Answer:
(a) The two balls collide [tex]2\; \rm s[/tex] after launch.
(b) The height of the collision is [tex]4\; \rm m[/tex].
(Assuming that air resistance is negligible.)
Explanation:
Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.
The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].
Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].
On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].
Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:
[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].
For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].
For the ball thrown downwards:
Initial velocity: [tex]v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 40\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Similarly, for the ball thrown upwards:
Initial velocity: [tex]v_0 = \left(12\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 0\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Equate the two expressions and solve for [tex]t[/tex]:
[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].
[tex]t = 2[/tex].
Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.
Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:
[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].
In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].
The time the two balls collide is 0.4 seconds while the height at which they collide is 4m
The given parameters are :
Initial Velocity U = 8m/s
Height H = 40m
For the second ball, the initial velocity = 12m/s
a.) For the first ball, the height attained at the point of collision will be
h = ut + 1/2gt^2
h = 8t + 1/2 x 10t^2 ........ (1)
For the second ball, the height attained at the point of collision will be
h = 12t - 1/2 x 10t^2 .........(2)
Since the height will be the same for the two balls, equate the two equations
8t + 10t^2 = 12t - 10t^2
Collect the like term
8t - 12t = -5t^2 - 5t^2
-4t = -10^2
10t = 4
t = 4/10
t = 0.4s
b.) Substitute time t in any of the equation to find the height
h = 12(0.4) - 0.5 x 10(0.4)^2
h = 4.8 - 0.8
h = 4m
Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m
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What is the correct organization of living things, from smallest to largest?
Cells - Tissues - Organs - Organ Systems - Organism
Organs - Tissues - Cells - Organ Systems - Organism
Cells - Organs - Tissues - Organism - Organ Systems
Cells - Organism - Tissues - Organ Systems - Organs
A racecar makes 24 revolutions around a circular track of radius 2 meters in
162 seconds. Find the racecar's frequency
Answer:
[tex]0.15\: \mathrm{Hz}[/tex]
Explanation:
The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.
Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.
Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).
The radius of the track is irrelevant in this problem.
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
WHat does that mean?
Which ray diagram demonstrates the phenomenon of absorption?
An illustration with a vector pointed right going through an opening in a boundary and splitting into 3 vectors. One up and to the right, one straight and one down to the right.
An illustration with a vector pointed right going through an opening in a boundary and turning into a transverse wave on the other side.
An illustration with a vector striking a boundary at an angle and a second vector coming off the boundary at the exact same angle.
Answer:
B
Explanation:
on edge
The illustration with a vector pointed right going through an opening in a boundary and turning into a transverse wave on the other side demonstrates the phenomenon of absorption, so, option B is correct.
What is absorption?Absorption, in wave motion, is the process by which a wave's energy is transferred to matter when the wave travels through it. The energy of an electromagnetic, acoustic, or other wave is related to the square of its amplitude, which is the maximum displacement or movement of a point on the wave.
The amplitude of a wave continuously diminishes as it travels through a substance. The medium is described as being transparent to a specific type of radiation if just a tiny portion of the energy is absorbed, whereas it is described as opaque if all the energy is lost.
To know more about absorption:
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A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving with
the same speed and in the same direction?
True or False do Eclipses, tides, season, and moon phases ALL have to do with the positions of the Earth, Sun, and Mars.
A student asks the following question:
"If all things with mass have a gravitational field, why doesn't this glue bottle and
stapler, sitting on the counter, stick together because of gravitational forces?"
Which classmate answers correctly?
Ashton says that the gravitational fields between the bottle and the stapler
cancel out because of Newton's 3rd Law.
O Natalie says that all things with mass have a gravitational field, but the force is
very weak and cannot be perceived around small objects.
Xavier says the bottle and the stapler are way too small to have a gravitational
field.
Katherine says the bottle and the stapler have a strong gravitational field, and
would move towards each other quickly if there were no friction on the counter.
Answer:
Natalie says that all things with mass have a gravitational field, but the force is very weak and cannot be perceived around small objects.
Explanation:
The force due to gravity is proportional to the mass of the object and inversely proportional to the square of the distance between objects. The Earth is so massive that the force due to its gravity is much greater than the force between objects on the counter.
If there were no friction, the objects might move toward each other, depending on what other masses were near them tending to cause them to move in other directions.
Natalie's explanation is about the best.
__
Additional comment
The universal gravitational constant was determined by Henry Cavendish in the late 18th century using lead balls weighing 1.6 pounds and 348 pounds. His experiment was enclosed in a large wooden box to minimize outside effects. While these masses are somewhat greater than those of a glue bottle and stapler, the experiment shows the force of gravity between "small" objects can be measured.
at what speed does the kg ball move ?
Answer: Choice A) 2 meters per second
=======================================================
Explanation:
The smaller ball has momentum of
p = m*v
p = (1 kg)*(4 m/s)
p = 4 kg*m/s
All of this momentum transfers into the larger ball because the smaller ball comes to a complete stop.
For the larger ball, we have p = 4 and m = 2. Let's find v.
p = m*v
4 = 2*v
4/2 = v
2 = v
v = 2 m/s which is why the answer is choice A
The larger ball moves at a speed of 2 meters per second. The speed is cut in half compared to the smaller ball because the larger ball has more inertia (aka more mass), and therefore it takes more energy to move it. If you apply the same energy to each, then the smaller object moves faster.
1.How much work does it take to get a 2Kg ball moving 15m/s if it starts from rest?
2. If a force of 235N was added to the ball, through what distance would this force have to act to give the ball a velocity of 15m/s
Two spherical objects have masses of 100 kg and 200 kg. Their centers are
separated by a distance of 40 cm. Find the gravitational attraction between
them.
Answer:
8.34 x 10⁻⁶N
Explanation:
Given parameters:
Mass 1 = 100kg
Mass 2 = 200kg
Distance of separation = 40cm = 0.4m
Unknown:
Gravitational force of attraction between them = ?
Solution:
To solve this problem, we use the expression below which is derived from the Newton's law of universal gravitation:
Fg = [tex]\frac{G x mass 1 x mass 2}{d^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
d is the separation
Now;
Fg = [tex]\frac{6.67 x 10^{-11} x 100 x 200}{0.4^{2} }[/tex] = 8.34 x 10⁻⁶N
which changes will increase the rate of reaction during combustion
Answer:
reducing temperature of the surrounding
Explanation:
combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
A car is traveling at a constant speed of 20 m/s for 3 seconds. Then the driver puts on the brakes. The total distance the car travels is 100 m. What is the total time the car was moving?
Answer:
15 seconds
Explanation:
If car was moving at 20m/s for 3 sec.
if car traveled 100m = 15 sec total
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.
Answer:
the tack's tangential speed is 5.59 m/s
Explanation:
Given that;
R = 0.331 m
wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so
ω = 2.69 rev/s × 2π/1s = 16.9 rad/s
using the relation of angular speed with tangential speed
tangential speed v of the tack is expressed as;
v = R × ω
so we substitute
v = 0.331 m × 16.9 rad/s
v = 5.59 m/s
Therefore, the tack's tangential speed is 5.59 m/s
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr
Plz help this is so confusing
Answer:
5 Km/h
Explanation:
From the question given above, the following data were obtained:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed is simply defined as the distance travelled per unit time. Mathematically, it can be represented as:
Speed = distance travelled /time.
With the above formula, we can obtain the speed at which the duck is travelling as follow:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed = distance travelled /time.
Speed = 10 / 2
Speed = 5 Km/h
Thus, the duck is travelling at a speed of 5 Km/h
A ray of monochromatic light is incident on a plane mirror at and angle of 30. The angle of reflection for the light is
1)15
2)30
3)60
4)90
Answer:
30 degrees
Explanation: reflection, same angle
For a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
Reflection occurs when radiation bounces off from a surface. Light is an electromagnetic wave and it can be reflected. According to the laws of reflection, the angle of incidence is equal to the law of reflection.
Hence, for a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
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A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the turnaround marker to the starting marker at 16 m/s. What is her average speed for the whole race?
Answer:
12.31 m/s
Explanation:
If we recall from the previous knowledge we had about speed,
we will know that:
speed = distance/ time.
As such:
The average speed of the rider bicycle is
average speed = total distance/ total time
Mathematically, it can be computed as:
[tex]v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}[/tex]
[tex]v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}[/tex]
[tex]\mathbf{v_{avg} =12.31 \ m/s}[/tex]
Can a single atom be considered a molecule?
A:only if the atom is found in water
B:no, it takes two or more atoms bonded to create a molecule
C:only if it is an oxygen atom floating in the air
D:yes, all atoms are made up of many different molecules
The second law of thermodynamics imposes what limit on the efficiency of a heat engine?
A. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy it extracts from a hot reservoir.
B. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy extracted as useful work.
C. A heat engine must deposit some energy in a cold reservoir.
Answer:
C. A heat engine must deposit some energy in a cold reservoir.
Explanation:
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."
This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.
Then we have the equation:
Q = W + q
From this we can conclude that the correct option is:
C. A heat engine must deposit some energy in a cold reservoir.
There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.
C. A heat engine must deposit some energy in a cold reservoir.
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir". This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir. Then we have the equation: Q = W + q There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.Therefore, option C is correct.
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A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
A pingpong ball has 2 kg/s of momentum when
thrown 8 m/s. Find the mass of the ball.
Answer:
0.25 kg
Explanation:
p = mv
2 = m(8)
2/8 = m(8)/8 *cancels
m = 1/4 OR 0.25 kg
what are ribosomes?
I'm tired. But I have insomnia. Big ugh moment. <.<.
Answer:
Ribosomes are organelles the make protein for the cell.
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Answer:
2 m/s²
Explanation:
From the given information:
The first mass m_1 = 0.6 kg
The second mass m_2 = 0.3 kg
The magnitude for the acceleration of 0.3 kg is:
a = net force/ effective mass
Mathematically, it can be computed as follows:
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]
[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]
a ≅ 2 m/s²
To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
There are two different isotopes; X and Y, both contain the same number of radioactive substances. If sample X
has a longer half-life than Y, compare their rate of radioactive decay.
O A. Rate does not depend on half-life
B. Both of their rates are equal
O C. X has a smaller rate than Y
O D. X has a greater rate than Y
Answer:
Half life refers to the time for 1/2 of the radioactive atoms to decay.
Suppose that X has a half life of 10 days and Y has a half life of 20 days
If both start out with 1000 radioactive atoms then after 20 days
X would have 250 radioactive atoms and Y would have 500 atoms
The rate of decay is greater for the shorter half life:
In the example given X must have the smaller rate of decay because it has a longer half life.