A 1000 kg car moving east at 5 m/s collides with a 2000 kg car moving east at 2 m/s. After the collision, the 2000 kg car moves east at 5 m/s. a. Determine the total momentum before collision. b. Find the final velocity and direction of the 1000 kg car.
1. A 1000 kg car moving east at 5 m/s collides with a 2000 kg car moving east at 2 m/s. After the collision, the 2000 kg car moves east at 5 m/s. a. Determine the total momentum before collision. b. Find the final velocity and direction of the 1000 kg car.
Answersisonthepictureabove.
The 1000 kg car is moving to the opposite of east, meaning it is going west.
Hopeithelps
2. A 2 kg cart has a momentum of 16 kg m/s. What is its velocity?
Jawaban:
8m/s
Penjelasan:
Diketahui
m=2kg
p=16 kg m/s
ditanya
v=...?
Jawab
p=m.v
v=p/m
=16/2
=8 m/s
#Semoga Membantu(. ❛ ᴗ ❛.)
#Terimakasih jika berkenan jadiin best answer(◍•ᴗ•◍)❤
#Tetap Semangat Belajar!!!(✿^‿^)
3. 11. M-C-0-S-T-A-HThe right arrangement is ....
Jawaban:
Stomach
Penjelasan:
Maaf Kalau Salah
S-T-O-M-A-C-HMaaf kalau salah
4. Tentukan jenis dari matriks – matriks dibawah ini ( jika memenuhi lebih dari satu, tuliskan semua ) ! [tex]A=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right] , B=\left[\begin{array}{ccc}1&0&0\\0&0&0\\1&0&1\end{array}\right] , C=\left[\begin{array}{ccc}1&0&2\\0&1&2\\0&0&0\end{array}\right] , D=\left[\begin{array}{ccc}1&2&2\\0&0&0\\0&0&1\end{array}\right][/tex]
Penjelasan dengan langkah-langkah:
A = matriks identitas, matriks persegi/bujur sangkar
B = matriks persegi/bujur sangkar
C = matriks persegi/bujur sangkar
D = matriks persegi/bujur sangkar
5. calculate the kinetic energy possessed by: a. 3 kg objects move at 10m/s. b. the 2,500 kg car moving at a speed of 72km/h. c. 80g ball moving at 15m/s! i will make the best answer for anyone who answers this question. thank you
weight=massa=kg
velocity=kecepatan=m/s
kinetic energy=joule=(1/2)m.v²
a. (1/2).3.10²
=300/2
=150 joule
b.(1/2).2500.20² (72km/h=72000m/3600s=20m/s)
=2500.200
=50000 joule
c. (1/2).0,08.15² (80g=0,08kg)
=(1/2).0,08.225
=9 joule
6. m¹ masa=5000 kg before velocity=25m/s rightm²mass=1.500 kgbefore velocity=0m/scari after velocity dari kedua m
Jawaban:
berarti kedua nya 6500 kg
maaf kalau salah
7. Persamaan kuadrat yang akar-akarnya 5 dan -2 adalah . . . [tex]\mathbf{\left(A.\right)\ \ x^{2}+7x+10=0}[/tex] [tex]\mathbf{\left(B.\right)\ \ x^{2}+3x-10=0}[/tex] [tex]\mathbf{\left(C.\right)\ \ x^{2}-7x+10=0}[/tex] [tex]\mathbf{\left(D.\right)\ \ x^{2}-3x-10=0}[/tex] [tex]\mathbf{\left(E.\right)\ \ x^{2}+3x+10=0}[/tex] Note = Pake Penjelasan / Jalannya #sinogen #matematika #soal #ez
Persamaan kuadrat yang akar-akarnya 5 dan -2 adalah x² - 3x - 10 = 0
PENDAHULUANDefinisi dari persamaan kuadrat yaitu suatu persamaan yang variabelnya memiliki pangkat tertinggi sama dengan dua.
Bentuk umum persamaan kuadrat:
ax² + bx + c = 0 dengan a, b, dan c bilangan real a ≠ 0.
Menyelesaikan persamaan kuadrat dpt dilakukan dgn cara:
a. Dengan pemfaktoran
b. Melengkapkan kuadrat sempurna
c. Menggunakan rumus abc
Persamaan kuadrat ax² + bx + c = 0 dapat ditentukan nilai diskriminan dengan rumus:
D = b² - 4ac
Rumus menentukan persamaan kuadrat jika diketahui akar-akarnya:
x² - (x1 + x2)x + (x1 × x2) = 0
PEMBAHASANx² - (x1 + x2)x + (x1 × x2) = 0
x² - (5 + (-2))x + (5 × (-2)) = 0
x² - (5 - 2)x + (-10) = 0
x² - 3x - 10 = 0
Kesimpulan:Jadi, jawabannya adalah Opsi D
Pelajari Lebih Lanjut:1. Materi tentang nilai diskriminan:
https://brainly.co.id/tugas/17038127
2. Materi menentukan akar persamaan kuadrat:
https://brainly.co.id/tugas/16155306
3. Bentuk persamaan kuadrat:
https://brainly.co.id/tugas/3681461
___________________
Detail Jawaban:Kelas : 9 SMP
Mapel : Matematika
Materi : Bab 9 - Persamaan dan Fungsi Kuadrat
Kode : 9.2.9
Kata kunci : Persamaan kuadrat
maaf fotonya kebalik
semoga membantu kak sinogen
8. Jika A adalah matriks 2x2 yang memenuhi [tex] \left[\begin{array}{ccc}1&2\end{array}\right] [/tex]A = [tex] \left[\begin{array}{ccc}1&0\end{array}\right] [/tex] dan [tex] \left[\begin{array}{ccc}4&6\end{array}\right] [/tex]A = [tex] \left[\begin{array}{ccc}0&2\end{array}\right] [/tex] , maka hasil kali [tex] \left[\begin{array}{ccc}2&4\\2&3\\\end{array}\right] [/tex] A adalah . . . A. [tex] \left[\begin{array}{ccc}1&0\\0&2\\\end{array}\right] [/tex] B. [tex] \left[\begin{array}{ccc}2&0\\0&2\\\end{array}\right] [/tex] C. [tex] \left[\begin{array}{ccc}2&0\\0&1\\\end{array}\right] [/tex] D. [tex] \left[\begin{array}{ccc}1&0\\0&2\\\end{array}\right] [/tex] E. [tex] \left[\begin{array}{ccc}0&2\\1&0\\\end{array}\right] [/tex]
A₂ₓ₂ =
[ a...b]
[ c...d]
[ 1 2 ] A = [ 1 0]
a+2c =1
b+2d = 0
[4 6] A = [0 2]
4a + 6c = 0
4b + 6d = 2
.
a + 2c = 1 ....kali 3
4a + 6c = 0
.
3a + 6c =3
4a + 6c = 0
kurangkan
-a = 3 maka a = - 3
.
a+2c = 1
-3 + 2c = 1
2c = 4
c= 2
b + 2d = 0 atau b = - 2d subs ke 4b + 6d = 2
4(-2d) + 6d = 2
-2d = 2 maka d = -1
b = -2d
b = -2(-1)
b= 2
[2...4] [-3 ..2]
[2...3] [ 2 .-1]
=
[2..0]
[0..1]
9. 15. S-0- G-0-E. The right order is ...a.sheepC.horseb. goosed. snake
Jawaban:
Jawabannya b. goose
Penjelasan:
Semoga membantu:)
SEMANGAT BELAJARNYA
Jawaban:
B. GOOSE
Penjelasan:
G - O - O - S - E
Goose memiliki artiangsa.
note : hi this is dewi, i'm sorry if there are any error in answering this question. thanks and please MAKE IT AS BRAINLEST ANSWER ^^
#SmartWithDewi
10. Example: The velocity of a particle moving along the x axis varies in time according to the expression vx = (40 - 5t2) m/s, where t is in seconds. a. Find the average acceleration in the time interval t = 0 to t = 2.0 s b. Determine the acceleration at t = 2.0 s. Example: The velocity of a particle moving along the x axis varies in time according to the expression vx = (40 - 5t2) m / s, where t is in seconds. a. Find the average acceleration in the time interval t = 0 to t = 2.0 s b. Determine the acceleration at t = 2.0 s.
Jawaban:
a. - 10 m/s²
b. - 20 m/s²
Penjelasan:
Vx = (40 - 5t²)
a.
saat t₁ = 0 --> Vx₁ = 40 m/s
saat t₂ = 2 --> Vx₂ = 40 - 5. 4 = 20 m/s
Average acceleration
ar = ΔV/Δt = ( Vx₂ - Vx₁ ) / ( t₂ - t₁ )
ar = ( 20 - 40 )/( 2 - 0 )
ar = - 10 m/s² ( perlambatan )
b.
acceleration at t = 2.0 s.
a = dV/dt = - 10. t
a = - 10. 2 = - 20 m/s²
11. A car is moving with an initial velocity of 20 m/s and then moving with constant acceleration of 4 m/s? The velocity of a car after 20 seconds is ... O a. 120 m/s O b. 150 m/s O C. 80 m/s O d. 100 m/s
Jawaban:
given : v1 = 20 m/s
a = 4 m/s^2
t = 20 s
ask :v2 .... ?
answer :
a = (v2 - v1)/ tv2 = at + v1 = 4×20 + 20 = 100 m/sGood luck....
Jawaban:
d. 100m/s
Penjelasan:
know
Vo=20m/s
a=4m/s²
t=20s
asked
vt=?
solution
The velocity of a car after 20 seconds:
vt=Vo+at
vt=20+4(20)
vt=20+80
vt=100m/s
12. a car reaches a velocity of 20 m/s with an acceleration of 2 m/s^2 . How far will it travel while it is accelerating if it is (a). initially at rest (b). initially moving at 10 m/s ?
◾ Materi : Kinematika Gerak Lurus
◾ Sub materi : GLBB
◾ Mapel : Fisika
Diketahui :
Vt = 20 m/s
a = 20 m/s²
Ditanya :
s = ?
a) Vo = 0
b) Vo = 10 m/s
Penyelesaian :
*Cari jarak tempuh jika diam
Vt² = Vo² + 2as
20² = 0² + 2(2)s
400 = 4s
s = 100 meter
*Cari jarak tempuh jika Vo = 10 m/s
Vt² = Vo² + 2as
20² = 10² + 2(2)s
400 = 100 + 4s
300 = 4s
s = 75 meter
Jadi, jarak tempuh untuk Vo diam Dan Vo 10 m/s adalah 100 meter dan 75meter
semoga membantu
# sharing is caring #
-vin
13. the pinion gear a rolls on the gear rack B and C is moving to the right at 8ft/s and C is moving to the left at 4ft/s, determine the angular velocity of the pinion gear and velocity of its center A
Jawaban:
Maayo ra ko pero oo tanga ka
14. Jika [tex]\left[\begin{array}{ccc}m&0\\0&n\end{array}\right][/tex][tex]\left[\begin{array}{ccc}2\\4\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}n\\5+3m\end{array}\right][/tex] , nilai m-n adalah
Kalikan didapat kesamaan:
2m = n
4n = 4 + 3m
substitusi n = 2m , maka:
4(2m) = 4 + 3m
8m = 4 + 3m
5m = 4
m = 4/5
n = 2(4/5)
= 8/5
m - n = 4/5 - 8/5 = - 4/5
15. A car is moving at 72 km/h=............m/s and it is declerated by 2 m/s in 10 s.Find it's speed in km/h.Write the way please
Jawaban:
648km/h
Penjelasan:
v=72 km/h=20m/s
a=2m/s
t=10s
vt=vot-at
vt=20x10-2x10
=200-20
=180m/s
vt=180x(10⁻³x3600)
=18x36=648km/h
16. o-0-n-s-p the right arrangement is
The right arrangement is spoon. s-p-o-o-n.
17. An object has a mass 5 kg and 2 forces F1 = 10N to the left and F2 = 30N to the right. the acceleration at the block is ... A. 4 m/s² to the left B. 4 m/s² to the right C. 5 m/s² to the right D. 5 m/s² to the left
dik
F = 20 N
M = 5 kg
dit = a = F:M
20 : 5 = 4
b.4m/s² to the right
Resultan F=30-10=20 ke kanan
massa=5 kg
F=ma
a=F/m
a=20/5= 4 m/s² to the right
18. 21. 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 .... a.25 J b. 24 J c. 23 J d. 22 J e. 21 J
Jawaban:
21 J (E)
Penjelasan:
Known :
m = 2 kg
Vo (initial velocity) = 2 m/s
Vt (final velocity) = 5 m/s
Ask : W
Answer :
[tex]W =\Delta KE\\W = KE_{final}-KE_{initial}\\W = \frac{1}{2} m v_{final}^2- \frac{1}{2}mv_{initial}^2\\W = \frac{1}{2} \times 2 \times 5^2- \frac{1}{2} \times 2 \times 2^2\\W = 25-4\\W= 21 $ joule \\\\Then, the total work applied to the body is 21 J (E)[/tex]
19. 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
20. jika [tex] \left[\begin{array}{ccc}m&0\\0&n\end{array}\right] [/tex] [tex] \left[\begin{array}{ccc}2\\4\end{array}\right] [/tex] [tex] \left[\begin{array}{ccc}n\\5+3m\end{array}\right] [/tex] , nilai m-n adalah
Kalikan didapat kesamaan:
2m = n
4n = 4 + 3m
substitusi n = 2m , maka:
4(2m) = 4 + 3m
8m = 4 + 3m
5m = 4
m = 4/5
n = 2(4/5)
= 8/5
m - n = 4/5 - 8/5 = - 4/5
0 komentar:
Posting Komentar