the ball loses 19% of its kinetic energy, which means the remaining kinetic energy is 81% (100% - 19%).
When the ball reaches the floor for the first time, it has converted all its potential energy into kinetic energy. So, the initial kinetic energy (K₁) is equal to the potential energy (PE) at the initial height:
K₁ = PE = mgh
Now, let's consider the ball's motion from the initial height to the first collision point. The ball undergoes free fall, so we can use the equations of motion:
h = (1/2)gt²
t = sqrt(2h/g)
Using this time, we can calculate the initial kinetic energy (K₁):
K₁ = mgh = m * 10 m/s² * 45 m
Since the ball loses 19% of its kinetic energy at each collision, the remaining kinetic energy is 81%:
K₂ = K₁ * 0.81
The ball then rebounds elastically from the floor, conserving both kinetic energy and speed. Therefore, the speed just after striking the floor for the second time (v₂) is equal to the speed just before the first collision (v₁):
v₂ = v₁
To find the speed just before the first collision (v₁), we can use the equation of motion:
v = gt
Substituting the time (t) we found earlier, we have:
v₁ = g * sqrt(2h/g)
Now, we can substitute the known values and calculate the speed just after striking the floor for the second time:
v₁ = 10 m/s² * sqrt(2 * 45 m / 10 m/s²)
v₂ = v₁
By evaluating the expression, we find:
v₁ ≈ 30.0 m/s
v₂ ≈ 30.0 m/s
Therefore, the speed of the ball just after striking the floor for the second time is 30.0 m/s.
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An 75-kilogram skier slides on waxed skis along a horizontal surface of snow at constant velocity while pushing with his poles. What is the horizontal component of the force pushing him forward? (1) 0.05 N (2) 40 N (3) 37 N (4) 4 N
[Use: Ff = µ Fn; remember Fn = Fg = mg]
The horizontal component of the force pushing him forward is 37 N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Frictional force due to ice, F = μN
Normal force N = mg = 75*9.8 = 735 N
Frictional force = 0.05*735 = 36.75 N
Coefficient of friction, μ for ice is 0.05
The horizontal component of the force pushing him forward is 37 N.
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Need Help Please and Thank You). Question 1: Explain how longitudinal waves and transverse waves are similar to each other and different from each other. Question 2) Compare and contrast reflection and refraction. Question 3) What does it mean that energy is neither created nor destroyed? ( Please No Plagiarism and Be 100% positive with your answer. Will Mark Brainliest. If you write anything down that doesn't contain to my question you will be reported. Need two answers.
1) The longitudinal waves move in parallel direction to the source of vibration. The transverse waves move in perpendicular direction to the source of vibration . They are similar in the sense that energy is transferred in the form of waves.
2) Reflection occurs when waves bounce from a surface back toward the source. Refraction is when waves are deflected when the waves go through a substance.
3) The Law of Conservation of Energy
states that energy can neither be created
nor destroyed, it can only be transferred or
changed from one form to another.
When a skater pulls her arms in, it
reduces her moment of inertia from
2.12 kg m² to 0.699 kg-m². If she was
initially spinning 3.25 rad/s, what is
her final angular velocity?
The skater's final angular velocity is approximately 9.86 rad/s.
The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, the skater has an angular momentum of:
L_initial = I_initial * ω_initial
Substituting the given values:
L_initial = 2.12 kg m² * 3.25 rad/s
The skater's final angular momentum remains the same, as angular momentum is conserved:
L_final = L_initial
The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:
L_final = I_final * ω_final
0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s
Solving for ω_final:
ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²
Hence, the skater's final angular velocity is approximately 9.86 rad/s.
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What is an indicator?
A. An organic compound that changes color in acids and bases
B. A strong acid
C. The amount of ions in solution
D. The concentration of acids or bases
Gold-198 is a radioactive isotope, and it has a half-life of about 2.5 days.
You have 100 grams of Gold-198. How many grams remain after 20 days?
3.125 grams
1.5625 grams
0.78125 grams
0.390625 grams
Answer: 0.390625 grams
Explanation:
A half-life of an element is the amount of time that it takes for half the mass of the element to decay.
Gold-198 having a half-life of 2.5 days therefore means that every 2.5 days, the mass is cut in half.
If there are 20 days, find out how many half-life periods there are:
= 20 / 2.5
= 8 periods
The half life is:
= Original mass * 0.5^number of half-life periods
= 100 * 0.5⁸
= 0.390625 grams
At time t = 0 a car has a velocity of 16 m/s. It slows down with an acceleration given by a = –0.50t, in m/s2 for t in seconds. At the end of 4.0 s it has traveled:
A: 12 m
B: 0 m
C: 14 m
D: 25 m
E: 59 m
Can this be solved without integrals?
The distance travelled by the car at the end of 4 seconds is 59 m ≈ 60 m
Option E is the correct answer.
What is the distance travelled by the car?
The distance travelled by the car at the end of 4 seconds is calculated by applying the following kinematic equation as shown below.
s = vt + ¹/₂at²
where;
s is the distance travelled by the carv is the initial velocity of the cart is the time of motion of the cara is the acceleration of the carThe acceleration of the car at the end of 4 seconds is calculated as follows;
a = -0.5t, m/s²
The unit of acceleration is already given as m/s², the value of t can be assumed to be 1
a = -0.5 (1)
a = - 0.5 m/s²
The distance travelled by the car at the end of 4 seconds is calculated as follows;
s = 16(4) - ¹/₂(0.5)(4)²
s = 60 m
The closest option here is option E, which is equal to 59 m, approximately 60 m.
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What is the speed of a car that travels 50m in 25min? *
Answer:
the speed is 2
Explanation:
speed= distance/ time
speed= 50/25
= 2
hope this helps!
when a student comned his dry hair for a long time, his hair began to stand up. Which of these has MOST LIKELY happened
His hair got charged due to continuous friction causing them to get attracted to other objects and thus standing up.
Answer:The comb and hair have become charged with opposite charges.
Explanation:When a comb is run through your hair charges pass between your hair and the comb, so the comb becomes charged either positively or negatively, and the hair oppositely charged. When the comb is brought close to paper an opposite charge is induced in the paper, and the opposite charges attract.
Maria read on an internet blog that infrared light is dangerous to humans. According to the blog, infrared light exposure is responsivle for a number of detrimental effects in humans. Which of these can actually be caused by exposure to infrared light?
a-overheating
b-skin cancer
c-radiation sickness
d-memory less
Of the options listed, the only effect that can be caused by exposure to infrared light is overheating (option a).
Infrared light is a form of electromagnetic radiation that is invisible to the human eye but can be detected as heat. When exposed to high levels of infrared light, such as in close proximity to a powerful infrared source, it can lead to overheating of the body or objects. Skin cancer (option b) is not directly caused by infrared light. It is primarily associated with overexposure to ultraviolet (UV) radiation from the sun or artificial sources like tanning beds. UV radiation falls in the higher energy range of the electromagnetic spectrum, while infrared radiation has lower energy. Radiation sickness (option c) is caused by exposure to high-energy ionizing radiation, such as gamma rays or X-rays. Infrared light does not possess enough energy to cause ionization and is therefore not capable of inducing radiation sickness. Memory loss (option d) is not a known effect of exposure to infrared light. Memory loss can be attributed to various factors, such as neurological conditions, head injuries, or aging, but not specifically to infrared light exposure. In summary, while exposure to high levels of infrared light can lead to overheating, it does not cause skin cancer, radiation sickness, or memory loss.
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What is the resistance (in Ω) of a 17.5 m long piece of 17 gauge copper wire having a 1.150 mm diameter?
In order to calculate the resistance of this wire, we can use the formula below:
\(R=\frac{\rho d}{A}\)Where p is the resistivity, d is the length and A is the cross-sectional area of the wire.
The resistivity of the copper is 1.72 * 10^-8 ohms*m, so we have:
\(\begin{gathered} R=\frac{1.72\cdot10^{-8}\cdot17.5}{\frac{\pi\cdot(1.15\cdot10^{-3})^2}{4}}\\ \\ R=\frac{30.1\cdot10^{-8}\cdot4}{\pi\cdot1.3225\cdot10^{-6}}\\ \\ R=\frac{120.4\cdot10^{-8}}{4.1547563\cdot10^{-6}}\\ \\ R=28.98\cdot10^{-2}\text{ ohms}\\ \\ R=0.2898\text{ ohms} \end{gathered}\)Which term defines the distance from crest to crest
Answer:
The horizontal distance between two adjacent crests or troughs is known as the wavelength.
Answer: Wavelength
Explanation:
From crest to crest, it is one full wavelength
5. It takes
to get a job done.
Answer:
determination
Explanation:
i) How many grams of CO₂ is formed when 73.9g of propane is burnt? C3H8+ O2 CO2 + H₂O
73.9g CO₂ is formed when 73.9g or 1 mole of propane is burnt
C₃H₈+ 5O₂ ⇄ 3CO₂ + 4H₂O
1 mole C₃H₈ gives 3 moles of CO₂
44 g C₃H₈ gives 44 g CO₂
73.9g C₃H₈ gives 73.9g CO₂
Quantifying the amount of a substance is made easier with the help of the mole concept. It is understood that even a single gram of a pure element has a sizable number of atoms when dealing with particles at the atomic (or molecular) level. In this regard, the mole notion is frequently used. The "mole," which is a count of a large number of particles, is the most commonly used unit of measurement.
The letter NA is typically used to indicate the value 6.02214076*1023, also known as the Avogadro constant. Atoms, molecules, monoatomic and polyatomic ions, as well as other particles, are among the elementary entities that can be represented in moles (such as electrons).
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Sources of error could have come from friction that may result in energy losses during the collision, The gliders may not be perfectly elastic, which means that some energy may be lost during the collision
are these errors random or systematic errors?
Sources of error could have come from small amounts of friction, and glider 2 could have not been totally at rest. These errors are systematic errors.
Are these errors random or systematic?
Answer:
The first set of errors, which include friction and imperfect elasticity, are systematic errors because they arise from consistent factors that affect the measurements in a predictable way. These errors will be present in every trial of the experiment and will cause a consistent deviation from the true value.
The second set of errors, which include small amounts of friction and the initial velocity of glider 2, are also systematic errors because they arise from consistent factors that affect the measurements in a predictable way. These errors will also be present in every trial of the experiment and will cause a consistent deviation from the true value.
Explanation:
A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is the coefficient of kinetic friction?
A block is given an initial speed of 3m/s up a 25° incline. Coefficient of friction
Answer:
a) μ = 0.475 , b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - \(W_{y}\) = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
The epiphyseal, or growth, plates close at relatively younger ages in which of the following groups?
The epiphyseal, or growth, plates close at relatively younger ages in boys and girls. That is option E.
What are growth plates ?The growth plate which is also called the epiphyseal plate is defined as the area of tissue near the ends of long bones in children and teens that determines the future length and shape of the mature bone.
The growth plate is made up of the following layers such as:
proliferation zone,hypertrophic cartilage zone,calcified cartilage zone, and ossification zone.The epiphyseal plate is known to close at relatively younger age for boys and girls because of Estrogen and testosterone release at puberty which initiates closure of the epiphyseal plates.
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Complete question:
The epiphyseal, or growth, plates close at relatively younger ages in which of the following groups?
a. boys
b. girls
c. early maturers
d. a and c
e. b and c
A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
of 20 m along level ground (Figure 3). The total weight of sled and load is 14,700
2
N. The tractor exerts a constant 5000 N force at an of 36.9
◦ angle of above the
horizontal. A 3500 N friction force opposes the sled’s motion. Find the work
done by each force acting on the sled and the total work done by all the forces.
(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
Work done by the applied forceThe work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
Work done by frictional forceW = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
Net work done by all the forces on the tractorW(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
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15 points!
a. Calculate the electric potential energy stored in a 1.4 x 10-7 F capacitor
that stores 3.40 x 10-6 C of charge at 24.0 V.
Answer:
\(4.12\times 10^{-5}\ J\).
Explanation:
Given that,
Capacitance, \(C=1.4\times 10^{-7}\ F\)
Charge stored in the capacitor, \(Q=3.4\times 10^{-6}\ C\)
We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :
\(E=\dfrac{Q^2}{2C}\)
Put all the values,
\(E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J\)
So, the required electric potential eenergy is equal to \(4.12\times 10^{-5}\ J\).
Ftension = 120 N
10 kg
Fg
What is the weight (not mass) of the box?
O 10 kg
0 19 N
98 N
O 98 kg
PLZ HELP
Answer:
98N
Explanation:
El peso se mide en kg y la fuerza no afecta
A bullet of mass m is fired horizontally into a wooden block of mass M lying on a table. The bullet remains in the block after the collision. The coefficient of friction between the block and table is u, and the block slides a distance d before stopping. Find the initial speed v0 of the bullet in terms of M, m, u, g, and d
Answer:
\(\displaystyle \frac{M + m}{m}\, \sqrt{2\, x\, u\,g}\).
Explanation:
This question can be solved in the following steps:
Using SUVAT equations, find the velocity of the block right after the collision, and thenUsing the conservation of momentum, find the velocity of the bullet before the collision.Assume that the table is level. The normal force on the block would be equal to the weight of the block in magnitude \((M + m)\, g\), but opposite in direction. As the block slows down, the only unbalanced force on the block would be friction \((-u\, (M + m)\, g)\) (negative since this force is opposite to the direction of motion.)
The acceleration of the block would be:
\(\begin{aligned} a &= \frac{(\text{net force})}{(\text{mass})} \\&= \frac{-u\, (M + m)\, g}{M + m} \\ &= (-u\, g)\end{aligned}\).
Apply the following SUVAT equation to find the velocity \(v_{i}\) of the block right after the collision:
\(\displaystyle {v_{2}}^{2} - {v_{1}}^{2} = 2\, a\, x\),
Where:
\(v_{2} = 0\) is the velocity after the acceleration,\(v_{1}\) is the velocity at the beginning of the acceleration, which is right after the collision, \(a = (-u\, g)\) is the acceleration, and\(x = d\) is the displacement during the acceleration.Rearrange and solve for \(v_{1}\), the velocity right after collision:
\(\begin{aligned}v_{1} &= \sqrt{{v_{2}}^{2} - 2\, a\, x} \\ &= \sqrt{0^{2} - 2\, (-u\, g)\, x} \\ &= \sqrt{2\, x\, u\, g}\end{aligned}\).
Apply the conservation of momentum to find the velocity of the bullet before the collision. Right after the collision, sum of momentum would be:
\((M + m)\, \sqrt{2\, x\, u\, g}\).
Right before the collision, sum of momentum would be:
\(m\, v_{0}\).
By the conservation of momentum:
\(m\, v_{0} = (M + m)\, \sqrt{2\, x\, u\, g}\).
Rearrange and solve for \(v_{0}\):
\(\displaystyle v_{0} = \frac{M + m}{m}\, \sqrt{2\, x\, u\, g}\).
The initial speed v0 of the bullet in terms of M, m, u, g, and d is identified by the equation v0 = (M + m) * \(\sqrt{((2 * u * g * d * m) / (M + m))} /m\).
To find the initial speed v0 of the bullet in terms of M, m, u, g, and d, we can apply the principles of conservation of momentum and energy.
First, let's consider the conservation of momentum. Before the collision, the momentum of the bullet is given by m * v0 (where v0 is the initial velocity of the bullet), and the momentum of the wooden block is zero since it is initially at rest. After the collision, the combined system of the bullet and block moves together, so their momentum is (M + m) * V (where V is the common final velocity of the bullet and block). Since momentum is conserved, we have:
m * v0 = (M + m) * V
Next, let's consider energy conservation. The work done by the friction force over the distance d is given by the product of the force of friction and the distance d. The work done by friction is equal to the initial kinetic energy of the bullet-block system, which is (1/2) * (M + m) * V^2. Thus, we have:
(1/2) * (M + m) * V² = u * (M + m) * g * d
Now we can solve these two equations simultaneously to find the initial velocity v0. Rearranging the first equation, we have:
v0 = (M + m) * V / m
Substituting this expression for v0 into the second equation, we get:
(1/2) * (M + m) * [(M + m) * V / m]² = u * (M + m) * g * d
Simplifying and solving for V, we obtain:
V = \(\sqrt{((2 * u * g * d * m) / (M + m))}\)
Finally, substituting this expression for V back into the first equation, we can find v0:
v0 = (M + m) * \(\sqrt{(2 * u * g * d * m) / (M + m)}\) / m
Therefore, the initial speed v0 of the bullet in terms of M, m, u, g, and d is given by the above equation.
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A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m3/s.
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answer: c. 40m
Explanation:
See picture
WILL GIVE BRAINLIEST PLS HELP
Design a repeatable experiment using various seismograph stations around the
globe to verify the hypothesis that the Earth has a liquid outer core and a solid
inner core. Describe how you would set up the experiment, what equipment would
be needed, and what information you would gather. What evidence would prove
that the outer core is liquid? What evidence would prove that the inner core is not
liquid? How would you use repeatability to show whether the hypothesis is valid or
not?
Answer:
Explanation:
To verify the hypothesis that the Earth has a liquid outer core and a solid inner core, we can use seismographs to study seismic waves that pass through the Earth's interior. The experiment can be set up as follows:
1. Select multiple seismograph stations around the globe to record seismic waves.
2. Choose a location for an earthquake to occur. The earthquake should be large enough to generate seismic waves that travel through the Earth's interior and be located far away from the selected seismograph stations.
3. Record the seismic waves generated by the earthquake at the various seismograph stations.
4. Analyze the seismic waves to determine how they interact with the Earth's interior. Specifically, we will study how the seismic waves pass through the Earth's outer and inner core.
5. Repeat the experiment using earthquakes of different magnitudes and at different locations, and record the resulting seismic waves.
Equipment needed for the experiment include seismographs, computers for data analysis, and earthquake monitoring systems. Seismographs can be installed in various locations around the globe to record the seismic waves generated by the earthquake. Data from these seismographs can be collected and analyzed using computer software to determine how the seismic waves interact with the Earth's interior.
Evidence that proves the outer core is liquid includes the observation of seismic waves that cannot travel through the liquid outer core, resulting in a shadow zone on the opposite side of the Earth from the earthquake. This shadow zone indicates that the seismic waves are refracted or absorbed by the liquid outer core. In contrast, evidence that proves the inner core is not liquid includes the observation of seismic waves that are reflected and refracted by the inner core boundary. This is due to the fact that the inner core is solid and has a different density and composition than the outer core.
To use repeatability to show whether the hypothesis is valid or not, we can repeat the experiment using earthquakes of different magnitudes and at different locations, and record the resulting seismic waves. If the results from multiple experiments are consistent with the hypothesis, then we can have greater confidence that the hypothesis is valid. If the results from multiple experiments are inconsistent, then we would need to investigate further to determine the cause of the inconsistency and revise the hypothesis accordingly.
In a change of an ideal gas from an initial state at 7 kPa and 3 m3 to a final state of 4 kPa and 6 m3, the internal energy change is -20 kJ. Calculate the change in enthalpy in kJ.
Given that the internal energy change is -20KJ the change in enthalpy is : -17 KJ
Given data :
P1 = 7 kPa
P2 = 4 kPa
V1 = 3 m³
V2 = 6 m³
Determine the change in enthalpyApplying the relationship below
Change in enthalpy = change in internal energy + ( P2V2 - P1V1 )
= -20 + ( 4*6 - 7*3 )
= -20 + ( 24 - 21 )
= - 17 KJ
Hence we can conclude that the change in enthalpy is : -17 KJ
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Callisto is a moon of Jupiter
(mass= 1.90 x 1027 kg), which orbits
the planet with a period of 16.9 days.
What is the radius of its orbit?
[?] x 10¹ m
Coefficient (green)
Exponent (yellow)
Enter
8.27 x 1013 meres is the orbital radius.
Additional details:-Jupiter's mass, 1.9 x 1027 kg, and the time interval, 16.9 days, are equal to 1.46 x 106 seconds. The radius is needed, thus r. Solution
The moon must be held in its orbit by a gravitational force equal to the centripetal force between Jupiter and the moon.
6.67 x 10⁻¹¹ N/m²kg
2 x 1.9 x 10/27 x 1.46 x 10'6 / 4 r = 6.85 x 102'7 G = 6.67 x 10'11 N/m2kg2 r = 8.27 x 10'7
What distinguishes Callisto, a huge moon orbiting Jupiter, from all other large moons in the solar system?The second-largest moon in Jupiter's orbit and the third-largest moon in the solar system is called Callisto. Of all the objects in our solar system, its surface has the most craters.
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A motor lifts a weight of 5N up to the height of 2m in 4 seconds. What will be the power of motor?
2.5 watt
Explanation:
we know
power = work done /time
work done gainst gravity is = mgh
as mg = weight = force = 5 N
and height = 2 m
so work done = mgh = 5*2 = 10N
therefore ,
Power = 10/4 = 2.5 watt
How do we sip water from glass using straw?can we sip water with straw on moon?
A roller coaster is at a peak of 20m and has a mass of 900kg. What is the potential energy of the roller coaster?
O 100000 J
10000 J
O 9.8 J
O 176400 J
The potential energy of the roller coaster is 176,400 J (joules).
The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.
In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).
Using the formula, we can calculate the potential energy:
PE = mgh
= (900 kg)(9.8 \(m/s^2\))(20 m)
= 176,400 J
Therefore, the potential energy of the roller coaster is 176,400 J (joules).
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Which statementsabout a neutral atom are correct? Check all that apply.
1. A neutral atom is composed of bothpositively and negatively charged particles.
2. Positively charged protons are located in the tiny, massivenucleus.
3. The positively chargedparticles in the nucleus are positrons.
4. The negatively chargedelectrons are spread out in a "cloud" around thenucleus.
5. The electrons areattracted to the positively charged nucleus.
6. The radius of the electroncloud is twice as large as the radius of the nucleus.
Answer:
1, 2, 4, 5 are correct
Explanation:
1) This is true because In a neutral atom, the number of positive charges (protons) is equal to the number of negative charges (electrons).
2) This is true because the mass of the atom which is made up of the protons and neutrons, is located in the tiny nucleus.
3) This is not true because the positively charged particles in the nucleus are called protons.
4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.
5) This is true because opposite charges attract each other. And electron is a negative charge.
6) This is not true because the radius of the electron cloud is normally 10,000 times larger than the radius of the nucleus.