Answer:
36,000 J
Explanation:
Work is equal to Force times Distance:
W = Fd
W = 60N*600m
36,000 = 60N*600m
7. A drag car takes off at the green light heading in a straight line. The car goes
from a complete stop to 45m/s in 9 seconds. What is its acceleration?
Answer:
405
Explanation:
If the car stops at 45m/s in 9 seconds then it took the car 405 meters to stop
Which car will experience a greater KE, a car traveling at 16 m/s or a half as massive car traveling at 32 m/s
Answer:
The car that is hald as massive
Explanation:
We can use the eqation for kinetic energy to solve this problem
KE = 1/2mv^2 (where m is mass and v is velocity/speed)
Lets give the original car a mass of 100kg and use the equation
KE = 1/2(100kg)(16m/s)^2 = 12800J
and now lets find kinetic energy of the half as massive car
KE2 = 1/2(50kg)(32m/s)^2 = 25600J
From this, we find that the hald as massive car will experience a greater kinetic energy while travling at double the velocity
What would be the difference in your welght if you were four times farther from the center of Earth than you are now?
you would die.
the core of the earth is as hot as the sun surface so you would be dead
How long does it take a car traveling at 45km/h to travel 100.0 m?
Answer:
8 seconds
Explanation:
Since the carspeed is in km/h, we need equal units, so we will make 100.0m 0.1000km.
Then we need to find how long it takes the car to travel 0.1km
We can use the formula distance=speed * time and get
0.1=45 * time
Therefore we get .002222... hours
Multiplying this by 3600 (to get seconds, 60x60), we get 8 seconds
Answer:
8m/s
Explanation:
speed = 45km/h
Distance = 100m
we have to find time = ?
Formula for speed is = Distance/ Time
Here Distance is given in 'm' so we need to convert speed value in 'm'
So to convert km/h in m formula is divide 45km/h by 3.6 then the km/h value gets converted in m so now the value is 12.5 m/s
So now,
speed = Distance/Time
we have to find Time
Then,
Time = Distance/ Speed
= 100/12.5
= 8m/s
Initial Velocity is 27.5 m/s. Time is 42 seconds. Final Velocity is 4.5 m/s. Solve for acceleration.
Answer:
-0.5476
Explanation:
a=(Vf-Vi)/t
a=(4.5-27.5)/42
a=-0.547619047
acceleration≈-0.5476
David delivers meals to elderly people once a week. He uses a cart to move the meals. The cart has four smooth wheels. Which type of friction acts between the cartwheels and the sidewalk?
Answer: rolling friction
Explanation: I think it is the answer
HELP ASAP WHOEVER ANSWERS FIRST WITH BE BRAINIEST PLEASE HELP!
The answer is : A₃ = A₁ - A₂ = 3.0 - 2.0 = 1.0A
Ok done. Thank to me :>
[tex] \huge \rm༆ question ༄[/tex]
Calculus proof of second equation of motion ~
Newton's second equation of motion :-
S=ut+1/2at^2 [where, u is the initial velocity, a is the acceleration and t is the time interval]
This Equation simply finds a relation between distance travelled by a particle (classically) under uniform acceleration.
So let's see what pieces of information (bundles of equations) do we have with us, initially.
We have, a very primary equation with us,
dS/dt = v… (I)
(Considering motion in a straight line only)
And we also have the equation
dv/dt = a…(II)
Simply replacing the v in eqn (II) by eqn (I), we find
d2S/dt^2 = a…(III)
This is what we need to solve. It's easy.
You know,
d2S/dt^2 = d/dt(dS/dt) = a
⟹ dS/dt = ∫adt = at+c1
Since, dS/dt is the velocity of the particle,
Therefore, at t = 0, dS/dt|t = 0 = u
⟹ u = a∗0 + c1 = c1
⟹ c1 = u
Therefore, dS/dt = u + at
Thus, S = ∫(udt + atdt)
⟹ S = ut + 1/2at^2 +c^2
If say, the particle is already having a displacement S0 the moment you start measuring it's motion. Then, at t = 0, S = S0
This makes S = S0 +ut + 1/2at^2
Since, in most of the practical cases, we start measuring a motion when the particle starts displacing (i.e., when S0=0 ),
We get
S = ut + 1/2at^2
Hope it helps :)
Please help with this ski question to find Delta y
Using a balance, determine the mass of the ball and basket of the ballistic pendulum. Fire the ballistic pendulum
and measure the maximum height the pendulum reaches. Use the conservation of energy (initial KE + initial GPE =
final KE + final GPE; KE = 12mv2; GPE = mgh) to calculate the initial speed of the pendulum after the ball has
collided with the pendulum basket. Next, use the conservation of momentum (total initial p = total final p; p = mv)
to determine the initial speed of the ball before it collided with the pendulum basket.
=
=
The conservation of energy and moment allows to find the results for the velocity of the pendulum and the bullet in the system are:
The velocity of the pendulum (bullet + basket) is: v₀ = 1.98 m The speed of the bullet, depending on the mass are:m (kg) v (m / s)
0.05 41.58
0.10 21.78
0.15 15.18
0.20 11.88
0.25 9.90
The conservation of mechanical energy is one of the most important principles of physics, it establishes that if there is no friction force, the energy is constant at all points. Mechanical energy is the sum of the kinetic energy plus the potential energies.
In the attachment we can see a diagram of the system, let's start by finding the speed of the pendulum when it leaves, for this we use the conservation of energy.
Starting point. In the lowest part of the movement.
Em₀ = K = ½ (m + M) v₀²
Final point. At the top of the movement.
Em_f = U = (m + M) g y
Energy is conserved because there is no friction.
Em₀ = Em_f
½ (m + M) v₀² = (m + M) g y
[tex]v_o = \sqrt{2gy}[/tex]
They indicate a table with several measurements of the masses and the period, let's use the relationship of the simple harmonic motion.
y = y₀ cos wt
The period and the angular velocity are related.
w = 2π / T
we substitute
y = y₀ cos ( [tex]2 \pi \frac{t}{T}[/tex] )
Let's analyze how long it takes to reach the point of maximum height, the period is the time of a complete oscillation, therefore from the lowest point to the highest point we have ¼ of oscillation, consequently the time to the highest point.
t = T / 4
y = y₀ cos ( [tex]2 \pi 4[/tex])
y = y₀
Therefore the point of maximum amplitude coincides with the maximum height and must be average by the student, suppose that the height is
y₀ = 20 cm = 0.20 m
Let's calculate the initial velocity.
v₀ = [tex]\sqrt{2 \ 9.8 \ 0.20 }[/tex]
v₀ = 1.98 m / s
They ask for the speed of the bullet before striking the basket of mass
M = 1 Kg.
The conservation of the momentum that for an isolated system the momentum is constant in all the instants. Let's form the system by the bullet and the basket.
Initial instant. Before the crash
p₀ = m v
Final moment. Right after the crash.
P_f = (m + M) v₀
The momentum is conserved because the system is isolated.
p₀ = p_f
m v = (m + M) v₀
v = m + M / m v₀
they have tabulated various mass for the bullet, we calculate the speed of each bullet.
m = 0.05 kg
v = [tex]\frac{0.05+1}{0.05} \ 1.98[/tex]
v = 41.58 m / s
m = 0.10 kg
v = [tex]\frac{0.10+1}{0.10} \ 1.98[/tex]
v = 21.78 m / s
m = 0.15 kg
v = [tex]\frac{0.15+1}{0.15y}[/tex]
v = 15.18 m / s
m = 0.20 kg
v = 11.88 m / s
m = 0.25 kg
v = 9.9 m / s
In conclusion with the conservation of energy and the momentum we can find the results for the speed of the pendulum and the bullet in the system are:
The velocity of the pendulum (bullet + basket) is: v₀ = 1.98 m The speed of the bullet, depending on the mass are:m (kg) v (m / s)
0.05 41.58
0.10 21.78
0.15 15.18
0.20 11.88
0.25 9.90
Learn more about conservation of energy and momentum here: brainly.com/question/25849204
Physicists call any change in energy an impulse true or false?
True or false organisms only compete with their own species?
Answer: yes
Explanation:
Animals of different species typically compete with each other only for food, water and shelter. But they often compete with members of their own species for mates and territory as well.
a 1.0 kg ball falls from rest a distance of 15 m. what was its change in potential energy?
Answer:
147 J
Explanation:
PE = mgh
PE = (1.0)(9.8)(15)
PE = 147 joules
please help quick i’ll mark you brainly
The equation below shows how to calculate work done by an applied force. W equals the work done in joules (J), F equals the applied force in newtons (N), and d equals the distance in meters (m) over which the force is applied.
W = F × d
This equation can be arranged to calculate the applied force instead:
F = W ÷ d
A scientist was performing some experiments to test the designs of some simple machines. One of her experiments involved a pulley. During the experiment, a total of 2,000 J of work was done to lift a crate 5 m straight up off the floor.
Ignoring the effects of friction, determine the applied force needed to lift the crate.
A.
2,005 N
B.
400 N
C.
10,000 N
D.
1,995 N
Answer:
The solutions to this exercise is B. 400N
A loaded truck collides with a car causing huge damage to the car. Which of the following is true about the collision? *
A. The force on the truck is greater than the force on the car
B. The force on the car is greater than the force on the truck
C. The force on the truck is the same in magnitude as the force on the car
D. During the collision the truck makes greater displacement than the car
E. During the collision the truck has greater acceleration than the car
The force on the truck is the same in magnitude as the force on the car.
The impulse experienced by an object during collision is directly proportional to the applied force and time of collision of the objects.
J = Ft
where;
J is the impulse experienced by the objectt is the timeThe increase in the force applied to an object causes an increase in the impulse experienced by the object.
Also, according to Newton's third law of motion, the force exerted on the loaded truck and the car are equal in magnitude but opposite in direction.
Thus, we can conclude that, the force on the truck is the same in magnitude as the force on the car.
Learn more about Newton's third law of motion here: https://brainly.com/question/25998091
Can any1 tell if my answer is right
Answer:
The correct answer would be C. 5.0kg
Explanation:
The mass of an object never changes unless parts of the object are taken away. In other words, although the gravitational force is different on the moon then on the earth the mass of the object would remain the same.
How did Thomson's model get its name?
Answer:
thats actually a really good question. sadly i don't know the answer to it. but im sure that there are others who can!
Helppppp 30pointsssssssssssssssssss
Answer:
so sorry i don't know the answer so sorry
The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?(1 point)
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface. How does the velocity of Satellite A compare to the velocity of Satellite B?(1 point)
A) It depends on the masses of the satellites.
B) The velocity of Satellite A is greater than the velocity of Satellite B.
C) The velocity of Satellite B is equal to the velocity of Satellite A.
D) The velocity of Satellite A is less than the velocity of Satellite B.
What is the direction of the net force acting on a satellite as it orbits Earth at a constant speed?(1 point)
A) in the direction the satellite is moving
B) toward the center of Earth
C) away from the center of Earth
D) opposite the direction the satellite is moving
What happens to the gravitational force and orbital velocity of a satellite as the satellite transfers to an orbit that is closer to Earth?(1 point)
A) The gravitational force decreases and the velocity increases.
B) The gravitational force increases and the velocity increases.
C) The gravitational force decreases and the velocity decreases.
D) The gravitational force increases and the velocity decreases.
Newton's second law and universal gravitation law allow to find the results for questions about the motion of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
Newton's second law establishes a relationship between force, mass, and the acceleration of bodies.
F = m a
In the case of the satellite the force is given by the law of universal gravitation.
[tex]F = - G \frac{Mm}{r^2 }[/tex]
Where G is the constant of universal gravitation. M and m the mass of each object and r the distance between them.
In this case the satellite is in a circular orbit, therefore the acceleration is centripetal.
[tex]a = \frac{v^2}{r}[/tex]
We substitute.
[tex]G \frac{Mm}{r^2} = m \frac{v^2}{r}[/tex]
[tex]\frac{GM}{r} = v^2[/tex]
Let's analyze the answers to find the correct one.
1) They indicate the height of the space station r = 4.08 10⁵ m and ask the speed.
[tex]v= \sqrt{ \frac{6.67 \ 10 ^{-11} 5.9 \ 10^{24}}{4.08 \ 10^6 } }[/tex]
v = 9.82 10³ m / s
The correct answer is D.
2) Satellite A has an orbit of hₐ = 500 km and satellite b an orbit of
h_b = 800 km
The distance from the center of the earth to each satellite is:
rₐ = R + hₐ
r_b = R + h_b
rₐ = 6.37 106 + 500 10³ = 6.87 10⁶ m
r_b = 6.37 10⁶ + 800 10³ = 7.17 10⁶ m
Let's find the ratio of the speeds
[tex]\frac{v_a}{v_b} = \sqrt{ \frac{r_b}{r_a} } \\\frac{v_a}{v_b} = \sqrt{ \frac{7.17}{6.87} }[/tex]
[tex]\frac{v_a}{v_b}[/tex] = 1,022
we see that the speed of satellite a is slightly greater than the speed of satellite b.
Let's analyze the claims.
A) False. The speed does not depend on the mass of the satellites.
B) True. The velocity of a is slightly greater than the velocity of b.
C) False. The speed of a is greater.
D) False. The speed of a is greater.
3) As the orbit is circular, the force must be radial, that is, it points towards the center of the earth.
Let's analyze the claims.
A) False. The speed modulus does not change, therefore there is no acceleration in the direction of the satellite.
B) True. Aim for the center of the Earth, change the direction of the velocity.
C) False. Aim for the scepter of the earth.
D) false. The modulus of velocity is constant and the direction changes towards the center of the earth, therefore the force must go towards the center of the earth.
4) The force in the law of universal gravitation increased as the distance decreased.
When a satellite approaches the earth its speed must increase since the speed is proportional in inverse of the square root of the distance.
Let's examine the claims.
A) False. The attraction force increases.
B) True. You agree with the explanation.
C) False. The gravitational force increases.
D) False. Speed increases.
In conclusion, using Newton's second law and the universal law of gravitation we can find the results for the questions about the movement of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
Learn more about the law of universal gravitation and circular motion here: brainly.com/question/24851258
Answer:
1. 7.66 × 10^3 m/s
2. The velocity of Satellite A is greater than the velocity of Satellite B.
3. toward the center of Earth
4. The gravitational force increases and the velocity increases.
Explanation:
NEED HELP ASAP 40 POINTS L!!!! Jaden is interested in learning more about the basics of astrono ny and looking more closely at the sky all by himself. He wantS to get a basic starter telescope to help him with this endeavor, and his teacher SuggestS an organization that might be able to help him with this. What organization is Jaden's teacher likely referring to? The Little Astronomers Celestial Sightseeing Group The Beginning Astronomy Organization Astronomers Without BordersS
Question 5. One 1,5V battery runs a portable MP3 player that draws 5,7mA of current
for about six hours before it runs out.
a) What is the resistance If the portable MP3 player?
b) How much power is dissipated in the portable MP3 player?
c) Find the energy delivered by the portable MP3 player?
Why would poor clusters of galaxies be more likely to have irregular shapes then rich
clusters
These Milky Way companion galaxies are easily visible from dark locations in the Southern Hemisphere. Prime examples of erratic galaxies are the Large and Small Magellanic clouds (left and right, respectively).
What clusters of galaxies likely to have irregular shapes?In comparison to a rich cluster, the poor cluster typically has a slightly more erratic shape. A number of smaller galaxies orbit each major spiral. The Small and Large Magellanic clouds are the two most well-known examples of atypical galaxies.
When two galaxies collide, irregular galaxies frequently result. This unusual Cartwheel Galaxy was created when a tiny galaxy slid through the centre of a massive spiral galaxy.
Therefore, Rich clusters are other clusters that include hundreds to thousands of galaxies. A weak cluster can't cling to its members strongly because of its low bulk.
Learn more about galaxies here:
https://brainly.com/question/8628958
#SPJ2
It requires a 70.4 N force (parallel to the inclined plane) to pull a 5.86 kg box up a 58.1° inclined plane with a rope at a constant speed. (a) What is the coefficient of kinetic friction between the inclined plane and the box?
(b) If the rope were to break, what acceleration would the box experience as it slid down the ramp?
Answer:
0.667; 4.965
Explanation:
Look at the picture I attached for the force analysis.
a) The coefficient=Friction/Normal Force. Because it's at constant speed, the force of friction + mgsin58.1° (because it's on an inclined plane and has split forces) is equal to the applied force (70.4N). Normal force is not equal to weight force though, because the box is on an inclined plane; it's equal to mgcos58.1°.
b) If it were to break, then the box no longer has an applied force, and the direction of friction has changed to up the inclined plane. F=m/a, so acceleration = mgsin58.1°- Friction/mass
The magnitude of acceleration due to gravity is
Select one:
a. 9.8 m/s
b. 8.9 m/s?
c. 4.6 m/s?
d. 9.8 m/s2
Answer:
The magnitude of acceleration is 9.8 m/s
Explanation:
Its not always in m/s^2 but in m/s due to the unit of distance and time
an object has a kinetic energy of 14 j and a mass of 17 kg how fast is the object moving
[tex]\\ \sf\Rrightarrow kE=\dfrac{1}{2}mv^2[/tex]
[tex]\\ \sf\Rrightarrow 14=\dfrac{1}{2}17v^2[/tex]
[tex]\\ \sf\Rrightarrow 17v^2=28[/tex]
[tex]\\ \sf\Rrightarrow v^2=28/17[/tex]
[tex]\\ \sf\Rrightarrow v^2=1.64[/tex]
[tex]\\ \sf\Rrightarrow v\approx 0.13m/s[/tex]
How would you explain conduction?
Answer:
the meaning of conduction is the process by which heat or electricity is directly transmitted through a substance when there is a difference of temperature or of electrical potential between adjoining regions, without movement of the material.
Explanation:
Answer:
the process by which heat or electricity is directly transmitted through a substance when there is a difference of temperature or of electrical potential between adjoining regions, without movement of the material.
Explanation:
Put the waves in order from shortest to longest wavelength
Answer:
b, a, c
Explanation:
The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.
Which data set has the largest range?
O A. 1, 5, 2, 4, 3, 5
O B. 5, 6, 4, 5, 6, 4
O C. 8, 9, 1, 2, 3, 4
O D. 3, 9, 5, 7, 3, 6
Answer:
D 3, 9, 5, 7, 3, 6
Explanation:
range is the difference between the highest and the lowest value and 9 is the highest while 3 is the lowest, if u subtract them u will get 6. while for the first u will get 4 and the second u will get 2. so the answer is D
is atmosphere pressure at high altitudes is less than the pressure at ground, true or false
Answer:
The answer is true I guess!!!
This is because the more you go up to less air it is and the pressure also gets less!!!!!!!!
Explanation:
I HOPE THAT I WAS HELPFUL TO YOU!
10. What do you think the effect is of jet airplanes on global warming?
Answer:
Like other emissions resulting from fossil fuel combustion, aircraft engines produce gases, noise, and particulates, raising environmental concerns over their global effects and their effects on local air quality.[2]Jet airliners contribute to climate change by emitting carbon dioxide (CO2), the best understood greenhouse gas, and, with less scientific understanding, nitrogen oxides, contrails and particulates. Their radiative forcing is estimated at 1.3–1.4 that of CO2 alone, excluding induced cirrus cloud with a very low level of scientific understanding. In 2018, global commercial operations generated 2.4% of all CO2 emissions.
Between 1940 and 2018, aviation CO2 emissions grew from 0.7% to 2.65% of all CO2 emissions.[1]
Jet airliners have become 70% more fuel efficient between 1967 and 2007, and CO2 emissions per Revenue Ton-kilometer (RTK) in 2018 were 47% of those in 1990. In 2018, CO2 emissions averaged 88 grams of CO2 per revenue passenger per km. While the aviation industry is more fuel efficient, overall emissions have risen as the volume of air travel has increased. By 2020, aviation emissions were 70% higher than in 2005 and they could grow by 300% by 2050.
Aircraft noise pollution disrupts sleep, children's education and could increase cardiovascular risk. Airports can generate water pollution due to their extensive handling of jet fuel and deicing chemicals if not contained, contaminating nearby water bodies. Aviation activities emit ozone and ultrafine particles, both of which are health hazards. Piston engines used in general aviation burn Avgas, releasing toxic lead.
Aviation's environmental footprint can be reduced by better fuel economy in aircraft or Air Traffic Control and flight routes can be optimised to lower non-CO2 effects on climate from NO
x, particulates or contrails. Aviation biofuel, emissions trading and carbon offsetting, part of the ICAO's CORSIA, can lower CO2 emissions. Aviation usage can be lowered by short-haul flight bans, train connections, personal choices and aviation taxation and subsidies. Fuel-powered aircraft may be replaced by hybrid electric aircraft and electric aircraft or by hydrogen-powered aircraft.