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A Person is carrying 6kg in one hand and 5kg in another.Calculate the resultant force applied by him
A student lifts a backpack straight up with a force of 53.5 N for a distance of 0.65 m. How much work is done on the backpack?
Happy Christmas Eve!!
Answer:
W = 34.775J
Explanation:
W = F × d
W = 53.5N × 0.65m
W = 34.775J
W = work = unknown
f = force = 53.5N
d = distance = 0.65m
Lower chamber of Congress has how many members?
Answer:435 Representatives
Explanation:The lower chamber of Congress, in which the number of representatives per state is determined by the state's population, with 435 Representatives total. Members of the House of Representatives serve two-year terms, so they are up for reelection every two years.
A 2-kg object is pulled to the left with a force of 30 N and to the right with a force of 9 N. What is the acceleration of the object?
An object that weighs 75 N is pulled horizontally to the right with a force of 50 N. The force of friction on this object is 30 N to the left. What is the acceleration of the object?
The force on the object is 50 N to the right and 30 N to the left.
⇒ The net force on the object is, =50−30=20 N to the right.
The weight of the object is, =75 N.
⇒ The mass of the object is ==759.8 kg.
⇒ The acceleration of the object is ==20759.8=2.613 m/s 2.
The acceleration of the object is 10 [tex]m/s^2[/tex].
What is Acceleration?Acceleration is defined as the rate of change of velocity of an object with respect to time which are vector quantities i.e. both magnitude and direction. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.
It can be expressed as:
a= v/t
where, a is the acceleration
v is the velocity
t is the time taken
It can also be expressed as F=ma
where, F= force applied
m= mass of the object
So, for above given information
The force on the object is 9N to the right and 30N to the left. The net force on the object is 21N to the left
The mass of the object is 2kg
So, a= 21N/ 2 kg = 10.5 [tex]m/s^2[/tex]
Thus, the acceleration of the object is 10 [tex]m/s^2[/tex].
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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
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
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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
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 gradesChoose all the answers that apply.
Light _____.
A. cannot travel through a vacuum
B. slows down when traveling through denser substances
C. speeds up when traveling through more dense substances
D. travels in transverse waves
E. can be absorbed and transformed into heat
Answer:
B
Explanation:
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.
Write any three facts in the support of molecular theory
Answer:
Every gas is made up of very small particles called molecules. Molecules of a gas are separated from each other by large distances so that volume is negligible as compared to total volume of gas. The force of gravitation on the molecules is also negligible.
Hope that helps. x
:
please someone help me understand this question
[tex]\\ \sf\longmapsto F=\dfrac{Gm_1m_2}{r^2}[/tex]
m1 and m2 has same units let both be m[tex]\\ \sf\longmapsto F=\dfrac{Gm^2}{r^2}[/tex]
[tex]\\ \sf\longmapsto G=\dfrac{Fr^2}{m^2}[/tex]
SI units:-
F=Newton (N)r=metre(m)mass=m=kg[tex]\\ \sf\longmapsto G=\dfrac{Nm^2}{kg^2}[/tex]
Option C is correct
What is the average speed of the car ??
Answer:
90 km / hr
Explanation:
V = S / t = 60 * 3.6 km / 2.4 hr = 90 km / hr
Hope that helps <3
What is the smallest value of the force F such that the 8.0 kg block will not slide down the wall? The coefficient of static friction between the block and the wall is 0.4.
Select one:
a. 10N
b. 100N
c. 200N
d. 98N
The smallest value of the force that will make the block not to slide down is 10 N.
We'll begin by calculating the normal reaction. This can be obtained as follow:
Mass (m) = 8 KgAcceleration due to gravity (g) = 10 m/s²Normal reaction (N) =?N = mg
N = 8 × 10
N = 80 N
Finally, we shall determine the frictional force.
Coefficient of friction (μ) = 0.4Normal reaction (N) = 80 NFrictional Force (F) =?F = μN
F = 0.4 × 80
F = 32 N
Since the frictional force is 32 N, therefore, a force lesser than the frictional force will make the block not to slide down.
From the options given above, only option A has a force that is lesser than the frictional force.
Therefore, the correct answer to the question is Option A. 10 N
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Please 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
4) What is the kinetic energy of o 5 kg object moving at a velocity of 25 m/s?
Ek = 1/2 mv^2
= 1/2 × 5 × (25)^2
= 1562.5 J
Describe the water cycle. Be sure to include the following terms in your response: evaporation, condensation, and precipitation.your respone:
Evaporation, or turning liquid to a gas, is first. Then it goes into the atmosphere and codensates, or turns back to a liquid, these water droplets form into clouds and come down as precipitation, rain, sleet, snow etc. Then the water cycle starts again
A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of 350 m/s [E] The dart comes to rest in 0.0050 s.
a. What is the acceleration of the dart?
b. How far does the dart penetrate into the backstop?
Answer:u=350m/sec
v=0m/sec
t=0.005sec
a=v-u/t
=0-350/0.005
= -350/0.005
= -350×1000/5
= -70×1000
= -70000 m/sec^2
Explanation: i think this will help you thankyou
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.
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The internal energy of a material is determined by _____.
its temperature
the kinetic energy of its molecules
the potential energy of its molecules
the combined amount of kinetic and potential energy of its molecules
The internal energy of a system is identified with the random, disordered motion of molecules; the total (internal) energy in a system includes potential and kinetic energy.
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.
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The planets never travel in a straight line as they orbit the Sun. According to Newton's second law of motion, this must mean that _________. the planets are always accelerating a force is acting on the planets the planets have angular momentum the planets will eventually fall into the Sun
Explanation:
The planets never travel in a straight line as they orbit the Sun. According to Newton's second law of motion, this must mean that: Your Answer: The planets have angular momentum.'
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According to Newton's second law of motion, this must mean that force is acting on the planets. The correct option is (b).
As per Newton's second law of motion, an object's acceleration is inversely related to its mass and directly proportional to the net force exerted on it.
The curving courses of the planets in their solar system show that they are constantly changing their direction, as it causes them to accelerate.
The planets are being accelerated in the direction of the Sun as a result of the Sun's gravitational pull on them. The planets continually speed up toward the Sun due to the centripetal force required to maintain them in their orbits, which is caused by the gravitational attraction between the Sun and the planets. As a result, the planets orbit the Sun in elliptical patterns rather than in a straight line.
Hence, According to Newton's second law of motion, this must mean that force is acting on the planets. The correct option is (b).
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The complete question is:
The planets never travel in a straight line as they orbit the Sun. According to Newton's second law of motion, this must mean that _________.
(a) the planets are always accelerating
(b) a force is acting on the planets
(c) the planets have angular momentum
(d) the planets will eventually fall into the Sun
An astronomical unit (A.U.) is 1 point A) a term for defining the luminosity of a star B) the average distance from the Earth to the sun C) the average distance of any given planet to the sun D) equal to a light year
Answer:
B) the average distance from the Earth to the Sun
Explanation:
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
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.
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what do i do on this rubixs cube
Answer:
You solve it
Explanation:
Answer:
you just spin it and spin it and spin it and spin it
and spin it
and spin it
and spin it
and spin it
until you get mad then throw it at the wall
Explanation:
ya then you look up a vid that is called rubix cube meme and there should be a pink animal its funny you should watch it!!!!!!!!!
How charged objects can affect other objects without touching them. ?
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
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.
Give two properties of an object that can be affected by a
physical change.
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center, as shown above. The coin doesn’t slip and the time it takes for the coin to make a complete revolution is 1.5 s.
A)The figure below shows the disk and coin as viewed from above. Draw and label vectors on the figure below to show instantaneous acceleration and linear velocity vectors for the pin when it is at the position shown below.
B)Determine the linear speed of the coin
C)The rate of rotation of the disk is gradually increased. The coefficient of stats if friction between the coin and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.
D)If the experiment in part c were repeated with a second, identical coin glued to the top of the first coin, how would this affect the answer to part c? Explain.
A) Figure attached below
B) The linear speed of the coin = 0.59 m/s
C) Linear speed as coin begins to slip = 0.83 m/s
D) The tangential speed will remain the same as seen in part C
Given data :
mass of coin = 0.0050 kg
Distance of coin from the center of disk = 0.14 m
Time to make a complete revolution = 1.5 s
A) Diagram showing the vectors on the figure is attached below
B) Determine the Linear speed of the coinLinear speed of coin = 2 * π * ( 0.14 ) / 1.5
= 0.59 m/s
C) Determine the linear speed of the coin when it just begins to slipgiven that: friction between coin and disk = 0.50
Friction becomes maximum when coin begins to slip
Maximum frictional force (Fmax) = uV
where V = mg
∴ Fmax = u*mg ---- ( 1 )centripetal force = [tex]\frac{mv^{2} }{r}[/tex] ---- ( 2 )Equating equations ( 1 ) and ( 2 ) to determine the linear speed ( v )
v² = u*r*g
∴ v = √(u*r*g ) = √( 0.5 * 0.14 * 9.8 )
= 0.83 m/s
D) If the experiment is repeated with a second coin glued to the top of the first coin the tangential speed will remain the same
Hence we can conclude that The linear speed of the coin = 0.59 m/s Linear speed as coin begins to slip = 0.83 m/s , The tangential speed will remain the same as seen in part C
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