Answer:
Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.
G.P.E = m*g*h
K.E = (m*v^2)/2
where
m = mass of toy car (kg)
g = gravity (m/s^2)
h = heigh of your car from the bottom (m)
v = velocity of the toy car as it reaches the bottom (m/s)
Equate K.E to G.P.E
G.P.E = K.E
m*g*h = (m*v^2)/2
make v the subject of the formula
v = (2*g*h)^(1/2)
Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v
v = (2*9.81*2)^(1/2)
v = 6.264 m/s
Convection currents occur when _________ energy transfers between two parts of a fluid
Answer:
heat
Explanation:
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
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:
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:
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
calculate the speed of longitudinal waves in aluminum (assuming the elastic modulus = 6.89 x 10^4 MPa)
The expression for the speed of waves in materials allows us to find the result for the speed of sound in aluminum is:
Sound speed is: v = 5050 m / s
The speed of a wave in a material is determined by the relationship between its volumetric modulus and its density, it is given by the expression.
v = [tex]\sqrt{ \frac{B}{\rho} }[/tex]
where v is the speed of the wave in the material (sound), B is the volume modulus and ρ the density.
They indicate that the volumetric or elastic modulus of aluminum is;
B = 6.89 10⁴ Mpa ( [tex]\frac{10^6 \ Pa}{1 \ MPa}[/tex] ) = 6.89 10¹⁰ Pa
The density of aluminum is tabulated ρ = 2.7 10³ kg / m³
We calculate.
v = [tex]\sqrt{ \frac{6.89 \ 10^{10} }{2.7 \ 10^3 }}[/tex]
v = 5.05 10³ m / s
In conclusion using the expression for the speed of waves in materials we can find the result for the speed of sound in aluminum is:
v = 5050 m / s
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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
plz answer the question.
Answer:
a
Explanation:
sana po makatulong <3♡♡
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 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]
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.
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
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. *
what is the meaning of word thermodynamics
Answer:
physics that deals with the mechanical action or relations of heat.
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
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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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 :)
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:
A stone is dropped from the edge of a roof, and hits the ground with a velocity of -150 feet per second. How high (in feet) is the roof
Answer:
how long does it take? we need it to answer ure question
Explanation:
cause we don't know how many feet until we know how long it was falling
If a stone is dropped from the edge of a roof and hits the ground with a velocity of -150 feet per second, then the height of the roof would have been 1148 feet.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem, a penny is dropped from a building that is 95 m tall, the initial velocity of the penny is zero, and the acceleration acting is due to the acceleration due to gravity,
By using the second equation of the motion for the vertical motion ,
v² = ( 2 × g ×h )
150² = 2 × 9.8 × h
h = 22500 / 19.6
= 1148 feet
Thus, the height of the roof would have been 1148 feet.
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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.
What is the best description of the distribution of the galaxies that lie within about 200 Mpc of Earth
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]
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 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
Give an example of intense aerobics activity. Prompt must be accurate.
Answer:
Explanation:
An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.
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:
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
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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jshshwjs sbwiwiw910mw s x djjskskekwkq
Answer:
jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb
The figure shown above is the circuit diagram for a simple dc power supply. Identify the type of rectifier circuit represented in the figure and explain the operation of the circuit with reference to the function of each component within the circuit.
Answer:
D1 FG 12 15×AG+5T×G7+3F