Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between both objects.
so, now some numbers change
Fgravitynew = G*((1/3)*mass1*2*mass2)/(2D)² =
= G*((2/3)*mass1*mass2)/(4D²) =
= (2/3)* (G*(mass1*mass2)/D²) / 4 =
= ((2/3)/4) * G*(mass1*mass2)/D² =
= (2/12) * Fgravity = Fgravity/6
the new gravitational force will be 16/6 = 8/3 units.
Air is pumped into the tyre to inflate it.
This increases the temperature and the pressure of the air in the tyre.
Use ideas about molecules to explain why the air pressure in the tyre increases. *
Convection currents occur when _________ energy transfers between two parts of a fluid
Answer:
heat
Explanation:
What is the best description of the distribution of the galaxies that lie within about 200 Mpc of Earth
A mass vibrates back and forth from the free end of an ideal spring of spring constant 20 N/m with an amplitude of 0.30 m. What is the kinetic energy of this vibrating mass when it is 0.30 m from its equilibrium position?
Hi there!
We can begin by using the work-energy theorem in regards to an oscillating spring system.
Total Mechanical Energy = Kinetic Energy + Potential Energy
For a spring:
[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]
A = amplitude (m)
k = Spring constant (N/m)
x = displacement from equilibrium (m)
m = mass (kg)
We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:
ME = KE + PE
ME - PE = KE
Thus:
[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]
Plug in the given values:
[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]
We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.
Thus, the object would have no kinetic energy since KE = 1/2mv².
A rollercoaster car passes the hill which is 5.5m above the ground at speed 9.3m/s, and rolls over the second hill which is 2.5m above the ground, and heads toward the third hill which is 4.0 m higher than the first one. If the track is frictionless,
a. What maximum height will the car climb on the third hill? [h max = 9.9m, so car will climb the entire 9.5m hill]
b. Will the speed of the car on top of the hill 3 be lower or higher than its speed on the top of the hill one? [lower]
c. Calculate the speed of the car when it is 1m lower than the top of the third hill. [5.3m/s]
Would somebody kindly go over the questions :D
Answer:
Explanation:
Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.
Let m be the mass of the car and ground level is the origin.
on the 5.5 m hill, total energy is
E = PE + KE
E = mgh + ½mv²
E = m(9.8)(5.5) + ½m(9.3)² = 97m J
a) The maximum height will occur when the total energy is all potential energy.
E = mgh
h = E/mg
h = 97m/m(9.8) = 9.9 m
As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.
b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.
c) KE = E - PE
KE = 97m - m(9.8)(9.5 - 1.0)
KE = 97m = 83.3m
KE = 13.7m = ½mv²
v² = √(2(13.7)
v = 5.2345...
v = 5.2 m/s
2) A rolling disk, mass m and radius R, approaches a step of height R/2 with velocity v. (i) Taking the corner of the step as the pivot point, what is the initial angular momentum of the disk
The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.
So, E = E'
1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'
where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.
Substituting the values of the variables into the equation, we have
1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'
1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2
mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2
mR²ω²/4 + 1/2mv² = mgR/2
R²ω²/4 = gR/2 + 1/2v²
R²ω²/4 = (gR + v²)/2
ω² = 2(gR + v²)/R²
ω² = √[2(gR + v²)/R²]
ω = √[2(gR + v²)]/R
Since angular momentum L = Iω, the rolling disk's initial angular momentum is
L = 1/2mR² ×√[2(gR + v²)]/R
L = mR√[2(gR + v²)]/2
the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
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A 1-kg mass at the Earth's surface weighs how much
Answer:
the answer is weight=10N
Answer:
[tex]\boxed {\boxed {\sf 9.8 \ Newtons}}[/tex]
Explanation:
Weight is also called the force of gravity. This force acts on all objects at all times, pulling them down toward the center of the Earth.
It is calculated by multiplying the mass by the acceleration due to gravity.
[tex]F_g=mg[/tex]
The mass of the object is 1 kilogram. This scenario is occurring on Earth, so the acceleration due to gravity is 9.8 meters per second squared.
m= 1 kg g= 9.8 m/s²Substitute the values into the formula.
[tex]F_g= 1 \ kg *9.8 \ m/s^2[/tex]
Multiply.
[tex]F_g= 9.8 \ kg*m/s^2[/tex]
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 9.8 kilogram meters per second squared is equal to 9.8 Newtons.
[tex]F_g= 9.8 \ N[/tex]
A 1 kilogram mass at Earth's surface weighs 9.8 Newtons.
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 4.7 cm from its equilibrium position?
A)The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
1. force constant k
2. total energy E
3. mass m
4. maximum velocity vmax
5. amplitude A
6. potential energy U at x
7. kinetic energy K at x
8. position x from equilibrium
B)What is the kinetic energy of the object on the spring when the spring is compressed 4.7 cm from its equilibrium position?
C)What is the potential energy U of the toy when the spring is compressed 4.7 cm from its equilibrium position?
Hi there!
Part A:
The only quantities explicitly given to us are:
3. mass (m)
4. Maximum velocity (vmax)
5. Amplitude (A)
8. Position x from equilibrium
Part B:
To solve, we must begin by calculating the force constant, 'k'.
We can use the following relationship:
[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]
We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:
[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]
[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]
Now, we can find the velocity at 4.7cm (0.047m) using the equation:
[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]
Plug this value into the equation for kinetic energy:
[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]
Part C:
The potential energy of a spring is given as:
[tex]U = \frac{1}{2}kx^2[/tex]
Find 'k' using the derived quantity above:
[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]
Now, calculate potential energy:
[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]
How do light travels
Answer:
Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.
Explanation:
The symbol delta x (x) is used to find what value?
Answer:
Explanation:
Δx means a change in the magnitude of the x variable, often used in reference to a number line on the horizontal axis of a plot.
Circuit connections can either be series or parallel. In a_____connections, there is only one path of electrons, loads that are connected have the same current passing through them.
Answer:
circuit
Explanation:
A car travelling at 79.3 Km/h on a highway has 4.22x10 5 J of kinetic energy.
a. What is the mass of the car?
b. If brakes are applied with a force of 2100 N, what distance will it take for the car to slow down to a speed of 56 Km/h?
Answer:
[tex]1.74\times10^3 kg; 100m[/tex]
Explanation:
Step a: mass of the car. Let's grab the definition of kinetic energy: [tex]K= \frac12 mv^2[/tex]. We have K, we have v (which we should convert in meters per second, dividing by 3.6) to get:[tex]4.22\times10^5 = \frac12m(22.03)^2 \rightarrow m= 2\times4.22 / 495.22 \times 10^5 = 1.74 \times 10^3 kg[/tex]
Point a is done.
Now for the (b)reaking part. (I'm sorry, it's an horrible joke, but I couldn't resist)
In theory we have the mass, we have the force, so we could find the acceleration, find how long it takes to slow down, and then find the distance traveled. Too long. Let's do things more easily: when the car slows down to 56 km/h it will have a different kinetic energy. The difference in kinetic energy is the work done by the breaking force ofer the slowing distance.
[tex]K_f-K_i=W[/tex] A quick note on signs: if you look carefully the final kinetic energy will be less than the initial value, thus the work will be negative: it means it's correct, since the work is against the motion, slowing it down. Let's get calculating, first by converting 56 kmh in m/s (15,56 m/s), then finding the final kinetic energy:
[tex]K_f =\frac12 (1.74\times10^3) (15.56)^2 =2.11 \times 10^5 J[/tex]
The difference will be the work done by the force, or
[tex](2.11 - 4.22) \times 10^5 = \vec F\cdot \vec x=Fx[/tex] where we are assuming that force and displacement have the same line of actions to simplify the dot product.
[tex]2.11\times 10^5 = 2100x = 1.00\times 10^2 m[/tex]
describe the motion of objects that are viewed from your reference frame both inside and outside while you travel inside a moving vehicle
Answer:
The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.
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:
What type of equilibrium maintains body position during sudden motion?
dynamic
rotational
static
balanced
I think static is the correct answer
A flywheel with a diameter of 0.692 m is rotating at an angular speed of 208 rev/min. (a) What is the angular speed of the flywheel in radians per second
[tex]\omega = 21.8\:\text{rad/s}[/tex]
Explanation:
We know that there are [tex]2\pi[/tex] radians in one revolution and 60 seconds in one minute so we can easily convert the rev/min unit to rad/s using the following conversion factors:
[tex]208\:\dfrac{\text{rev}}{\text{min}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}}×\dfrac{1\:\text{min}}{60\:\text{s}}[/tex]
[tex]\;\;\;\;\;=21.8\:\text{rad/s}[/tex]
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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After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure.The horizontal steel beam had a mass of 88.50 kg
per meter of length, and the tension in the cable was =11650 N
. The crane was rated for a maximum load of 500 kg
. If =5.580 m
, =0.522 m
, =1.350 m
, and ℎ=2.070 m
, what was the magnitude of L
(the load on the crane) before the collapse? The acceleration due to gravity is =9.810 m/s2
The magnitude of the load L on the crane before the collapse is 3211.81 N
To determine the magnitude of the load on the crane (L), we will need to make use of the equilibrium conditions of the torque.
It is always an ideal process to list out all the parameters given as this will let you understand how you can determine the answer to the question from the given parameters.
From the given information;
The tension in the cable = 11650 NThe length (d) = 5.580 mThe mass of the horizontal steel beam (M) = 88.50 kg/m (d)= 88.50 kg/m × 5.580 m= 493.83 kgDistance (s) = 0.522 mx = 1.350 m and h = 2.070 mAcceleration due to gravity = 9.81 m/s²From the question;
the angle at which the crane is positioned can be determined by taking the tangent of the angle θ. i.e.
[tex]\mathbf{tan \ \theta = \dfrac{h}{d-s}}[/tex]
[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{h}{d-s} \Big )}[/tex]
[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{2.070 }{5.580 - 0.522} \Big )}[/tex]
[tex]\mathbf{\theta =22.26^0}[/tex]
Consider the equilibrium conditions of the torques with respect to the magnitude of the load at point P.
∴
[tex]\mathbf{Tsin \theta (d-s) - W_L (d-x) -(Mg) (\dfrac{d}{2}) = 0}[/tex]
By making the magnitude of the load [tex]\mathbf{W_L}[/tex] the subject of the formula, we have:
[tex]\mathbf{W_L = \dfrac{Tsin \theta (d-x) -(Mg) (\dfrac{d}{2})}{ (d-s) } }[/tex]
[tex]\mathbf{W_L = \dfrac{(11650 )sin (22.26) (5.580-1.350) -(88.50\times 9.81) (\dfrac{5.580}{2})}{ (5.580-0.522) } }[/tex]
[tex]\mathbf{W_L = 3211.81 \ N }[/tex]
Therefore, we can conclude that the magnitude of the load is 3211.81 N
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Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two different tectonic plates, each moving in different directions, border the western side of the North American Plate. Use the map to identify the two tectonic plates that border the North American Plate to the west.
Answer:
Remember, NORTH ^, EAST >, SOUTH v, WEST <
Explanation:
It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.
I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)
Answer:
The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.
Explanation:
Edmentum
jshshwjs sbwiwiw910mw s x djjskskekwkq
Answer:
jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb
I NEED THE ANSWER ASAPP
Answer:
Explanation:
a) The spring force will equal the weight.
b) If up is positive
kx - mg = 0
mg = kx kx = 25 N
c) m = kx/g = 25/10 = 2.5 kg
Velocity and Acceleration Quick Check
C
D
E
During which of the labeled time segments is the object moving forward but slowing down?
(1 point)
Ο Α
0 С
OD
ОВ
Answer:
Explanation:
1 Object C has an acceleration that is greater than the acceleration for D.
2 B
3 17M
4 The velocity is zero.
5 a straight line with negative slope
just took it
Four small 0.600-kg spheres, each of which you can regard as a point mass, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane at point O.
Answer:
.192 kg x m^2
Explanation:
I= mass of a times radius of a squared + mass of b times radius of b squared +...
I= .6 kg x .4m^2 + .6 kg x .4m^2
= .192 kg x m^2
Hope this helps :)
A 100 N crate is being pulled at a constant velocity by a rope a 30 degrees to the horizontalas depicted in the diagramFind the force of friction Show your work and explain your reasoning in two to sentences
Answer:
Explanation:
As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.
Ff = Fcosθ
if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.
μN = Fcosθ
μ(mg - Fsinθ) = Fcosθ
μmg = Fcosθ + μFsinθ
100μ = F(cos30 + μsin30)
F = 100μ / (cos30 + ½μ)
Ff = 100μcos30 / (cos30 + ½μ)
A 100 N create is being pulled at a constant velocity by a rope a 30 degrees to the horizontal as depicted in the diagram given in question the force of friction Ff = 100μcos30 / (cos30 + ½μ).
What is force?
A force in physics is an effect that has the power to alter an object's motion. 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.
As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.
Ff = Fcosθ
if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.
μN = Fcosθ
μ(mg - Fsinθ) = Fcosθ
μmg = Fcosθ + μFsinθ
100μ = F(cos30 + μsin30)
F = 100μ / (cos30 + ½μ)
Ff = 100μcos30 / (cos30 + ½μ)
the force of friction Ff, is 100μcos30 / (cos30 + ½μ).
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What is most likely the amount of energy available at a trophic level of primary consumers if the amount of energy available to secondary consumers in that food web is 200 kilocalories?
0 kilocalories
20 kilocalories
200 kilocalories
2,000 kilocalories
Answer:
200 kilocalories
Explanation:
Question 3. A wire 25.0cm long lies along the z-axis and carries a current of 9.00A in the +z-direction. The a magnetic field is uniform and has components Bx = -0.242T, By= -0.985, and B2=-0.336. a. Find the components of the magnetic force on the wire? b. What is the magnitude of the net magnetic force on the wire?
a.
The components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N
The force on a current carrying conductor in a magnetic field is given by F = iL × B where i = current = 9.00 A, L = 25.0 cmk = 0.25 mk (since the conductor is along the z-direction). B = magnetic field. Since B has component Bx = -0.242T, By= -0.985, and Bz = -0.336, B = -0.242i + (-0.985j) + (-0.336)k = -0.242i - 0.985j - 0.336)k.
So, F = iL × B
F = 9.00 A{(0.25 m)k × [-0.242Ti + (-0.985Tj) + (-0.336T)k]T}
F = 9.00 A{(0.25 m)k × (-0.242T)i + (0.25 m)k × (-0.985Tj) + (0.25 m)k × (-0.336T)k]}
F = 9.00 A{-0.0605mT)k × i + (-0.24625 mT)k × j + (-0.084 m)k × k]}
F = 9.00 A{-0.0605mT)j + (-0.24625 mT) × -i + (-0.084 mT) × 0]}
F = 9.00 A{-0.0605mT)j + (0.24625 mT)i + 0 mT]}
F = -0.5445 AmT)j + (2.21625 AmT)i + 0 AmT]}
F = -0.5445j + 2.21625i + 0 k
F = (2.2163i - 0.5445j + 0 k) N
So, the components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N
b.
The magnitude of the net force on the wire is 2.282 N
The net force F = √(Fx² + Fy² + Fz²)
F = √[(2.2163 N)² + (-0.5445 N)² + (0 N)²)
F = √[(4.912 N)² + 0.2964 N)² + (0 N)²)
F = √[5.2084 N)²
F = 2.2822 N
F ≅ 2.282 N
So, the magnitude of the net force on the wire is 2.282 N
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A 0.035-kg bullet is fired vertically at 214 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
Explanation:
conservation of momentum during the collision
0.035(214) + 0.15(0) = 0.185v
v = 40.486 m/s
The kinetic energy after impact will convert to gravity potential energy
(ignoring air resistance)
mgh = ½mv²
h = v²/2g
h = 40.486² / (2(9.8))
h = 83.6303...
h = 84 m
Find the dimension of the gravitational constant in this equation F=Gm1m2/r¹r²
The gravitational force acting between the two bodies is given by:
F=G
r
2
m
1
m
2
G=
m
1
m
2
Fr
2
The dimension of the force is [MLT
−2
]
=
[M][M]
[MLT
−2
][L
2
]
=M
−1
L
3
T
−2
I need help been struggling on this question
Answer:
440 m
Explanation:
S=(u+v) t / 2
S = (11+33) × 20/2
S= 44× 20/2
S=440 m
plz answer the question.
Answer:
a
Explanation:
sana po makatulong <3♡♡