Explanation:
At a height of 93 m, the gravitational potential energy is given by :
P = mgh
Where, m is the mass of a penny
We can find its mass.
[tex]m=\dfrac{P}{gh}\\\\m=\dfrac{3}{9.8\times 93}\\\\m=0.00329\ kg[/tex]
We need to find how much potential energy will have transformed into kinetic energy at the half way point (46.5m). It can be calculated as :
[tex]E=mgh'\\\\E=0.00329\times 9.8\times 46.5\\\\E=1.499\ J[/tex]
Hence, a penny will have transferred 1.499 J of potential energy into kinetic energy.
planets A & B are near each other but there is a large difference in their temperatures using the data in the table explain how the atmosphere of these two planets can influence the average temperature
Answer: Planet A is closer to the Sun and has much greater atmospheric pressure. This suggests that planet A has a thicker atmosphere. Planet A's atmosphere is also mostly made of carbon dioxide, which is a greenhouse gas. This gas retains heat, raising the surface temperature of planet A.
Explanation:
Planet A is closer to the Sun and has much greater atmospheric pressure. This suggests that planet A has a thicker atmosphere. Planet A’s atmosphere is also mostly made of carbon dioxide (CO2), which is a greenhouse gas. This gas retains heat, raising the surface temperature of planet A.
Hope this one help.
In 2006, NASA’s Mars Odyssey orbiter detected violent gas eruptions on Mars, where the acceleration due to gravity is 3.7 m/s2. The jets throw sand and dust about 62.0 m above the surface. Scientists estimate that the jets originate as high-pressure gas speeds through vents just underground at about 130 km/h. How much energy per kilogram of material is lost due to nonconservative forces as the high-speed matter forces its way to the surface and into the air? (Express your answer to two significant figures.)
Answer:
The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.
Explanation:
We can estimate the unit energy losses of gas eruption by Principle of Energy Conservation and Work-Energy Theorem:
[tex]U_{g,1} + K_{1} = U_{g,2}+K_{2}+W_{loss}[/tex] (Eq. 1)
Where:
[tex]U_{g,1}[/tex] - Gravitational potential energy of gas eruptions at surface, measured in joules.
[tex]U_{g,2}[/tex] - Gravitational potential energy of gas eruptions at highest height, measured in joules.
[tex]K_{1}[/tex] - Translational kinetic energy of gas eruptions at surface, measured in joules.
[tex]K_{2}[/tex] - Translational kinetic energy of gas eruptions at highest height, measured in joules.
[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.
We clear the component associated with energy losses in (Eq. 1):
[tex]W_{loss} = U_{g,1}-U_{g,2}+ K_{1}-K_{2}[/tex]
And we expand it afterwards:
[tex]W_{loss} = m\cdot g\cdot (z_{1}-z_{2}) + \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2a)
[tex]w_{loss} = g\cdot (z_{1}-z_{2})+\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2b)
Where:
[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.
[tex]w_{loss}[/tex] - Unit energy losses due to nonconservative forces, measured in joules per kilogram.
[tex]g[/tex] - Gravitational acceleration, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Bottom and top height, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Gas eruption speeds at surface and highest heights, measured in meters per second.
If we know that [tex]g = 3.7\,\frac{m}{s^{2}}[/tex], [tex]z_{1} = 0\,m[/tex]. [tex]z_{2} =62\,m[/tex]. [tex]v_{1} = 36.111\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the unit energy losses are:
[tex]w_{loss} = \left(3.7\,\frac{m}{s^{2}} \right)\cdot (62\,m-0\,m)+\frac{1}{2} \cdot \left[\left(36.11\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]w_{loss} = 881.40\,\frac{J}{kg}[/tex]
The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.
What is a point of view with regard to motion called in physics?
O reference grid
O displacement
O distance
O frame of reference
Answer:
Frame of reference
Lexington walked to her friends house.lexie walked 2miles west then 5 miles south then 3 miles east then 4 miles north then 2 miles east and finally 1 mile north what is levies distance and displacement.
procedure for determining the thermal conductivity of a solid involves embedding thermocouple in a thick slab of the material and measuring the response to a prescribed change in temperature at one surface. Consider an arrangement for which the thermocouple is embedded 10 mm from a surface that is suddenly brought to a temperature of 100degreeC by exposure to boiling water. If the initial temperature of the slab was 30degreeC and the thermocouple measures a temperature of 65degreeC, 2 minutes after the surface is brought to 100degreeC, what is the thermal conductivity. The density of the material is 2200 kg/m3 and the specific heat is 700 J/M- Find: What is the thermal conductivity of the material
Answer:
The thermal conductivity [tex]k = 1.4094 W/ m\cdot K[/tex]
Explanation:
From the question we are told that
The depth of the thermocouple from the surface is x = 10 mm = 0.01 m
The temperature is [tex]T_f = 100 ^o C[/tex]
The initial temperature is [tex]T_i = 30 ^o C[/tex]
The temperature of the thermocouple after t = 2 minutes( 2 * 60 = 120 \ seconds) is [tex]T_t = 65 ^o C[/tex]
The density of the material is [tex]\rho = 2200 kg/m^3[/tex]
The specific heat of the solid [tex]c_s = 700 J/kg \cdot K[/tex]
Generally the equation for semi -infinite medium is mathematically as
[tex]\frac{T_s - T }{T_i - T} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]\frac{65 - 100 }{30 - 100} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]0.5 = erf [\frac{0.01}{2 \sqrt{\alpha * 120} } ][/tex]
Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]
[tex]\frac{0.01}{ 2 (\sqrt{\alpha * 120 } )}[/tex] this is from the Gaussian function table
[tex]0.0 1 = 0.954 * (\sqrt{\alpha * 120 } )[/tex]
=> [tex]\sqrt{\alpha * 120 } = \frac{0.01 }{0.954 }[/tex]
=> [tex]\alpha = 9.1525 *10^{-7} \ m^2 /s[/tex]
Generally the thermal conductivity is mathematically represented as
[tex]k = \alpha * \rho * c_s[/tex]
[tex]k = 9.1525 *10^{-7} * 2200 * 700[/tex]
[tex]k = 1.4094 W/ m\cdot K[/tex]
Which of the the following distance vs time graphs represents an object the is moving at constant non zero velocity
As mentioned in problem 1.2 from Chapter 1, if you stand on one leg the load exerted on the hip joint is 2.4 times your body weight. Assuming a simple cylindrical model for the implant, calculate the corresponding stress (in MPa) on a hip implant in a individual with a hip implant with a cross-sectional area of If the hip implant is made of Ti6Al4V (124 GPa elastic modulus), what is the strain for the given loading conditions?
The question is incomplete. Here is the complete question.
As mentioned in problem 1.2 from Chapter 1, if you stand on one leg the load exerted on the hip jpint is 2.4 times your body weight. Assuming a simple cylindrical model for the implant, calculate the corresponding stress (in MPa) on a hip implant in a 175lb individual with a hip implant with cross-sectional area of 5.6cm². If the implant is made of Ti6AI4V (124 GPa elastic modulus), what is the atrain for the given loading conditions?
Answer: σ = 3.336 MPa
Strain = 2.69.[tex]10^{-5}[/tex]
Explanation: Stress is the measurement of how hard a weight works to change a material's shape. Its symbol is σ and is calculated as
[tex]\sigma = \frac{F}{A}[/tex]
where
F is force
A is cross sectional area
For the hip implant, force is 2.4x of an 175lb individual:
F = 2.4 * 175
F = 420lb
As the question is asking for Pa and 1 Pa = 1 N/m², transform lb in Newtons:
F = 420lb * 4.448N/lb
F = 1868 N
And cm² in m²:
A = [tex]5.6.10^{-4}[/tex] m²
Calculating Stress:
[tex]\sigma=\frac{1868}{5.6.10^{-4}}[/tex]
[tex]\sigma=333.6.10^{4}[/tex] Pa
or
σ = 3.336 MPa
Stress on a hip implant of an individual with weight 175lb is 3.336 MPa.
Strain is the measure of the deformation of a material due to a load. It can be calculated as:
strain = [tex]\frac{stress}{modulus}[/tex]
But first, change MPa into GPa:
3.336 MPa = [tex]3.336.10^{-3}[/tex] GPa
strain = [tex]\frac{0.003336}{124}[/tex]
strain = [tex]2.69.10^{-5}[/tex]
Ti6Al4V is a titanium alloy used in implants. So, the strain the alloy does on a hip implant of a 175lb individual is [tex]2.69.10^{-5}[/tex].
Why is it important for scientists to find better ways to store solar and wind energy?
Answer:
Energy storage plays an important role in this balancing act and helps to create a more flexible and reliable grid system. For example, when there is more supply than demand, such as during the night when low-cost power plants continue to operate, the excess electricity generation can be used to power storage devices.
Explanation:
I hope this helps you out and if your feeling generous plz mark brainliest it helps me a lot thank you:)
It important for scientists to find better ways to store solar and wind energy because solar and wind energy have variable outputs.
Lately, the world have begun to gradually gravitate away from non renewable energy sources such as fossil fuels and research efforts have been concentrated on renewable energy sources such as solar and wind energy.
However, the energy output from these sources are variable. Therefore, it is necessary to device ways to store energy from these sources in order to improve overall energy supply.
Learn more: https://brainly.com/question/17319210
Which platform has touch controls?
A.
consoles
B.
arcades
C.
personal computers
D.
mobiles
Answer:
I think it's D
Explanation:
Most mobiles have touch screens
At serve, a tennis player aims to hit the ball horizontally, as shown in the figure.
(a) What minimum speed is required for the ball to clear the 0.90-m-high net about 15.0 m
from the server if the ball is “launched” from a height of 2.50 m?
(b) Where will the ball land if it just clears the net (and will it be “good” in the sense that it
lands within 7.0 m of the net)?
(d) How long will it be in the air?
(a) The minimum speed required for the ball to clear the net is 26.3 m/s.
(b) The horizontal distance of the ball when it clears the net is 11.3 m
(c) The total time spent in the air by the ball is 1 s.
Time of motion
The time of motion of the tennis ball is calculated by using the following kinematic equation as shown below;
[tex]h = h_0 + v_0_yt - \frac{1}{2} gt^2\\\\0.9 = 2.5+ 0 - \frac{1}{2} (9.8) t^2\\\\0.9 = 2.5 - 4.9t^2\\\\4.9t^2 = 1.6\\\\t^2 = \frac{1.6}{4.9} \\\\t^2 = 0.327\\\\t = \sqrt{0.327} \\\\t = 0.57 \ s[/tex]
Minimum speedThe minimum speed of the ball is calculated as follows;
[tex]v_x = \frac{X}{t} \\\\v_x = \frac{15}{0.57} \\\\v_x = 26.3 \ m/s[/tex]
Time of motion from top of the high netThe time of motion from top of the high net is calculated as follows;
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.9}{9.8} } \\\\t = 0.43 \ s[/tex]
Horizontal range = 0.43 x 26.3 = 11.3 m
Total time in air = 0.57 s + 0.43 s = 1.0 s
Learn more about horizontal velocity here: https://brainly.com/question/24949996
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA's idea is basicallyan electric slingshot that consists of 4 electrodes arranged in a horizontal square with sides of length d at a height h above the ground. The satellite is then placed on the ground aligned with the center of the square. A power supply will provide each of the four electrodes with a charge of Q/4 and the satellite with a charge -Q. When the satellite is released from rest, it moves up and passes through the center of the square. At the instant it reaches the square's center, the power supply is turned offand the electrodes are grounded, giving them a zero electric charge. To test this idea, you decide to use energy considerations to calculate how big Q will have to be to get a 100 kg satellite to a sufficient orbit height. Assume that the satellite startsfrom 15 meters below the square of electrodes and that the sides of the square are each 5 meters. In your physics text you find the mass of the Earth to be 6.0 x 1024kg.
Answer:
The answer is "[tex]q=0.0945\,C[/tex]".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).
[tex]U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}[/tex]
It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:
[tex]U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}[/tex]
Potential energy shifts:
[tex]= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\[/tex]
[tex]=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J[/tex]
Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.
[tex]=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\[/tex]
[tex]=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\[/tex]
[tex]\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C[/tex]
This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
state and explain the changes in stability of the beaker when the water freezes to ice
Answer:
if the question is referring to what happens when ice freezes you could say that the water molecules have lass energy so they don't move around as much
A mass of 15 kg is resting on a horizontal, frictionless surface. Force 1 of 206 N is applied to it at some angle above the horizontal, force 2 has a magnitude of 144 N and is applied vertically downward, force 3 has a magnitude of 5 N and is applied vertically upwards, and force 4 has a magnitude of 42 N and is applied in the -x direction to the object. When these forces are applied to the object, the object is moving at 20 m/s in the x direction in a time of 3 seconds. What is the normal force acting on the mass in Newtons
Answer:
N = 136.77 N
Explanation:
This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.
Let's use trigonometry to decompose the force F1
cos θ = F₁ₓ / F₁
sin θ = F_{1y} / F₁
F₁ₓ = F₁ cos θ
F_{1y} = F₁ sin θ
now let's apply Newton's second law to each axis
X axis
F₁ₓ - F4 = m a
Y axis
N + F3 + F_{1y} -F₂ -W = 0
the acceleration can be calculated with kinematics
v = v₀ + a t
since the object starts from rest, the initial velocity is zero v₀ = 0
a = v / t
a = 20/3
a = 6.667 m / s²
we substitute in the equation
F₁ₓ = F₄ + m a
F₁ₓ = 42 + 15 6,667
F₁ₓ = 142 N
F₁ cos θ = 142
cos θ = 142/206 = 0.6893
θ = cos⁻¹ 0.6893
θ = 46.42º
now let's work the y axis
N = W + F₂ - F₃ - F_{1y}
N = 15 9.8 + 144 -5 - 206 sin 46.42
N = 286 - 149.23
N = 136.77 N
PLEASEEEEE THIS A TIMED TESTTTTTTTTT
Answer:
C
Explanation:
Answer:
that car got obliterated almost like the day my uncle said come with me
Explanation:
the outcome was not good
An object falls freely from rest on a planet
where the acceleration due to gravity is
29 m/s ^2
After 3.8 s, what will be its speed?
Answer in units of m/s.
Answer:
v=u+gt , initially u=0 and g acting in the direction of movement of body.
v=0+9.8×2
v=19.6m/s
Explanation:
sorry i dont have exact answer but hope this above equation will help you ....♡
______ is a disease in which the body does not produce or probory use inslin
Answer:
Diabetes
Explanation:
Insulin is an important hormone that helps the body convert sugar, starches, and other foods into energy.
A
6N
11 N
What is the net force ?
Answer:
66n
Explanation:
4376
What type of energy transfer causes a sunburn?
A. Radiation, because particles of hot air move in space
B. Radiation, because electromagnetic waves transfer heat
C. Conduction, because hot air comes in contact with the skin
D. Conduction, because invisible light carries heat from the sun
If your right you get 100 points and a brainliest
Answer:
Hi!!
It should be A, ratation
(sorry that I didnt spell that write)
Have a great day!
To solve this we must be knowing each and every concept related to energy and its different types. Therefore, the correct option is option A among all the given options.
What is energy?In physics, energy is the capacity to accomplish work. It can be potential, kinetic, thermal, electric, chemical, radioactive, or in other forms.
There is also heat and work—energy inside the process of being transferred by one body to the other. Energy is always classified according to its type once it has been transmitted. As a result, heat transported could become thermal energy, whereas labor done may emerge as mechanical energy. Radiation is a t type of energy that transfer energy that causes a sunburn because particles of hot air move in space.
Therefore, the correct option is option A among all the given options.
To know more about energy, here:
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A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a density of 800 kg/m3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m
Answer:
The answer is below
Explanation:
Pascal's law states that pressure is exerted in all parts of a static fluid equally.
The area of the narrower arm with a radius of 0.05 m (5 cm) is given as:
Area of narrow arm = π(0.05)²
The area of the wider arm with a radius of 0.18 m (18 cm) is given as:
Area of narrow arm = π(0.18)²
The ratio if the wider arm area to the narrow arm area = π(0.18)² / π(0.05)² = 12.96
To move the 12000 N car, the amount of force needed = 12000/12.96 = 926 N
9. A plane starts at rest & accelerates along the ground before takeoff. It
moves 1600m in 18s. Calculate the acceleration rate of the plane. *
Answer:
9.877 m/s^2
Explanation:
The acceleration can be computed from ...
d = (1/2)at^2
(1600 m) = (1/2)a(18 s)^2
a = (1600/162) m/s^2 ≈ 9.877 m/s^2
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters. What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
Answer:
The change in potential energy of the mass as it goes up the incline is 0.343 joules.
Explanation:
We must remember in this case that change in the potential energy is entirely represented by the change in the gravitational potential energy. From Work-Energy Theorem and definition of work we get that:
[tex]U_{g}= m\cdot g\cdot \Delta y[/tex]
Where:
[tex]U_{g}[/tex] - Gravitational potential energy, measured in Joules.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]\Delta y[/tex] - Change in vertical height, measured in meters.
This work is the energy needed to counteract effects of gravity at given vertical displacement.
If we know that [tex]m = 0.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta y = 0.07\,m[/tex], the change in the potential energy of the mass as it goes up the incline is:
[tex]U_{g} = (0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.07\,m)[/tex]
[tex]U_{g} = 0.343\,J[/tex]
The change in potential energy of the mass as it goes up the incline is 0.343 joules.
The change in the potential energy (in Joules) of the mass as it goes up the incline is 0.343 J.
Calculation of the change in the potential energy:We know that
Potential energy = m*g*h
Here m means the mass = 0.5 kg
g means the gravity = 9.8
And, the h means the height = 7cm = 0.07m
So, the change in the potential energy should be
=0.5*9.8*0.07
=0.343 J
hence, we can conclude that the change in the potential energy (in Joules) of the mass as it goes up the incline is 0.343 J.
Learn more about energy here: https://brainly.com/question/13203990
A baseball player is running at a constant velocity on a level field and tosses a baseball straight up. When the ball comes back down, it will land
Answer:
]
Explanation:
If a 3.5 kg object is accelerating at 0.80 /s^2, the net force F causing this motion is ____ N.
A) 2.2
B)2.8
C)3.4
D)4.0
Answer:
The answer is option BExplanation:
To find the force acting on an object we use the formula
force = mass × accelerationFrom the question
mass = 3.5 kg
acceleration = 0.80 m/s²
We have
Force = 3.5 × 0.8
We have the final answer as
2.8 NHope this helps you
Can someone please help me out?
Explanation:
If you want to get speed, u have to divided distance over time
The lowest speed will lose
On a drive from your home to town, you
wish to average 60 mph. The distance from
your home to town is 78 miles. However,
at 39 miles (half way), you find you have
averaged only 45 mph.
What average speed must you maintain in
the remaining distance in order to have an
overall average speed of 60 mph?
Answer in units of mph.
On a drive from your home to town, you wish
to average 33.6 mph. The distance from your
home to town is 90 miles. However, at 45 miles
(half way), you find you have averaged only
25.2 mph.
What average speed must you maintain in
the remaining distance in order to have an
overall average speed of 33.6 mph?
Answer in units of mph
let the speed be x
total average = 33.6
PLEASE HELP GIVING BRAINLIEST!!
how you brother and sister
12) If a man weighs 900 N-on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 m/s?
Answer:
Force=Mass*acceleration
on earth, acceleration=9.81 m/s^2
900 N=Mass*9.81 m/s^2
Mass=91.74 Kg
F=Mass*acceleration(Jupiter)
F=91.74Kg*25.9m/s
F=2376.066 N on Jupiter
Plz mark me as brainliest if u found it helpful
If an object is on a 27º frictionless incline, what will be the acceleration of the object on the incline?
A pendulum can be formed by tying a small object, like a tennis ball, to a string, and then connecting the other end of the string to the ceiling. Suppose the pendulum is pulled to one side and released at t1. At t^2, the pendulum has swung halfway back to a vertical position. At t^3, the pendulum has swung all the way back to a vertical position. Rank the three instants in time by the magnitude of the centripetal acceleration, from greatest to least. Most of the homework activities will be Context-rich Problems.
Answer:
1- t^3
2- t^2
3- t1
Explanation:
The acceleration produced in a body, while travelling in a circular motion, due to change in direction of motion is called centripetal acceleration. The formula of the centripetal acceleration is as follows:
ac = v²/r
where,
ac = centripetal acceleration
v = speed
r = radius
for a constant radius the centripetal acceleration will be directly proportional to the speed of object. The speed of pendulum will be lowest at t1 due to zero speed initially. Then the speed will increase gradually having greater speed at t^2 and the highest speed and centripetal acceleration at t^3. Therefore, the three instants in tie can be written in following order from greatest centripetal acceleration to lowest:
1- t^3
2- t^2
3- t1
