[tex] = \sf \frac{1}{2} m {v}^{2} [/tex]
Kinetic energy of the 1st bullet [tex] \sf = \frac{1}{2} \times m1 \times {(v1)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(40)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 1600J \\ \sf = 2.4J[/tex]Kinetic energy of the 2nd bullet [tex] \sf = \frac{1}{2} \times m2 \times {(v2)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(80)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 6400J \\ \sf = 9.6J[/tex] So, the 2nd bullet which has greater velocity has more kinetic energy.Therefore, the ratio of their kinetic energies[tex] \sf = \frac{2.4J}{9.6J} \\ \sf= \frac{24}{96} = \frac{1}{4} \\ \sf = 1 : 4[/tex]
Answer:
The kinetic energies of the bullets are 2.4 J and 9.6 J.
The bullet having greater velocity has more kinetic energy.
The ratio of their kinetic energies is 1 : 4.
Hope you could get an idea from here
Doubt clarification - use comment section.
The kinetic energy of the two bullets are 2.4 J and 9.6 J respectively.
The ratio of the kinetic energy of the bullets is 1:4.
What is kinetic energy?The kinetic energy of an object is the energy possessed by the object due to its motion.
The kinetic energy of the two bullets is calculated as follows;
[tex]K.E_1 = \frac{1}{2} \times 0.003 \times 40^2 = 2.4 \ J\\\\K.E_2 = \frac{1}{2} \times 0.003 \times 80^2 = 9.6 \ J[/tex]
Ratio of the kinetic energy of the bullets is calculated as follows;
[tex]K.E_1 : K.E_2 = 2.4: 9.6 \ = \ 1: 4[/tex]
Learn more about kinetic energy here: https://brainly.com/question/25959744
The relationship between Acceleration and Mass is described as:
O a. Equal
O b. Directly Proportional
O c. Inversely Proportional
O d. Opposite
O e. None of these
Explanation:
According to Newton's second law
[tex]\\ \sf{:}\dashrightarrow Force=Mass\times Acceleration [/tex]
[tex]\\ \sf{:}\dashrightarrow Mass=\dfrac{Force}{Acceleration}[/tex]
[tex]\\ \sf{:}\dashrightarrow Mass\propto \dfrac{1}{Acceleration}[/tex]
Option C
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
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:
If astronomers discovered a new planet and found its period of rotation around the Sun to be 105 years, how long would its semi-major axis length be as it orbited the Sun in AU?
From Kepler's third law, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C
Given that an astronomers discovered a new planet and found its period of rotation around the Sun to be 105 years.
According to Kepler's third law,
[tex]T^{2} \alpha r^{3}[/tex]
Where
T = Period ( in earth years) = time to complete one orbit
r = Length of the semi major axis in Astronomical unit.
[tex]T^{2}[/tex] = [tex]\frac{4\pi ^{2} }{GM} * r^{3}[/tex]
convert years to seconds
105 x 365 day x 24 hours x 3600 s
T = 3311280000 seconds
Mass of the sun M = 1.989 × 10^30 kg
G = 6.67 x [tex]10^{-11}[/tex]N m^2/kg^2
Substitute all the parameters into the formula
[tex]T^{2}[/tex] = 1.096 x [tex]10^{19}[/tex] = [tex]\frac{4\pi ^{2} }{6.67 * 10^{-11} * 1.989 * 10^{30} } * r^{3}[/tex]
1.096 x [tex]10^{19}[/tex] = 2.976 x [tex]10^{-19}[/tex] [tex]r^{3}[/tex]
[tex]r^{3}[/tex] = 1.096 x [tex]10^{19}[/tex] / 2.976 x [tex]10^{-19}[/tex]
[tex]r^{3}[/tex] = 3.68 x [tex]10^{37}[/tex]
r = [tex]\sqrt[3]{3.68 * 10^{37} }[/tex]
r = 3.33 x [tex]10^{12}[/tex] m
1 AU = 1.5 x [tex]10^{11}[/tex] m
r = 3.33 x [tex]10^{12}[/tex] / 1.5 x [tex]10^{11}[/tex]
r = 22.18 AU
Therefore, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C
Learn more about Kepler's laws here: https://brainly.com/question/4639131
Answer:
C. 22.3 AU
Explanation:
Not only is the above an unnecessarily complicated answer, it's not even fully correct, and definitely not what they want you to do.
T^2 = s^3, where T = orbital period and s = semi-major axis length.
Substitute T and you get 105^2 = s^3. Solve for s.
11025 = s^3
3√11025 = s
22.25663649 = s
Therefore, the answer is C. 22.3 AU
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.
If an input force of 202N is applied to the handles of the wheelbarrow in the sample problem how large is the output force that just lifts the load?
For this case, the relationship between the two forces is given by:
F1 = nF2
Where,
F1: output strengthF2: input forcen: mechanical advantageThen, replacing the values we have:
F1 = (2.2)(202)
Having the calculations we have:
F1 = 444.4N
Answer: The output force that only lifts the load is F1 = 444.4N.
The weight of a person in an elevator at rest = 500 N. Acceleration due to gravity is 9.8 m/s2. When lift accelerated, the tension force is 750 N. What is the acceleration of lift.
Answer:If the elevator accelerated downward then the tension force smallest then 500 N. Otherwise, if the elevator accelerated upward then the tension force larger then 500 N.
The tension force = 750 N because the elevator accelerated upward. Force acts upward has plus sign and force acts downward has minus sign.
T – w = m a
750 – 500 = 50 a
250 = 50 a
a = 250 / 50
a = 5.0 m s–2
Explanation:
Please help with this ski question to find Delta y
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!
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
A boxer throws a punch with a force of 1,400 N that lasts 0.02 s. What is the impulse of this punch? (1 point) 28 kg⋅m/s 28 kilograms times meters per second 280 kg⋅m/s 280 kilograms times meters per second 70,000 kg⋅m/s 70,000 kilograms times meters per second 7,000 kg⋅m/s
The impulse of the boxers punch is 28 kgm/s.
The given parameters;
applied force by the boxer, F = 1400 Ntime of force action, t = 0.02 sThe impulse of the boxers punch is calculated as follows;
[tex]J = Ft[/tex]
where;
F is the applied force (N)t is the time of force action (s)The magnitude of the impulse is calculated as follows;
[tex]J = 1400 \times 0.02 \\\\J = 28 \ kg m/s[/tex]
Thus, the impulse of the boxers punch is 28 kgm/s.
Learn more about impulse here: https://brainly.com/question/25700778
An eagle flying at a constant 120 km/h and has kinetic energy of 2,800 J. What is the mass of the eagle?
Answer:
The mass of the eagle is about 5 Kg
Explanation:
1/2 M= Ke/V^2
120 km/h = 33.3333m/s
1/2 M = 2,800/33.3333^2
1/2 M = 2,800/ 1111.10888889
1/2 M = 2.52000504001
(2) 1/2 M = (2) 2.52000504h001
= M = 5.04001008002
About 5 Kilograms
An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runnera''s average speed? (Round your answer to the nearest tenth of a meter per second. ) 0. 9 m/s 1. 2 m/s 8. 7 m/s 10. 1 m/s.
The average speed of the runner is 8.7m/s
Hence, Option C) 8.7m/s is the correct answer.
Given the data in the question;
Distance covered by runner; [tex]d = 200m[/tex]Time taken; [tex]t = 23s[/tex]Average speed; [tex]s =\ ?[/tex]
Speed is simply the rate at which a particle covers a given distance. It is expressed as:
Speed = Distance / Time
[tex]s = \frac{d}{t}[/tex]
We substitute our given distance into the equation;
[tex]s = \frac{200m}{23s} \\\\s = 8.7m/s[/tex]
The average speed of the runner is 8.7m/s
Hence, Option C) 8.7m/s is the correct answer.
Learn more: https://brainly.com/question/21503615
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
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
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
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
Helppppp 30pointsssssssssssssssssss
Answer:
so sorry i don't know the answer so sorry
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
Assertion: In domestic electric circuits, the metallic body is connected to the earth wire, which
provides a low-resistance conducting path for the current.
Reason: It ensures that any leakage of current to the metallic body of the appliance keeps its
potential to that of the earth, and the user may not get a severe electric shock. pls fast
Answer:
The answare is it ensures that any leakage of current to the metallic body of the appliance keeps its potential to that of the earth, and the user may not get a severe electric shock.
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
[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 :)
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.
Which two changes would decrease the electric force between two charged
particles?
- A. Decrease the charge of one of the particles.
B. Increase the charge of both particles.
C. Increase the charge of one of the particles.
D. Decrease the distance between the particles.
E. Increase the distance between the particles.
PLEASE HELP I KEEP FAILING!!!!
Answer:
it is b and e
Explanation:
if u look at the words twice you will notice that b and e are both saying the same meanings just in diff rent words way u need to look close on things like that and u will get passing gradesPlease I need help with this ❤️ ?
Answer:
Compound 'A' C5H12 does not react with phenyl hydrazine. Oxidation of 'A' with K2Cr2o7,/H" gives B' (c5H10o). Compound 'B' reacts with phenyl hydrazine but does not give Tollen's test. The
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
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
A strong weightless rope has a mass, m, hanging from the middle of it. The tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees. How much mass is hanging from the rope?
By using Lami's theorem, Mass m = 1.75 kg approximately
Given that a strong weightless rope has a mass, m, hanging from the middle of it. If the tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees to the horizontal.
By using Lami's theorem, we can get how much mass is hanging from the rope.
Let the angle between the rope = α = 180 - 40
α = 140 degrees
The angle between one of the rope and mass = β = 20 + 90
β = 110 degrees
The angle between the mass and the other rope = γ = 360 - (140 + 110)
γ = 360 - 250
γ = 110 degrees
W/ sinα = T/ sinβ = T/sinγ
W/ sinα = T/ sinβ
Substitute all the necessary parameters
W/sin140 = 25/sin 110
W / 0.643 = 25 / 0.939
W = 17.1 N
Weight W = mg
17.1 = 9.8m
mass m = 17.1/9.8
Mass m = 1.7455 kg
Mass m = 1.75 kg approximately
Therefore, 1.75 kg mass is hanging from the rope.
Learn more about resolution of forces here: https://brainly.com/question/1858958
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.
ONLY ANSWER IF YOU KNOW FOR SURE PLEASE :)
answer all three please, it should not take long
Answer:
Question one is b, Question two is b, and question three is b, im pretty positive that what it is