A ball of mass 2.0 kg is travelling at a speed of 12 m/s. It moves towards an object of mass 3.0 kg which is at rest.
The ball hits the object and sticks to it.
Which row gives the total momentum, and the speed of both objects immediately after the collision?
total momentum (kg m/s)
A- 0
B- 0
С- 24
D- 24
AND..
speed (m/s)
A- 4.8
B- 8.0
C- 4.8
D- 8.0
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The velocity for the entire trip is 0.4 m/s as It takes her 500 seconds to make the round trip and 60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s).
Path 1 = 400 m В B. A Path 2 = 200 m Path 3 = 300 m. Thus, option C is correct.
What is velocity?A particle's settling velocity known as the rate at which is travels through a still fluid. The specific gravity of the particles, their size, and their shape all have an impact on settling velocity.
A particle in still air will gravitationally settle and reach its terminal velocity fairly quickly. A particle's terminal velocity in a still fluid is referred to as the settling velocity (also known as the "sedimentation velocity").
Understanding variations in the hydraulic regime and interactions between sediment and fluid in the surf zone depends heavily on the particle settling velocity at the foreshore region. In contrast to sedimentation, which is the end product of the settling process, settling is the movement of suspended particles through the liquid.
Therefore, Thus, option C is correct.
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How are fission and fusion similar?
A. A small fraction of the reactant mass is converted into small amounts of energy.
B. A large fraction of the reactant mass is converted into small amounts of energy.
C. A large fraction of the reactant mass is converted into large amounts of energy.
D. A small fraction of the reactant mass is converted to large amounts of energy.
Answer:
C
Explanation:
Fusion and fission are similar in that they both release large amounts of energy. Nuclear fusion is a process in which two nuclei join to form a larger nucleus. Nuclear fission is a process in which a nucleus splits into two smaller nuclei.
Does the volume of a gas affect the rate of diffusion in a porous material
Answer:
No
Explanation:
The rate of diffusion defines the gas that passes through some area over the course of time.
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller diagram below.
1) Find the total energy for the roller coaster at the initial point.
2) Find the potential energy at point A using the PE formula.
3) Use the conservation of energy to find the kinetic energy (KE) at point B.
4) Find the potential energy at point C.
5) Use the conservation of energy to find the Kinetic Energy (KE) of the roller coaster at point C.
6) Use the Kinetic Energy from C, find velocity of the roller coaster at point C.
For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:
1) The total energy for the roller coaster at the initial point is 19820 J
2) The potential energy at point A is 19620 J
3) The kinetic energy at point B is 10010 J
4) The potential energy at point C is zero
5) The kinetic energy at point C is 19820 J
6) The velocity of the roller coaster at point C is 19.91 m/s
1) The total energy for the roller coaster at the initial point can be found as follows:
[tex] E_{t} = KE_{i} + PE_{i} [/tex]
Where:
KE: is the kinetic energy = (1/2)mv₀²
m: is the mass of the roller coaster = 100 kg
v₀: is the initial velocity = 2 m/s
PE: is the potential energy = mgh
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 20 m
The total energy is:
[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]
Hence, the total energy for the roller coaster at the initial point is 19820 J.
2) The potential energy at point A is:
[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]
Then, the potential energy at point A is 19620 J.
3) The kinetic energy at point B is the following:
[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]
[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]
Since
[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]
we have:
[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]
Hence, the kinetic energy at point B is 10010 J.
4) The potential energy at point C is zero because h = 0 meters.
[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]
5) The kinetic energy of the roller coaster at point C is:
[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]
[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]
Therefore, the kinetic energy at point C is 19820 J.
6) The velocity of the roller coaster at point C is given by:
[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]
[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]
Hence, the velocity of the roller coaster at point C is 19.91 m/s.
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Block 1, of mass m1, moves across a frictionless surface with speed ui. It collides elastically with block 2, of mass m2, which is at rest (vi=0). (Figure 1)After the collision, block 1 moves with speed uf, while block 2 moves with speed vf. Assume that m1>m2, so that after the collision, the two objects move off in the direction of the first object before the collision. What is the final speed vf of block 2?
The conservation of the momentum allows to find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1- \frac{m_2}{m_1} }[/tex]
the momentum is defined by the product of the mass and the velocity of the body.
p = mv
The bold letters indicate vectors, p is the moment, m the mass and v the velocity of the body.
If the system is isolated, the forces during the collision are internal and the it is conserved. Let's find the momentum is two instants.
Initial instant. Before crash.
p₀ = m₁ u₀ + 0
Final moment. After crash.
[tex]p_f = m_1 u_f + m_2 v_f[/tex]
The momentum is preserved.
p₀ = [tex]p_f[/tex]
[tex]m_1 u_o = m_1 u_f + m_2 v_f[/tex]
Since the collision is elastic, the kinetic energy is conserved.
K₀ = [tex]K_f[/tex]
½ m₁ u₀² = ½ m₁ [tex]u_f^2[/tex] + ½ m₂ [tex]v_f^2[/tex]
Let's write our system of equations.
[tex]m_1 u_o = m_1 u_f + m_2 v_f \\m_1 u_o^2 = m_1 u_f^2 + m_2 v_f^2[/tex]
Let's solve
[tex]u_f = u_o - \frac{m_2}{m_2} \ v_f \\u_f^2 = u_o^2 - \frac{m_2}{m_1} \ v_f^2[/tex]
[tex]( u_o - \frac{m_2}{m_1} v_f)^2 = u_o - \frac{m_2}{m_1} \ v_f^2 \\u_o^2 - 2 \frac{m_2}{m_1} \ u_o v_f + (\frac{m_2}{m_1} )^2 v_f^2 = u_o^2 - \frac{m_2 }{m_1} \ v_f^2[/tex]
[tex]2 \frac{m_2}{m_1} \ u_o = \frac{m_2}{m_1} v_f \ ( 1 - \frac{m_2}{m_1}) \\v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
In conclusion, using the conservation of momentum, we can find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
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Light enters a glass block at an angle of incidence of 46°. The light refracts at an angle of refraction of 26°. What is the refractive index of the glass?
A 0.57
B 0.61
C 1.64
D 1.77
Answer:
Choice C: Approximately [tex]1.64[/tex].
(Assuming that light entered into this glass block from air.)
Explanation:
Let [tex]n_{1}[/tex] denote the refractive index of the first medium (in this example, air.) Let [tex]\theta_{1}[/tex] denote the angle of incidence.
Let [tex]n_{2}[/tex] denote the refractive index of the second medium (in this example, the glass block.) Let [tex]\theta_{2}[/tex] denote the angle of refraction.
By Snell's Law:
[tex]n_{1} \, \sin(\theta_{1}) = n_{2}\, \sin(\theta_{2})[/tex].
Rearrange the equation to find an expression for [tex]n_{2}[/tex], the refractive index of the second medium (the glass block.)
[tex]\begin{aligned}n_{2} = \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1}\end{aligned}[/tex].
The refractive index of air is approximately [tex]1.00[/tex]. Substitute in the values and solve for [tex]n_{2}[/tex], the refractive index of the glass block:
[tex]\begin{aligned}n_{2} &= \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1} \\ &= \left(\frac{\sin(46^{\circ})}{\sin(26^{\circ})}\right)\times 1.00 \\ &\approx 1.64\end{aligned}[/tex].
A small sphere 0.70 times as dense as water is dropped from a height of 8 m above the surface of a smooth lake. Determine the maximum depth to which the sphere will sink. Neglect any energy transferred to the water during impact and sinking.
Answer:
18.66m
Explanation:
This was actually fun! Let set up the problem first: As you drop the sphere, it will accelerate till it hits the water with a given speed [tex]\dot z[/tex]. Once sinking the ball is subjected to two forces: its own weight [tex]m_sg = \rho_sVg[/tex] directed towards the bottom of the lake, and buoyancy (Archimedes law), ie a force upward equal to the volume of displaced water, [tex]m_wg=\rho_wVg[/tex]. At this point it's the classical "how high can I toss a ball before it falls down" with a water twist. Please note that I'm using z as the height of the sphere, and [tex]\dot z; \ddot z[/tex] represent the velocity (with one dot) and the acceleration (with two dots) in the z direction .Let's assume the sign of all quantities being positive if directed upwards, and negative if towards the bottom of the lake.
Let's first determine how fast the ball hit the water. For me the easiest way is saying "at 8m it has a given potential energy, and 0 kinetic energy. When it hits the water it loses all potential energy and has only kinetic energy". In numbers:
[tex]m_sgz=\frac12m_s\dot z_0^2[/tex] Let's divide by the masses, g is a known value ([tex]9.81 ms^{-2}[/tex]), z is 8 meters, and we get that [tex]\dot z_0 = 4\sqrt g =12.52 m/s[/tex]
At this point, let's determine the force acting on the sphere, courtesy of Newton second law and the debate we had earlier:
[tex]m_s \ddot z = -m_sg + m_wg\\\rho_sV \ddot z = -\rho_sVg + \rho_wV g[/tex]
At this point we can divide by the volume of the sphere, and make use of the fact that [tex]\rho_s = 0.7 \rho_w[/tex]
[tex]0.7\rho_w \ddot z = -.7\rho_w g+ \rho_wg \rightarrow 0.7 \ddot z = 0.3 g\\\ddot z = \frac {0.3}{0.7} g= \frac 37g = 4.2 ms^-2[/tex]
We're almost there. At this point we can write the law of motion for the sphere.
[tex]\dot z = \dot z_0 + \ddot z t[/tex] that we will use to find the time for the ball to stop sinking - ie reaches 0 velocity.
[tex]0 = -12.52 +4.2t \rightarrow t= 12.52/4.2 \approx 3s[/tex]
At this point we can use the other law of motion to find out the distance traveled
[tex]z= z_0 + \dot z_ot + \frac12 \ddot z t^2\\z= 0 -12.52 (3) + \frac12 (4.2)(3)^2 = -37.56+18.90 = -18.66 m[/tex]
as usual, please double and triple check all the calculations, it's almost 1.30 am here and I am not the most confident number cruncher even when fully awake.
2hrt 20 min +60 sec
Answer:
2hr 21 min
Explanation:
0 days 2 hours 20 minutes 0 seconds
+ 0 days 0 hours 0 minutes 60 seconds
= 0 days 2 hours 21 minutes 0 seconds
An element with a force of 0.00SN moves at a minimum: 0.2m / s². Calculate the mass of that element.
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How much work must be done to stop a 925-kg car traveling at 95 km/h?
The amount of work done to stop the car is equal to 322,099.85 Joules.
Given the following data:
Mass = 925 kilogramsVelocity = 95 km/hConversion:
Velocity = 95 km/h = [tex]\frac{95 \times 1000}{60 \times 60} = 26.39 \;m/s^2[/tex]
To determine the amount of work done to stop the car, we would apply the work-energy theorem:
Note: The amount of work done must balance the kinetic energy possessed by the car due to its motion.
From the work-energy theorem, we have:
[tex]Work\done = \Delta K.E\\\\Work\done = \frac{1}{2} MV^2[/tex]
Substituting the given parameters into the formula, we have;
[tex]Work\;done = \frac{1}{2} \times 925 \times (26.39)^2\\\\Work\;done =462.5 \times 696.4321[/tex]
Work done = 322,099.85 Joules
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Energy a substance or system has because of its motion Is what?
light energy
chemical energy
sound energy
electrical energy
thermal energy
mechanical energy
Answer:
Mechanical energy is the energy a substance or system has because of its motion. For example machines use mechanical energy to do work.
Find the work done while lifting a 115 kg sofa 1.5 m off the ground.
Find the weight first
P= 10m = 10 x 115 = 1150 N
work done while lifting a 115 kg sofa 1.5 m off the ground :
A= Ph = 1150 x 1.5 = 1725 J
P/s: I think the weight first is 115kg
ok done. Thank to me :>
3. What happens if an object has a negative acceleration?
Answer:
They will go backwards?
Explanation:
If positive they go forward but is negative they go backwards.
A ____ is a region where jets of gas from young stars impact and heat the gas surrounding the young star.
How much potential energy did the the milkshake have after it fell off of the counter, just before it hit the ground (at the bottom)? Make sure to write the correct units with
your answer!
Answer:
The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters.
The energy transferred per unit electric charge in a circuit is .
A current
B charge
C power
D potential difference
Answer:
(D)
E = V Q energy of charge Q thus V = E / Q
the small lightweight particles outside the nucleus of an atom are called___.
The neutron has no charge, and a mass of slightly over 1 amu. Some scientists propose the neutron is made up of a proton and electron-like particle. The electron is a very small particle located outside the nucleus.
Electron: -1
Neutron: 0
Proton: +1
Calculate the force applied if 100 N/m² pressure is exerted over the area of 0.2m²
Answer: 100 pascals times 0.2 m^2= 20 N force is applied.
Explanation:
You increase your walking speed from 1 m/s to 3 m/s in a period of 1 s.
What is your acceleration?
5 m/s2
3 m/s2
2 m/s2
4 m/s2
Answer:
2 m/s2
Explanation:
the answer is 2 m/s2
the following can increase the amount of friction in the surface, except ______.
a. glue
b. paper
c. lubricant
d. foil
Answer:
c. lubricant..........
The initial temperature of the mixture was +20 °C. The mixture froze at -1.5 °C.
A total of 165 kJ of internal energy was transferred from the mixture to cool and
freeze it.
specific heat capacity of the mixture = 3500 J/kg °C
specific latent heat of fusion of the mixture = 255 000 J/kg
Calculate the mass of the mixture.
Give your answer to 2 significant figures.
[6
M
Answe16000/3500(-1.5-2-)225000= 1.1
Explanation:
The mass of the mixture will be 2.19 ×10⁻³ kg. The unit of mass is kg and it is a scalar quantity.
What is mass?Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.
In the International System of Units (SI), the kilogram is the unit of mass.
The given data in the problem is;
The initial temperature of the mixture was, [tex]\rm T_i = 20 ^0 \ C[/tex]
The mixture froze at the temperature of, [tex]\rm T_F = -1.5 ^0 \ C[/tex]
The internal energy is,[tex]\rm E = \rm 165 \ kJ[/tex]
The specific heat capacity of the mixture is,[tex]\rm C = 3500 J/kg ^0C[/tex]
The specific latent heat of fusion of the mixture is,[tex]\rm l = 255 000 J/kg[/tex]
m is the mass of the mixture,
The latent heat is equal to the heat transfer. Because the net work done is zero.
The formula for the heat transfer is found as;
[tex]\rm Q = mCdt \\\\ m = \frac{Q}{Cdt} \\\\ m = \frac{165 }{3500(20-(-1.5)} \\\\ m = 2.19 \times 10^{-3} \ kg[/tex]
Hence the mass of the mixture will be 2.19 ×10⁻³ kg.
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The weight of a 1,000 kg car is ______________ (pick best answer)
Group of answer choices
9800 N
102.4 N
1000 N
10000 N
What is the weight of a car with the mass of 1000 kg?
Answer: it is 9800 N.
How is the input energy carried to the light bulb's energy stores?
Answer:
the electric energy travels through the plug or wire or whatever is connecting the energy source to the lightbulb, which conducts electricity, and to the light bulb. the electric energy is transfered into light energy and thermal energy
Explanation:
i hope i understood this question and answered it correctly XD
A car of mass 1500 kg is moving with a velocity of 30 m/s. If the velocity-time graph for this car
is a horizontal line parallel to the time axis, then the velocity of the car at the end of 25 s will be
Answer:
v=30m/s
Explanation:
m = 1500kg
v = 30m/s
t = 25s
The velocity time graph is parallel to the time axis,
that means The Velocity is constant. ('constant' means remains same).
Therefore, after 25s, v=30m/s
2. Given what you know about the acceleration of Earth's gravity (g = 9.8 m/s2), is this number accurate?
accurate. If not explain why you think it is not accurate. Pleaseee help mee
Answer:
it is correct
Explanation:
Though no rounded numbers can be defined as accurate, if we were going by people's discovery, and research, we can define that the number, g = 9.8m/s^2, is accurate
What time is the lunar eclipse tonight in california?
According to the table, what was the hiker's average speed for
the entire hike?
A: 2.0 km/h B: 5.7 km/h C: 6.7 km/h D: 10 km/h
Answer:
6.7 km/h (C)
Explanation:
According to the table, the hiker's average speed for the entire hike would be 6.7 kilometers/hour.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
The total distance traveled by the hiker = 4 kilometers + 6 kilometers + 4 kilometers west + 6 kilometers north
= 20 kilometers
The total time is taken by the hiker = 45 min + 60 min + 30 min+ 45 min
= 180 min
=3 hours
The average speed of the hiker = 20 kilometers / 3 hours
= 6.7 kilometers / hours
Thus, the hiker's average speed for the entire hike would be 6.7 kilometers/hour.
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help me solve please
Answer:
please check the image
Explanation:
I hope this image helps you please follow the steps. Thank you
Answer:
1kg/```````2
Explanation:
solution
what is the medium for the sound waves for whales to communicate with each other?
Answer:
SONGS CLICKS AND WHISTLE
Explanation:
Imagine you have a ball tied to the end of a string. You hold the other end of the string and swing it around. Suppose the string breaks, what direction will the ball travel