Explanation:
Fgravity = G*(mass1*mass2)/D²
so, if you double one of the masses, what does that do to our equation ?
Fgravitynew = G*(2*mass1*mass2)/D²
due to the commutative property of multiplication
Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity
so, the correct answer will be 2×45 = 90 units.
a man can jump 9meteres on the moon.how high can he jump on the earth.
Answer:
You can jump 1.5 feet on Earth.
Explanation:
Because the moons gravity is weaker that earth so it would be easier to jump further on the moon.
An object is travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the acceleration?
Free-fall Acceleration is -10 m/s^2
I also need the Formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
50 = 0 + 0(4) + ½a(4²)
50 = 8a
a = 50/8 = 6.25 m/s²
What is the meaning of eddy currents in electromagnetic series?
Answer:
currents which circulate in conductors like swirling eddies in a stream in electromagnetic series....
What is sixth state of matter?
A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landing is angled at 60 degree with the roof.
a) find the horizontal distance d it travels.
b) what is the magnitude of the balls initial velocity?
c) what is the angle (relative to the horizontal) of the balls initial velocity?
Answer: 20
Explanation:
A pendulum is made from a long rod of mass M and length L with a solid sphere (ball) of mass m and radius R attached to one end. As measured from the top of the pendulum (the end of the rod without the sphere), how far down the rod is the center of mass of the pendulum located
Answer:
Explanation:
If we assume the rod and sphere are of uniform construction so that their individual centers of mass are at their geometric centers, and that the rod end is attached to the surface of the sphere.
Balance moments about the rod free end of the assembly with its parts
(M + m)Cx = M(L/2) + m(L + R)
Cx = (M(L/2) + m(L + R)) / (M + m)
The photo shows a skateboarder pushing her foot against the ground as she rides down a hill.
How does this action cause the skateboarder’s speed to change?
Answer:
A
Explanation:
Down the hill, the net force increases if she pushes more forward.
Answer:
its a im just did the test
Explanation:
the c component of vector a is 5.3 units, and it’s y component is -2.3 units. the angle that vector a makes with the +x axis is closest to
110
160
23
340
250
Answer:
340
Explanation:
Sorry I don't know how to do this one yet, I just found the answer in a textbook.
The angle that vector a makes with the +x axis is closest to 23.
What is direction of a vector?The direction of a vector is represented tangent of angle equal to the ratio of the y component and the x component of the vector quantity.
tangent of angle = y/x
angle = tan⁻¹ (-2.3/5.3)
angle = 23.46°
Thus, the angle that vector makes with +x is 23.
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Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.
Answer:
Explanation:
0.120 rev/s(2π rad/rev) = 0.24π rad/s
At the highest point of the arc, gravity must supply the required centripetal acceleration. As the normal force is 1/4 of his normal weight, then 3/4 of gravity acceleration must be used as centripetal acceleration
0.75g = ω²R
R = 0.75(9.81) / (0.24π)²
R = 12.942198...
R = 12.9 m
The radius of the circle is equal to 12.95m which is rotating with an angular velocity of 0.120 rev/s.
What is vertical circular motion?A body spins in a vertical circle so that its motion at different points is different from the motion of the body is said to be vertical circular motion.
The velocity and tension vary in maximum magnitude from the lowest to the highest position because of the effect of the gravitational force of the earth.
Given, the angular velocity of the Ferris wheel, ω = 0.120 rev/s
ω = 0.120 rev/s × 2π rad/rev
ω= 0.7536 rad/s
If r is the radius of the circle and 'm' is the mass of the jack.
From newton's second law of motion, the net force will be equal to
mg - N = mrω²
mg - (mg/4) = mrω²
r = 3g/4ω²
r = 3×9.81 / (4× 0.7536)
r = 12.95 m
Therefore, the radius of the circle in which the jack travels is equal to 12.95m.
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Your question is incomplete, most probably the complete question was,
Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s. As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels?
PLSSS HELP IT'S VERY IMPORTANT FOR ME WILL MAKE AS BRAINLIEST
Fill in the blank
Valence electrons are______
"The answer is not electrons in the outer shells that are not filled!!"
Answer:
negatively charged particle
valence electron is a electron of an atom, located in the outermost shell (valence shell ) of the atom, that can be transferred to or shared with another atom.
A spherical ball of lead (density 11.3 g/cm 3) is placed in a tub of mercury (density 13.6 g/cm 3). Which answer best describes the result
The lead ball will float with about 17% of its volume above the surface of the mercury.
We know that density is defined as mass per unit volume of a substance. The density of a substance is an intrinsic property which can be used to identify a substance.
Given that Lead is less dense that mercury, we know that lead will float on mercury. Since the density of mercury is 13.6 g/cm3 and that of lead is 11.3 g/cm3, lead ball will float with about 17% of its volume above the surface of the mercury.
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Missing parts;
A spherical ball of lead (density 11.3 g/cm3) is placed in a tub of mercury (density 13.6 g/cm3). Which answer best describes the result?
A.The lead ball will float with about 83% of its volume above the surface of the mercury.
B.The lead ball will float with about 17% of its volume above the surface of the mercury.
C.The lead ball will float with its top exactly even with the surface of the mercury.
D.The lead will sink to the bottom of the mercury.
E.none of the above
What is the word that is used to describe someone that is using a drug?
A: Sober
B: Crazy
C: High
D: Slumped
Helppp
True or False: The basketball should be dribbled below the waist.
Velocity and Acceleration Quick Check
Item 1
Use this graph of velocity vs. time for two objects to answer the question.
Item 2
Item 3
С
Item 4
Item 5
D
velocity
time
Which statement makes an accurate comparison of the motions for objects C and D?
(1 point)
lol
Answer:it’s C
Explanation:
by using graph of velocity vs. time for two objects, Item 4 and Item 5 statement makes an accurate comparison of the motions for objects. thus option C is correct.
What is velocity ?
velocity is the rate of change of the position of the object with respect to reference and it is complicated but velocity is basically speeding a particular object in a specific direction.
Velocity is a vector quantity which means both magnitude (speed) and direction are combinedly define define velocity. The SI unit of velocity is meter per second (ms-1) and the magnitude or the direction of velocity of a body changes leads to acceleration.
Speed and velocity are the two closest term but the major difference between speed and velocity is that speed gives us an idea that the object with the faster rate of movement r where as velocity speed up as well as tells us the direction of the body
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The graph of an object's position over time is a horizontal line and y is not equal to 0. What must be true abou
motion? (1 point)
O The acceleration is constant and non-zero.
O The velocity is constant and non-zero.
0 The acceleration is negative
O The velocity is zero.
Answer:D: the velocity is zero
Explanation:
A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resistance of the dryer?
Answer:
22Ω
Explanation:
if V ⇒ voltage
I ⇒ current
R ⇒ resistance
V = IR
220 = 10 x R
220 / 10 = R
22 = R
What is the best description of the distribution of the galaxies that lie within about 200 Mpc of Earth
When a 25000-kgkg fighter airplane lands on the deck of the aircraft carrier, the carrier sinks 0.30 cmcm deeper into the water.
A cubical box with sides of length 0.368 m contains 1.980 moles of neon gas at a temperature of 298 K. What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35x10-26 kg.
The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
The given parameters;
length of the cubical box, L = 0.368number of moles of the gas, n = 1.98temperature, T = 298 Kmass of the gas, m = 3.35 x 10⁻²⁶ kgThe average kinetic energy of the gas molecules is calculated as follows;
[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]
The average speed of the gas molecules is calculated as follows;
[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]
The time of collision of the gas molecules with the walls of the container is calculated as follows;
[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]
The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;
[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]
Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
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what is the velocity of this graph between points a and b? 0.0m/s 2.5m/s 5.0m/s 6.0m/s?
Answer:
Pick c is the right one
Explanation:
5.0m/s
what would happen if gravity were to stop everywhere?
Answer:
everything will float up and go up to space and die
Explanation:
gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.
if fire needs oxygen to burn, where does the sun get oxygen if there is no oxygen in space?
The sun is not a burning fire.
It's much much much hotter than that.
The sun's energy is the result of continuous nuclear fusion in it's core. We know how to do that on Earth, but the only thing we've been able to use it for so far is hydrogen bombs and other thermonuclear weapons.
2 W' is the symbol of a) antimony b) gold c) polonium d) tungsten
Answer:
D. Tungsten
Explanation:
W - Wolfram
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.44 m/s2. Determine the orbital period of the satellite.
Explanation:
The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as
[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]
where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.
Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write
[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]
[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]
where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is
[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]
[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]
Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,
[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]
or
[tex]4\pi^2r = gT^2[/tex]
Solving for T, we get
[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]
We can further simplify the above expression into
[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]
Plugging in the values for r and g, we get
[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]
Read the sentence from the text. “They are as glossy as satin or sunlight reflecting off water!" What does the word glossy mean in the sentence? O A. pointed o B. shiny O C. small O D. strong
Answer:
b Shiny
Explanation:
Trust me it's right
I need your help with this question, it’s my final exam for physics
Answer:
●Bx=Bcos40
Bx=10 × 0.76
Bx=7.6
●By=Bsin40
By=10×0.64
By=6.4
Hope it will help you.
5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Can someone solve by showing the steps?
This question involves the concepts of work done and the frictional force.
a. Work done by the person is "692.82 N".
b. Work done by the frictional force is "490.5 N".
a.
Work done by the person can be given by the following formula:
[tex]W=FdCos\theta[/tex]
where,
W = work done by the person = ?
F = Force applied by the person = 80 N
d = distance traveled = 10 m
θ = angle between force and motion = 30°
Therefore,
[tex]W=(80\ N)(10\ m)Cos30^o[/tex]
W = 692.82 N
b.
Work done by the frictional force is given by the following formula:
[tex]W_f=fd\\W_f=\mu mgd[/tex]
where,
[tex]W_f[/tex] = work done by the frictional force = ?
μ = coefficient of friction = 0.5
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]W_f=(0.5)(10\ kg)(9.81\ m/s^2)(10\ m)[/tex]
[tex]W_f=490.5\ N[/tex]
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What is the birth rate of a population of 3000 chipmunks if 200 chipmunks are born each year?
o 0.022 births per chipmunk per year.
0.067 births per chipmunk per year.
O 0.82 births per chipmunk per year.
o 15 births per chipmunk per year.
Answer:
0.067 births per chipmunk per year
Explanation:
I took the test
a body of mass 15 kg accelerates from rest of the rate of 4.0 ms^-2. determine the distance with the body travel in 25 seconds
The distance traveled by the body in the given time is 1,250 m.
The given parameters;
mass of the body, m = 15 kgacceleration of the body, a = 4 m/s²time of motion, t = 25 sinitial velocity, u = 0The distance traveled by the body in the given time is calculated as follows;
[tex]s =ut + \frac{1}{2} at^2\\\\s = 0 \ + \ \frac{1}{2} (4)(25^2)\\\\s =1,250 \ m[/tex]
Thus, the distance traveled by the body in the given time is 1,250 m.
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How does a balanced chemical equation demonstrate the Law of Conservation of Mass? it shows that only physical changes follow the Law of Conservation of Mass it shows that only physical changes follow the Law of Conservation of Mass it shows that the properties of the elements stay the same after the reaction it shows that the properties of the elements stay the same after the reaction it shows that all compounds remain bonded after the reaction it shows that all compounds remain bonded after the reaction it shows that no atoms have been gained or lost during the reaction
Answer:
it shows that the properties of the elements stay the same after the reaction
it shows that the properties of the elements stay the same after the reaction
it shows that all compounds remain bonded after the reaction
it shows that all compounds remain bonded after the reaction
it shows that only physical changes follow the Law of Conservation of Mass
it shows that only physical changes follow the Law of Conservation of Mass
it shows that no atoms have been gained or lost during the reaction
it shows that no atoms have been gained or lost during the reaction