Answer:
the car movves briefly as you ran, however, it stops again after you ran in to the wall
Explanation:
Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.
If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.
i.e
[tex]m \times V_1 + M \times V_2 = 0[/tex] --- (1)
here;
m = mass of the boy
M = mass of the train
Thus;
[tex]m \times V_1 =- M \times V_2[/tex] --- (2)
As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.
When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV.
What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface?
Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.
Answer:
Explanation:
energy of photon having wavelength of 400 nm = 1237.5/400 eV
= 3.1 eV.
Maximum kinetic energy of photoelectrons = 1.1 eV .
Threshold energy Ф = 3.1 - 1.1 = 2 eV .
energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .
Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy
= 4 - 2 = 2 eV .
Required kinetic energy K₀= 2 eV.
A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it
Answer:
w = 0.319 rad / s
Explanation:
This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.
initial instant. Before the squirrel jumps
L₀ = m v r
final instant. After the trough and the squirrel are together
L_f = (I_fetter + I_ardilla) w
angular momentum is conserved
L₀ = L_f
m v r = (I_fetter + I_ardilla) w
w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]
the moment inercial ofbody is
I_thed = 2.00 kg m²
We approach the squirrel to a specific mass
I_ardilla = m r²
we substitute
w = m v r / ( I_[feefer + m r²)
let's calculate
w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )
w = 0.6426 / 2.0119
w = 0.319 rad / s
state newton first law of motion
Newton’s first law of motion states that there must be a cause—which is a net external force—for there to be any change in velocity, either a change in magnitude or direction. An object sliding across a table or floor slows down due to the net force of friction acting on the object.
got it off g lol..
Answer:
it state that everybody in the universe is state that" universe continues its state of rest or uniform motion in a straight path unless it is acted upon by external force."
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled? *
A)60km
B)60km[W]
C)17km[W]
D) 6.6km[W]
Answer:
B)60km[W]
Explanation:
The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].
Hope that helps!
Which of the following best describes what occurs in a fission reaction?
A.
Two low mass nuclei are joined to form one nucleus.
B.
Electrons are shared between the nuclei.
C.
A single nucleus divides into two or more nuclei and gives off energy.
D.
A chemical reaction occurs between the nuclei.
Answer:
C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.
Answer:
C.
A single nucleus divides into two or more nuclei and gives off energy.
hope it is helpful to you
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]
v₂ = 30 m/s
b)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]
m₁ = 12600 kg
c)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]
m₂ = 12600 kg
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
An object’s
✔ mass
will remain constant throughout the universe, but its
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
Answer:
mass, weight, strength of gravity increases, weight decreases
Explanation:
got it on edge
Answer:
An object’s
✔ mass
will remain constant throughout the universe, but its
✔ weight
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
✔ strength of gravity increases
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
✔ weight decreases
Explanation:
right on edge 22
An object is placed 12.0 cm from a thin diverging lens with a focal length of 4 cm. Which one of the
following statements is true concerning the image?
A. The image is virtual and 3.0 cm from the lens.
B. The image is real and 6.0 cm from the lens.
C. The image is virtual and 12 cm from the lens.
D. The image is real and 12 cm from the lens.
Answer:
soluble soluble soluble soluble
Explanation:
solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci
While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.
Answer:
the smallest separation between two objects is 0.8067 m
Explanation:
Given the data in the question;
Altitude h = 5.75 km = 5750 m
Diameter D = 4.0 mm = 0.004 m
λ = 460 nm = 4.6 × 10⁻⁷ m
Now, Using Rayleigh criterion for Airy disks resolution.
we know that, Minimum angular separation for resolving two points is;
θ = 1.22λ / D
so we substitute
θ = (1.22 × 4.6 × 10⁻⁷) / 0.004
θ = 5.612 × 10⁻⁷ / 0.004
θ = 1.403 × 10⁻⁴ rad
so minimum separation [tex]d_{min[/tex] = θh
so we substitute
[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m
[tex]d_{min[/tex] = 0.8067 m
Therefore, the smallest separation between two objects is 0.8067 m
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of 0.295 cm above its initial position. What is the velocity of the bullet on leaving the gun's barrel
Answer:
The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 2.47 g = 0.00247 kg
mass of the wooden block, m₂ = 2.43 kg
initial velocity of the wooden block, u₂ = 0
height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m
let the initial velocity of the bullet on leaving the gun's barrel = v₁
let final velocity of the bullet-wooden block system after collision = v₂
Apply the principle of conservation of linear momentum;
Total initial momentum = Total final momentum
m₁v₁ + m₂u₂ = v₂(m₁ + m₂)
0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)
0.00247v₁ = 2.4325v₂ -------(1)
The kinetic energy of the bullet-block system after collision;
K.E = ¹/₂(m₁ + m₂)v₂²
K.E = ¹/₂ (2.4325)v₂²
The potential energy of the bullet-block system after collision;
P.E = mgh
P.E = (2.4325)(9.8)(0.00295)
P.E = 0.07032
Apply the principle of conservation of mechanical energy;
K.E = P.E
¹/₂ (2.4325)v₂² = 0.07032
1.21625 v₂² = 0.07032
v₂² = 0.07032 / 1.21625
v₂² = 0.0578
v₂ = √0.0578
v₂ = 0.24 m/s
Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;
0.00247v₁ = 2.4325v₂
0.00247v₁ = 2.4325 (0.24)
0.00247v₁ = 0.5838
v₁ = 0.5838 / 0.00247
v₁ = 236.36 m/s
Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
Drag each statement to the correct location on the table.
Match the characteristics with the states of matter.
does not have
a definite shape
or volume
has definite volume
but does not have a
definite shape
has a definite shape
and volume
changes to liquid
on heating
changes to liquid
on cooling
changes to solid
on cooling
Solid
Liquid
Gas
mentum. All rights reserved.
Answer:
Solid:
- has definite shape and volume.
- change to liquid on heating.
Liquid:
- has definite volume but does not have definite shape .
- changes to solid on cooling.
Gas :
- does not have definite shape or volume.
- change to liquid on cooling
Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
Hint : ΔV = Ed
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F[tex]_{net[/tex] = 0
so, F[tex]_E[/tex] = F[tex]_B[/tex]
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
The density of 1 kilogram of gold is
Answer:
0.02 kg/cm³
Explanation:
2(A + B)
15. The resultant of A and B is perpendicular to A
What is the angle between A and B?
(a) cos
(b) cos
La
(c) sin
(d) sin
Answer:
θ = cos^(-1) (-A/B)
Explanation:
The image of the reauktant forces A & B are missing, so i have attached it.
Now, from the attached image, we will see that;
Angle between A and B is θ
Also;
A = Bcos(180° − θ)
Now, in trigonometry, we know that;
cos(180° − θ) = -cosθ
Thus;
A = -Bcosθ
cosθ = -A/B
Thus;
θ = cos^(-1) (-A/B)
A radioactive material produces 1160 decays per minute at one time, and 4.0 h later produces 170 decays per minute. whats the half life
Answer:
Half life is 3.23 hours
Explanation:
Given
Decay rate at starting = 1160 decays per minute
Decay rate after 4 hours = 170 decays per minute
As we know know
[tex]N = N_0 *e ^{\Lambda *T}[/tex]
Substituting the given values, we get -
[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]
Also
[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]
Substituting the given values we get -
[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled?
Answer:
A boat travels for three hours with a... A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in still water?
Explanation:
A car is moving with speed 30 m/s and acceleration 4 m/s2 at a given instant. (a) Using a second-degree Taylor polynomial, estimate how far the car moves in the next second.
Answer:
68 meters moved in the next seconds
Explanation:
Given
[tex]u= 30m/s[/tex]
[tex]a = 4m/s^2[/tex]
Required
Distance covered by the car in the next second
At a point in time t, the current distance is calculated as:
[tex]s(t) = ut + \frac{1}{2}at^2[/tex]
Substitute values for a and u in the above equation.
[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]
[tex]s(t) =30t + 2t^2[/tex]
Next, we generate the second degree Taylor polynomial as follows;
Calculate velocity (s'(t))
Differentiate s(t) to get velocity
[tex]s(t) =30t + 2t^2[/tex]
[tex]s'(t) =30 + 4t[/tex]
Calculate acceleration (s"(t))
Differentiate s'(t) to get acceleration
[tex]s'(t) =30 + 4t[/tex]
[tex]s"(t) =4[/tex]
When t = 0
We have:
[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]
[tex]s'(0) =30 + 4*0 = 30[/tex]
[tex]s"(0) = 4[/tex]
So, the second degree tailor series is:
[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]
To see the distance moved in the next second, we set t to 1
So, we have:
[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]
Solving s(1), s'(1) and s"(1)
We have:
[tex]s(1) =30*1 + 2*1^2 = 32[/tex]
[tex]s'(1) =30 + 4*1 = 34[/tex]
[tex]s"(1) =4[/tex]
Hence:
[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]
[tex]T_2(1) = 32 + 34 + 2[/tex]
[tex]T_2(1) = 68[/tex]
Saved Which of the following is NOT an important function of facial display? Multinio Choic
A. emotion
B. attractiveness
c. Primacy
d. identity
Answer:
C
Explanation:
Primacy means being first or important so thats not an important facial display as the others.
You need friction created by your tires and the road ____
to control your speed and direction.
Answer:
surface
Explanation:
You need friction created by your tires and the road surface
to control your speed and direction.
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Answer:
[tex]523269.9\ \text{N/m}[/tex]
Explanation:
q = Charge
r = Distance
[tex]q_1=25\ \text{C}[/tex]
[tex]r_1=3000\ \text{m}[/tex]
[tex]q_2=40\ \text{C}[/tex]
[tex]r_2=850\ \text{m}[/tex]
The electric field is given by
[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]
The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]
Which form of energy increases when a spring is compressed?
Answer:
When the spring compresses, elastic potential energy increases.
Answer:
the answer is b
Explanation:
elastic potential energy
According to the Traditional Square of Opposition: If "All S are P" is true, then is "Some S are P" true or false?
[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the concept of Logic.
Since in the above statement it is given that,
All S are P ==> True,
then obviously Some S are also P always, hence it is true.
Answer is True.
How much work will a 500 watt motor do in 10 seconds?
Answer:
50j
Explanation:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Answer:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Explanation:
10 POINTS!! SPACE QUESTION!
Answer:
The Gas Giants have more moons.
Explanation:
Mercury-0
Venus-0
Earth-1
Mars-2
Jupiter-66
Saturn-62
Uranus-27
Neptune-13
An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k = 450 N/m. The object is determined to have a velocity of 0.3 m/s when passing through the equilibrium.
1. Find the amplitude of the motion
2. Find the total energy of the object at any point of its motion
Answer:
1) The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object at any point of its motion is 16.892 joules.
Explanation:
1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ([tex]U_{e}[/tex]), in joules, is equal to maximum translational kinetic energy ([tex]K[/tex]), in joules:
[tex]U_{e} = K[/tex]
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]A[/tex] - Amplitude, in meters.
[tex]m[/tex] - Object mass, in kilograms.
[tex]v[/tex] - Speed of the object at equilibrium, in meters per second.
If we know that [tex]k = 450\,\frac{N}{m}[/tex], [tex]m = 0.25\,kg[/tex] and [tex]v = 0.3\,\frac{m}{s}[/tex], then the amplitude of the motion is:
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]k\cdot A^{2} = m\cdot v^{2}[/tex]
[tex]A = v\cdot \sqrt{\frac{m}{k} }[/tex]
[tex]A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }[/tex]
[tex]A \approx 0.274\,m[/tex]
The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object ([tex]E[/tex]), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ([tex]k = 450\,\frac{N}{m}[/tex], [tex]A \approx 0.274\,m[/tex])
[tex]E = U_{e}[/tex]
[tex]E = \frac{1}{2}\cdot k\cdot A^{2}[/tex]
[tex]E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}[/tex]
[tex]E = 16.892\,J[/tex]
The total energy of the object at any point of its motion is 16.892 joules.
which statement can best describe the energy transformations that occur when a pendulum swings back and forth?
a. gravitational energy is converted to spring energy and back.
b. gravitational energy is converted to kinetic energy and back
c. kinetic energy is converted to spring energy and back
d. none of the above.
Answer:
I think it is C, I'm not for sure.
The nucleus of a certain type of uranium atom contains
92 protons and 143 neutrons. What is the total charge of
the nucleus?
Answer:
charge = electrons + protons
=92+92
=184
A cart of mass m is moving with negligible friction along a track with known speed v1 to the right. It collides with and sticks to a cart of mass 4m moving with known speed v2 to the right. Which of the two principles, conservation of momentum and conservation of mechanical energy, must be applied to determine the final speed of the carts, and why
Answer:
conservation of linear momentum
We were told that two objects became stuck together hence we have to use the principle of conservation of momentum to obtain the final velocities of the carts.
What is conservation of momentum ?The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.
We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.
Learn more about conservation of momentum: https://brainly.com/question/11256472
Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________
a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.
Answer:
d. the star is a member and also a part of an eclipsing binary star system.
Explanation:
If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).
The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.
Thus, option D is correct.