Answer:
Explanation:
For an incline projection made at angle of θ , expression of horizontal range R can be given by the following equation .
R = u² sin 2θ / g
u is initial velocity of throw .
For range R to be maximum , the value of sin 2θ must be maximum . The maximum value of sin of an angle is 1 , so
For maximum R ,
sin2θ = 1 = sin90
2θ = 90 .
θ = 45⁰ .
So for projection made at 45⁰ , horizontal range is maximum .
A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
A 87 kg man has a total mechanical energy of 1780 J .If he is swinging downward and is currently 1.4 m above the ground, what is his speed? Use g = 10 m/s^2
Answer:
6.4m/s
Explanation:
The total mechanical energy of the man is 1780J.
This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.
Also, mass = 87kg
Kinetic energy = [tex]\frac{1}{2}[/tex] m v²
m is the mass
v is the velocity
1780 = [tex]\frac{1}{2}[/tex] x 87 x v²
v² = 40.9
v = 6.4m/s
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 horse and a dog have same momentum. Which of them have greater kinetic energy.
horse
dog
both have same K.E
insufficient information
Answer:
C. both have same K.E
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = Mass * Velocity [/tex]
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}MV^{2}[/tex]
Where, K.E represents kinetic energy measured in Joules.
M represents mass measured in kilograms.
V represents velocity measured in metres per seconds square.
Both momentum and kinetic energy are related to the velocity of an object or a body.
Since the horse and a dog have same momentum. Thus, they both have same kinetic energy.
Particle A with charge q and mass ma and particle B with charge 2q and mass
mb, are accelerated from rest by a potential difference AV and subsequently
deflected by a uniform magnetic field into semicircular paths. The radii of the
trajectories by particle A and B are R and 3R, respectively. The direction of
the magnetic field is perpendicular to the velocity of the particle. Determine
their mass ratio?
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
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.
Why is damage from sound waves is an issue on the launchpad but not in the air
(I would have done more points for answering but I'm almost out sry. just pls answer and help.)
The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.
When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.
Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.
Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
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how much power does it take to do 104 J of work in 8 sec?
Answer:
Given:
W=104J
t=8sec
Solve:
P=?
Equation:
P=W/t or Fd/t
P= 104J'8sec
P=13W
Planet Earth is held in orbit by the sun's gravitational force pulling on the Earth.
Which one of the following statements are true about that gravitational force?
The sun does not have a gravitational force because it is made of gas.
The earth pulls with more force since it is less massive than the sun.
The sun and the earth pull on each other with equal and opposite forces.
The sun pulls with more force since the sun is more massive than the earth.
Answer:
The sun pulls with more force since the sun is more massive than the earth.
Explanation:
That your answer because the Earth is getting pulled by the gravitational pull from the Sun
Kayla starts at -3, walks 5 blocks right and 3 blocks left. What is her displacement?
Answer: The displacement is 1 block.
Explanation:
Let's define:
The right is the positive side.
The left is the negative side.
Then if you start at position A, and you walk N blocks to the right, the new position is:
A + N
And if you start at position A, and you walk M blocks to the left, the new position is:
A - M.
In this case, we know that Kayla starts at -3 and she walks 5 blocks to the right.
Then her new position is:
-3 + 5 = 2
Now she walks 3 blocks to the left, then her new position is:
2 - 3 = -1
The displacement will be equal to the difference between the final position (-1) and the initial position (-2)
Then the displacement is:
D = -1 - (-2) = -1 +2 = 1
The displacement is 1 block.
a spring stretches from an initial height of 5 cm to a final stretch of 10 cm. the spring constant is 800 n/m.How much work was done on the spring?
what is the final force on the spring when it is at its 10 cm stretch?
explain why it is not appropriate to use the equation w=f//d when considering springs.
Answer:
.
Explanation:
F = kx so k = 800/((10-5)/100) = 16000 N/m
W = 1/2 kx^2 = 1/2 * 16000 * .05^2 = 20 J.
(sorry if it's wrong)
explain resolution of Force
Answer:
it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.
Explanation:
A tangent line drawn on a velocity-time graph has a rise of 19 m/s and a run of 4.0 m/s. How large is the acceleration? What type of acceleration Is this?
Answer:
Acceleration = 4.8 m/s²
Explanation:
Given:
Change in velocity = 19 m/s
Change in time = 4 s
Find:
Acceleration
Computation:
Acceleration = Change in velocity / Change in time
Acceleration = 19/4
Acceleration = 4.8 m/s²
Positive acceleration
7Which of the following terms describes how glaciers move?
A Quickly
B Gradually
C Aggressively
D Rapidly
Answer:
D is the answer I think (0 w 0 )
Explanation:
The glaciers move gradually. Hence, option (C) is correct.
What is glacier?A glacier is a long-lasting mass of heavy ice that is perpetually moving. When the ablation of snow is greater than the accumulation over a long period of time, frequently centuries, a glacier forms.
As it slowly flows and deforms under forces brought on by its weight, it gains distinctive features like crevasses and seracs. Cirques, moraines, and fjords are the result of the erosion of rock and debris from its substrate as it travels.
The considerably thinner sea ice and lake ice that form on the surface of bodies of water are not the same as glaciers, which form only on land and may flow into water bodies.
The huge ice sheets, commonly referred to as "continental glaciers," in the polar areas of the planet contain 99 percent of the planet's glacial ice.
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Solve each of the following problems to 3 sig figs and correct Sl units, showing all work.
1. A cart with a mass of 45.0 kg is being pulled to the right with a force of 250 N giving it an
acceleration of 1.30 m/s2. The wheels of the cart are locked and the cart must be dragged.
a) Draw a free body diagram of the cart.
b) Calculate the net force acting on the cart.
c) Create a force table and fill it in.
d) Find the coefficient of kinetic friction.
Answer:
every number to 3 sf = 1) 45.0 2) 250 3) 1.30
Explanation:
your welcome :)
PLEASEEEEEE HELPPPPPPP
Define resistance and discuss how it affects current.
Answer:
Resistance is the opposing of the flow of current through a conductor.
treatment for PTSD campus be described A.multumodal multi-phasic b. atheoretical C. stress inclusion D.eclectic
Answer:
Stress inclusion
Explanation:
A type of cognitive behavior therapy for PTSD is pressure inoculation preparation. CBT is a widely used form of antidepressant that can make you know wrong and/or pessimistic emotions that have influenced your actions and alter them.
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
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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
A ball rolling on the floor eventually comes to a stop. Which statement best describes this situation?
A) The energy on the macroscopic scale and the energy on the molecular scale both decreased.
B) The energy on the macroscopic scale increased, and the energy on the molecular scale decreased.
C) The energy on the macroscopic scale decreased, and the energy on the molecular scale increased.
D) The energy on the macroscopic scale and the energy on the molecular scale both remained constant.
If you have the rest please put them down :)
Answer:
Explanation:
C. The energy on the macroscopic scale decreased, and the energy on the molecular scale increased.
A. friction converts some energy on the macroscopic scale to energy on the molecular scale.
B. The mechanical energy of the system decreased because of friction.
B. The mechanical energy of the system at the beginning of the experiment is equal to the mechanical energy of the system at the end of the experiment.
B. the mechanical energy of the system decreased due to the transfer of energy on a macroscopic scale to energy on a molecular scale.
The ball rolling on the floor eventually comes to a stop which means that
the energy on the macroscopic scale decreased, and the energy on the
molecular scale increased.
What is Macroscopic scale?This is a scale in which an object can be seen with the eyes. When
an object is in motion, the ball rolls away and becomes less visible hence a
decrease in the macroscopic scale.
During motion, friction converts some macroscopic energy into molecular
scale which is why there was an increase in it.
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Amy has a mass of 50 kg, and she is riding a skateboard traveling 10 meters per second. What is her momentum?
5 kg·m/s
10 kg·m/s
50 kg·m/s
500 kg·m/s
Answer:
[tex]500 \: \mathrm{kg} \cdot \mathrm{m/s}[/tex]
Explanation:
The momentum of an object is given as [tex]p=mv[/tex]. Since Amy has a mass of 50 kg and is travelling 10 m/s, her momentum is [tex]p=mv=50\cdot 10 =\fbox{$500\: \mathrm{kg\cdot m/s}$}[/tex].
Answer:
500
Explanation:
A bottle rocket is fired off and has an acceleration of 14.5 m/s2 for the 2.25s until it burns out. If it starts at rest, what distance does it cover?
Answer:
S = 16.3125m
Explanation:
Given the following data;
Acceleration, a = 14.5m/s²
Time, t = 2.25secs
Since the bottle rocket starts from rest, its initial velocity is 0m/s.
To find the distance S, we would use the second equation of motion.
S = ut + ½at²
Substituting into the equation, we have
S = 0(2.25) + ½*14.5*2.25
S = 0 + 7.25*2.25
S = 16.3125m
Therefore, the bottle rocket covered a distance of 16.3125 meters.
How long ago did most Middle Eastern countries gain their independence?
A.
10-20 years ago
B.
50-100 years ago
C.
200-300 years ago
D.
400-500 years ago
Please select the best answer from the choices provided
A
B
C
D
Answer:
the correct answer is B
Explanation:
Answer:
.B
Explanation:
This is the answer on edge 2021
Have a good day!
Do it in order.
from smallest to largest
Answer:
The earth, The sun, the solar system and the milky way.
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
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
A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is required from the person in order to complete this task in 9 seconds?
__ Watts
Answer:
Power is 1061.67W
Explanation:
Power=force×distance/time
Power=65×9.8×15/9 assuming gravity=9.8m/s²
Power=3185/3=1061.67W
Which action will leave the dump trucks inertia unchanged?? PLEASE ANSWER FAST!!!
A. add gas
B. increase force applied to engine
Answer:
B.
Explanation:
David is driving a steady 28.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.90 m/s^2 at the instant when David passes.
Required:
a. How far does Tina drive before passing David?
b. What is her speed as she passes him?
Answer:
Explanation:
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1
For David u = 28.0 m/s where 'a' is set to nought
S = ut
S = 28t.......2
For Tina consider equation 1
Where acceleration = 2.90m/s^2 and u is set at nought
S = 1/2×2.90 m/s×t^2.......3
Equate 2 and 3
28t = 1.45t^2
Divide through by t
28 = 1.45t
t = 28/1.45
t = 19.31seconds
Now put the value of t into equation 3
S = 1/2×2.90 m/s×t^2.......3
= 1.45×20×20
= 580m
Tina must have driven 580meters before passing David
Considering the equation of linear motion : V^2 = U^2+2as
Where u is set at nought
V^2 = 2as
V^2 = 2×2.9×580
V^2 = 3364
V = √3364
V = 58m/s
Her speed will be 58m/s
(a) Tina should drive for 580 m, before passing the David.
(b) The speed of Tina during her passage through the David is 58 m/s.
Given data:
The initial velocity of the David is, u = 28.0 m/s.
The magnitude of acceleration is, [tex]a = 2.90 \;\rm m/s^{2}[/tex].
(a)
We can use the second kinematic equations of motion to obtain the distance covered by Tina, before passing the David. As per the second kinematic equation of motion,
[tex]s= u't + \dfrac{1}{2}at^{2}[/tex]
Here, u' is the initial speed of Tina and t is the time interval. Then,
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion as,
S = ut + 1/2at²...............................................................(1)
Also,
S = ut
S = 28t ...........................................................................(2)
For Tina consider equation 1
S = 1/2×2.90t²................................................................(3)
Equate 2 and 3
28t = 1.45t²
28 = 1.45t
t = 28/1.45
t = 19.31 seconds
Now put the value of t into equation (3)
S = 1/2×2.90 t².
= 1.45×20×20
= 580m
Thus, we can conclude that Tina should drive for 580 m, before passing the David.
(b)
Now, using the third kinematic equation of motion to obtain the speed of Tina during her passage through David as,
v² = u²+2as
Solving as,
v² = 28.0² + 2(2.90)(580)
v = √3364
v = 58m/s
Thus, we can conclude that the speed of Tina during her passage through the David is 58 m/s.
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