A small car with mass of 1000 kg is moving eastward with velocity of 30m/s and big car with mass of 2000 kg is moving southward with velocity of 20 m/s.The magnitude of momentum after colliding is..
1. A small car with mass of 1000 kg is moving eastward with velocity of 30m/s and big car with mass of 2000 kg is moving southward with velocity of 20 m/s.The magnitude of momentum after colliding is..
Penjelasan:
dik : m1 = 1000 kg
v1 = 30m/s
m2 = 2000 kg
v2 = 20 m/s
dit: momentum atau p.... ?
jawab:
karna tidak ada diketahui koefisien dan cuma momentum setelah tabrakan maka menggunakan hukum kekekalan momentum sehingga persamaan seperti ini
p = p'
(m1 x v1) + (m2 x v2) = p'
p' = (1000 x 30) + (2000 x -20)
p' = 30.000 - 40.000
p' = -10.000
jadi besarnya momentum adalah -10.000 atau 10.000... mohon maaf kalau ada yang salah
2. A particle moves to the right along the x-axis. After t second, the particle position at X1 = 3 meters and after moving 4 seconds, the particle position at x2 = 12 meters. What is the average speed of the moving particles?
Jawaban:
3 m/s
Penjelasan:
S=12
t=4
v=S/t
v= 12/4
v=3 m/s
3. the origin of surabaya comes from
Jawaban:
di Spain ini ya
Penjelasan:
aku bingung maaf ya
tolong jangan di hapus aku udah trauma ini cewe soalnya ya pliiis
Jawaban:
kami berasal dari surabaya
4. the origin of the sangkuriang is from..
Penjelasan:
bandungg
kasih jawaban yang terbaik ya
the origin of the Sangkuriang from Jawa Barat
5. where does the origin of Surabaya come from
Jawaban:
in Indonesia
Penjelasan:
and including the city in Hmm I don't know
6. A particle travels 28.5 km in 30 minutes. Find the speed of the particle in km/h
Jawab:
57 km/h
Penjelasan dengan langkah-langkah:
from the question, we know that:
D = 28,5 km
T = 30 minutes = 0,5 hour
Because the formula is Speed = [tex]\frac{distance}{time}[/tex]
so, Speed = [tex]\frac{28,5}{0,5}[/tex] = 57 km/h
Hope you find it helpful! Feel free to ask if there's something you don't understand from my explanation :)
7. fania plays with her doll from 05.33 p. m. to 06.33 p. m then starts to do her homework from 07.00 a. m until 08.00 a. m when the time fania starts homework?
Jawaban:
she stars homework from 07.00 a.m
Penjelasan:
semoga membantu
Jawaban:
she start do homework from 07.00 AM
8. The magnetic force of a material comes from the spinning of what atomic particle?
Jawaban:
Every substance is made up of tiny units called atoms. Each atom has electrons, particles that carry electric charges. Spinning like tops, the electrons circle the nucleus, or core, of an atom. Their movement generates an electric current and causes each electron to act like a microscopic magnet.
Read more on Brainly.in - https://brainly.in/question/40909693#readmore
9. An object moving with a speed of 67 m/s and has a kinetic energy of 500 J, what is the mass of the object?
Jawaban:
Ek = ½mv²
so,
m = 2.Ek / v²
= 2. 500 / 67²
= 0.222 kg
10. An object has a kinetic energy of 49 J and a mass of 2 kg, how fast is the object moving?
Energi Kinetik
Diketahui:
Ek = energi kinetik = 49 J m = massa = 2 kgDitanya:
v = kecepatan ? JawabanEk = ½ × m × v²
49 = ½ × 2 × v²
49 = 1 × v²
v² = 49/1
v² = 49
v = 7 m/s ✔
11. ana and ani are twins two have a body mass difference of 2,5 kg,ana's body mass is greater than ani's.ana's body mass is 25 kg and she sits at a distance of 1 m from the pivot of a seesaw.in order for the seesaw to balance,ani should sit at a distance of...m from the pivotplease bantu pake cara
Ana > Ani (beda = 2,5 kg)
Ana = 25 kg, Ani = 22,5 kg
Ana berjarak 1 m dari titik tumpu (total gaya = 25 N)
25/22,5 = 10/9 = [1,10 m]
Setau aku sih jawabannya kayak gitu.. Tapi maap misal kalo salah kak ^^
12. Where does the origin of Surabaya come from
Jawaban:Surabaya is the second largest city in Indonesia which is located in East Java Province. The city is led by a very famous female mayor, Tri Rismaharini.
Penjelasan:Smg Membantu
13. A body with mass of 2 kg is moving with velocity 2 m/s. A short time later the body is moving with velocity of 5 m/s. The total work applied to the body is ....Tolongin kaka" mau di kumpulin sebentar lagi
Jawaban :
Penjelasan:
W = (½ . m . Vt²) - (½ . m . Vo²)
W = (½ . 2 . 5²) - (½ . 2 . 2²)
W = 25 - 4
W = 21 J
14. When is the ray of light get "slowed down"?A. moving from optically denser to optically rarer.B. moving from optically rarer to optically denser.C. moving from water to air.D. none of these
Jawaban:
A. moving from Optically denser to optically rarer
semoga bermanfaat
Penjelasan:
Kapan sinar cahaya "diperlambat"?
A. bergerak dari optis lebih rapat ke optis lebih jarang.
B. bergerak dari optis lebih jarang ke optis lebih padat.
C. bergerak dari air ke udara.
D.tidak satupun
jawabannya:c maaf kalo salah
15. A mass of car 1750 kg starts from rest and moves along straight road with a constant acceleration, reaching a speed of 20 m/s after 5 second. The resultant force acting on the car during the first 5 seconds is..... 1750 N a.
Jawaban:
The resultant force acting on the car during the first 5 seconds can be calculated using the equation for force, which is Force = Mass x Acceleration.
The acceleration of the car can be calculated using the equation for acceleration, which is Acceleration = Change in velocity / Time.
Change in velocity = Final velocity - Initial velocity = 20 m/s - 0 m/s = 20 m/s
Time = 5 seconds
So, acceleration = 20 m/s / 5 seconds = 4 m/s^2
Force = Mass x Acceleration = 1750 kg x 4 m/s^2 = 7000 N.
Therefore, the resultant force acting on the car during the first 5 seconds is 7000 N.
Penjelasan:
16. The displacement x of a point moving in a straight line is given by; The instantaneous velocity of the particle at t = 4 second will be:
KINEMATIKA
▶ GLBB
x = 3t⁴ + 3t³ - 5t² - 2t - 5
v = dx / dt
v = 12t³ + 9t² - 5t - 2
instantaneous velocity of the particle at t = 4 second will be :
v = 12(4)³ + 9(4)² - 5(4) - 2
v = 890 m/s ✔
17. 1.(DO NOT ALIGHT FROM A MOVING VEHICLE( the warning means that we must not A. bring heavy things on a moving vehicle B.take alight from a moving vehicle C. Bring alight on a moving vehicle D. Gett of a moving vehicle
B.take alight from a moving vehicle
18. An object has mass of 6kg and 6 m from ground. Calculate the potential energy!
Jawaban:
mass (m) = 6 kg
height (h) = 6 m
gravitational acceleration (g) = 10 m/s²
Potential Energy (Ep) = m×g×h
Ep = 6 kg × 6 m/s² × 10 m
Ep = 360 kg m²/s² = 360 Joule.
19. fania plays with her doll from 05.33 p. m. to 06.33 p. m then starts to do her homework from 07.00 a. m until 08.00 a. m when the time fania starts homework?
Jawaban:
07:00AM
Penjelasan:
karena begitu
Jawaban:
07.00
Penjelasan:
Karena dia mulai pukul 07.00 dan selesainya jam 08.00
20. 20. The displacement x in metre of a particle of mass m kg moving in one dimension under the action of a force is related to the time t in second by the equation x= (t-3)². The work done by the force (in joules) in first six seconds is1) 18 m Joules 2) zero 3) 9m/2 Joules 4) 36 m Joules
Question:
The displacement x in metre of a particle of mass m kg moving in one dimension under the action of a force is related to the time t in second by the equation x= (t-3)². The work done by the force (in joules) in first six seconds isAnswer:
The work done is 18 m ( in Joules )Explaination:
Giventhat
The displacement x in metre of a particle of mass m kgThe particle is moving in one dimension under the action of a force is related to the time t in secondsThe force is given by the equation x= (t-3)²ToFind:
The work done by the force is first 6 secondsRequiredSolution:
Let's find out the velocity of the particle differentiating[tex]{ \leadsto \: { \boxed{ \pink{ \bf{Velocity = \frac{dx}{dt} }}}}}[/tex]
Calculating velocity :
[tex] : \implies \sf \: Velocity = \dfrac{dx}{dt} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \sf \: Velocity = \frac{d}{dt} (t - 3) {}^{2} [/tex]
Usingformula:
[tex] \leadsto \: { \pink{ \boxed{ \bf{ \frac{d}{dx} ( {x}^{n}) = nx {}^{(n - 1)} }}}}[/tex]
[tex]: \implies \sf \: Velocity = \frac{d}{dt} (t - 3) {}^{2} \\ \\ : \implies \sf \: Velocity =2(t - 3) \: \: \: \: \\ \\ \\ : \implies \sf \: Velocity =2t - 6 \: \: \: \: \: \: \: \: [/tex]
CalculatingAcceleration:
[tex]: \implies \sf Acceleration = \dfrac{dv}{dt} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \sf Acceleration = \frac{d}{dt} (2t - 6) \: \: \: \: \: \: \\ \\ : \implies \sf Acceleration = \frac{d(2t)}{dt} - \frac{d}{dt} (6) [/tex]
Usingformula:
[tex] \leadsto \: { \pink{ \boxed{ \bf{ \frac{d}{dx} ( {x}^{n}) = nx {}^{(n - 1)} }}}}[/tex]
[tex] : \implies \sf Acceleration = \frac{d(2t)}{dt} - \frac{d}{dt} (6) \\ \\ \\ : \implies \sf Acceleration = 2 \: \frac{d}{dt}( t )- 0 \: \: \: \: \: \\ \\ \\ : \implies \sf Acceleration = 2 - 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\: \implies \sf Acceleration = 2 \: m /s {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
UsingFormula:
[tex] \leadsto{ \pink{ \boxed{ \bf{Force = mass \times acceleration}}}}[/tex]
CalculatingForce:
[tex] \: : \implies \sf \:Force = mass \times acceleration \\ \\ \: : \implies \sf \:Force = m(2 ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \\ \\ : \implies \sf \:Force = 2m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
CalculatingDisplacement:
[tex] \: : \implies \sf Displacement = (t - 3) {}^{2} \\ \\ \: : \implies \sf Displacement = (6 - 3) {}^{2} \\ \\ \: : \implies \sf Displacement = {3}^{2} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: : \implies \sf Displacement = 9 \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
UsingFormula:
[tex]{ \leadsto} \: { \pink{ \boxed{ \bf{ Work done = force \times displacemnet}}}}[/tex]
[tex] \: {: \implies }\sf \:Work done = force \times displacement \\ \\ \: {: \implies }\sf \:Work done =2m \times 9 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: {: \implies }\sf \:Work done =18m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
Henceforth:
The workdone by the particle is 18m ( in joules )*Note:
If any confusion in the differentiation of the acceleration part of the question remember :
The differentiation of contestants is zero!!The differentiation of a variable woth power 1 will be one▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
0 Komentar