A bullet of mass 0.20 kg travels horizontally at a speed of 400 m/s. it strikes a stationary wooden block of mass 1.80 kg resting on a frictionless, horizontal surface. the bullet stays in the block. what is the speed of the bullet and the block immediately after the impact? 200 m/s 80 m/s 44 m/s 40 m/s (in ulanganku hari ini)
1. A bullet of mass 0.20 kg travels horizontally at a speed of 400 m/s. it strikes a stationary wooden block of mass 1.80 kg resting on a frictionless, horizontal surface. the bullet stays in the block. what is the speed of the bullet and the block immediately after the impact? 200 m/s 80 m/s 44 m/s 40 m/s (in ulanganku hari ini)
Jawaban:
40 m/s
Penjelasan:
Known :
Mb = 0.20 kg
Vb = 400 m/s
Mw = 1.8 kg
Vw = 0 m/s (Resting)
Solution :
Because bullet stays in the block, then the speed of the bullet and the block will be same :
Mb . Vb + Mw . Vw = (Mb + Mw) . V
0.2 . 400 + 0 = (0.2 + 1.8) . V
V = 40 m/s
2. What is the density of a substance that has a volume of 5,5 cm3 and a mass of 10g?
Jawaban:
1,81 g/cm³
Penjelasan:
Density=Mass/Volume
=10/5,5
=1,81 g/cm³
Density=Massa Jenis
Mass=Massa
Volume=Volume
3. A half metre rule AB is freely pivoted at 18cm from end A and balances horizontally when a body of mass 35g is hung 48cm from end B.Calculate the mass of the rule
sikap kayang Berdasarkan pernyataan tersebut melakukan gerak
4. A tennis ball is travelling at 50 m/s and has a kinetic energy of 75 J. Calculate the mass of the tennis ball.
velocity = V = 50 m/s
kinetic energy = Ek = 75 Joule
mass = m = .... ?
[tex]ek = \frac{1}{2} \times m\times {v}^{2} [/tex]
[tex]75 = \frac{1}{2} \times m \times {50}^{2} [/tex]
[tex]75 = \frac{1}{2} \times m \times 2500[/tex]
[tex]75 = 1250 \times m[/tex]
[tex]m = \frac{75}{1250} = 0.06 \: kg[/tex]
5. A wagon with 15 kg of mass moves at velocity of 0,25 m/s toward a concreate wall. At the same time, a bullet 10 g mass is fired at velocity 1.000 m/s toward the wall. Calculate: a. The momentum of the wagon b. The momentum of the bullet
mass wagon = 15 kg
v wagon = 0,25 m/s
mass bullet = 0,01 kg
v bullet = 1000 m/s
a. P wagon = mass wagon * v wagon
= 15 kg * 0,25 m/s
= 3,75 kg*m/s
b. P bullet = mass bullet * v bullet
= 0,01 kg * 1000 m/s
= 10 kg*m/s
Correctmeifiwrong
6. The density of water is 1.0g/cm3. This is useful as it means we can measure themass of water and easily work out its volume. A. What is the mass of 1cm³ of water? B. What is the mass of 20cm³ of water? C. What is the volume of 10g of water? D. What is the volume of 22.5g of water
Jawaban:
The density of water is 1.0g/cm3. This means 1 g of water mass equals to 1 cm³ of water volume.
a. 1 cm³= 1g
b. 20 cm³= 20g
c. 10 g= 10 cm³
d. 22.5g = 22,5 cm³
semangat belajar
#first
7. A kitten and a basket has a mass of 3 kg. A hanster and a similar basket has a mass of 1.4 kg. The mass of the kitten is 5 times the mass of the hamster. (a) what is the mass of the hamster? (b) what is the mass of the kitten?
hamster = x
kitten =y
basket=z
y=5x
x + z = 1,4
y+ z = 3. (+)
x-y = -1,6
x-5x=-1,6
6x = 1,6. x =0,267
y = 5x=5×0,267=1,335
8. the mass of a watermelon is 7/1 kg. the mass of a apples is 2/5 kg less than the mass of the watermelona. what is the mass of the applesb.what is the total mass of the apples and watermelon
Diketahui:
Berat semangka = 7/1 kg
Berat apel = 2/5 kg kurang dari semangka
Maka:
a. Berat apel = 7/1 - 2/5
= 35/5 - 2/5
= 33/5 = 6 3/5
b. Berat apel dan semangka = 7/1 + 6 3/5
= 35/5 + 33/5
= 68/5 = 13 3/5
Maaf klo salah
Semoga membantu
Jadikan jawaban terbaik ya...
9. 1. A bag of grapes has a mass of 4.55 kg. A bag of mangoes has a mass of 6.90 kg. What is the total mass of the two bags of fruits? *
Jawaban:
4.55 + 6.90 = 11.45
Penjelasan:
sorry if it's wrong, hope it helps^^
Jawaban:
4,55 + 6,90 = 11,45
Penjelasan:
total mass of the two bags of fruits is 11,45 kg
10. 23.A watermelon has a mass of 12.78 kg. Its mass is thrice the mass of a papaya. Find thetotal mass of the two fruits.
Jawaban:
mtk atau b inggris sih:)
11. a kitten and a basket has a mass of 3 kg . A hamster and a similar basket has a mass of 1.4kg . the mass of the kitten is 5 times the mass of the hamster .a. what is the mass of the hamster ? , b. what is the mass of the kitten ?
kitten = x ,hamster = y and baskest = z
x+z = 3 kg
y+z = 1,4 kg
z = x- y = 3 kg - 1,4 kg = 1,6 kg
x+y = 4,4 kg - 1,6 kg = 2, 8 kg
x = 5y
5y +y = 2,8 kg
6y = 2,8 kg
y = 2,8 kg /6
y = 0,47 kg
x = 5y
= 5 * 0,47 kg
= 2,35 kg
a. mass off hamster = 0,47 kg
b. mas off kitten = 2,35 kg
bonus...mass of basket = 1,6 kg
thank's
12. A ball travelling at 25 m/s and has a kinetic energy of 50 J. Calculate the mass of the ball !
velocity (v) = 25 m/s
KE = 50 J
mass (m) = ....?
KE = ½ × m × v²
50 = ½ × m × 25²
50 = ½ × m × 625
312.5m = 50
m = 50/312.5
m = 4/25
13. Which definition of concentration given below best describes the concept expressed in “in a solution of 10g of salt and 70ml of water, the concentration of the solution by percent mass is 12,5%2?”
jawabannya
1.The amount of water in the solution
2 A word denoting 70g
3 The mass of the solution
4 The amount of a component (solute) present.
5 The amount of solution
Terjemahan :
Jumlah 1. air dalam larutan 2 A 70g kata yang menunjukkan 3 Massa solusi 4 Jumlah komponen ( zat terlarut ) hadir . 5 Jumlah solusi
14. a bullet of mass 200 gram is fired at elevation angle of 30° and with initial velocity of 10 m/s. if the earth gravitational acceleration 10 m/s², the potential energy of the bullet at its highest point is...(sekalian cantumin arti soalnya ya,biar yang lain juga maksud:) )TERIMAKASIH UNTUK PENJAWAB^^
Sebuah peluru bermassa 200 gram dilemparkan dengan kecepatan 10 m/s pada sudut 30° dengan horizontalnya. Jika percepatan gravitasi 10 m/s² , energi potensial pada titik tertinggi adalah 2,5 Joule
PEMBAHASANBerat sebuah benda adalah gaya tarik gravitasi yang dialami oleh benda tersebut . Berat sebuah benda ini di pengaruhi oleh percepatan gravitasi yang dialami olehnya yang di rumuskan dengan :
W = m x g
dimana :
W = berat benda ( N )
m = massa benda ( kg )
g = percepatan gravitasi ( m/s² )
Gaya adalah tarikan atau dorongan. Gaya termasuk besaran vektor karena memiliki besar dan arah. Satuan Gaya adalah Newton ( N ).
Beberapa gaya dapat bekerja pada benda yang sama dan menghasilkan gabungan gaya atau lebih di kenal dengan istilah resultan gaya. Secara sederhana gaya yang searah akan menghasilkan total gaya yang semakin besar , sebaliknya gaya yang saling berlawanan arah akan menghasilkan total gaya yang semakin kecil.
Hukum yang terkenal mengenai gaya ini adalah hukum Newton tentang gerak.
Hukum ke-1 Newton menyatakan bahwa:
Jikalau resultan gaya yang bekerja pada benda adalah 0 N maka benda akan bergerak dengan kecepatan tetap atau benda akan diam
Hukum ke-2 Newton menyatakan bahwa :
∑F = m . a
dimana :
∑F = Resultan Gaya pada benda ( N )
m = massa benda ( kg )
a = percepatan benda ( m/s² )
Hukum ke-3 Newton menyatakan bahwa :
F aksi = - F reaksi
dimana :
F aksi = Gaya aksi benda 1 ke benda 2
F reaksi = Gaya reaksi benda 2 ke benda 1
Gerak lurus dapat di bedakan menjadi dua macam yakni:
Gerak Lurus Beraturan (GLB)Gerak ini terjadi saat benda bergerak lurus dengan kecepatan tetap. Rumus untuk gerak ini adalah :
v = s / t
dimana :
v = kecepatan benda ( m/s )
s = jarak tempuh benda ( m )
t = waktu tempuh ( s )
Gerak Lurus Berubah Beraturan (GLBB)Gerak ini terjadi saat benda bergerak lurus dengan kecepatan yang senantiasa berubah secara teratur. Dengan kata lain benda bergerak dengan percepatan tetap. Rumus untuk gerak ini adalah :
a = ( Vt - Vo ) / t
s = ( Vt + Vo ) . t / 2
s = Vo. t + (1/2) a . t²
( Vt )² = ( Vo )² + 2.a.s
dimana :
a = percepatan benda ( m/s² )
Vt = kecepatan akhir benda ( m/s )
Vo = kecepatan awal benda ( m/s )
t = waktu tempuh benda ( m/s )
s = jarak tempuh benda ( m )
Gerak parabola adalah gabungan antara gerak lurus beraturan (GLB) dan gerak lurus berubah beraturan (GLBB). Kita bisa meninjaunya secara terpisah yakni secara vertikal adalah GLBB karena ada percepatan gravitasi, dan secara horizontal adalah GLB karena tidak ada percepatan alias benda bergerak dengan kecepatan konstan.
Jangkauan maksimum (R) dan tinggi maksimum (H) yang bisa di capai benda :
R = vo² . sin 2Ф / g
H = vo² . sin² Ф / (2g)
dimana :
R = jangkauan maksimum ( m )
H = tinggi maksimum ( m )
vo = kecepatan awal ( m/s )
Ф = sudut elevasi ( derajat )
g = percepatan gravitasi ( m/s² )
Okay marilah kita gunakan prinsip ini untuk menyelesaikan soal yang di maksud.
Diketahui :
m = 200 gram = 0,2 kg
vo = 10 m/s
Ф = 30°
g = 10 m/s²
Ditanyakan :
Ep = ?
Penyelesaian :
H = vo² . sin² Ф / (2g)
H = 10² . sin² 30° / (2 . 10 )
H = 1,25 m
Ep = m . g . h
Ep = 0,2 x 10 x 1,25
Ep = 2,5 Joule
Pelajari lebih lanjut :[tex]\textbf{Percepatan}[/tex] : https://brainly.co.id/tugas/18400185
[tex]\textbf{Daya Gerak}[/tex] : https://brainly.co.id/tugas/17121561
---------------------------
Detil Jawaban :[tex]\textbf{Kelas:}[/tex] 8
[tex]\textbf{Mapel:}[/tex] Fisika
[tex]\textbf{Bab:}[/tex] Gerak
[tex]\textbf{Kode:}[/tex] 8.6.1
[tex]\textbf{Kata Kunci:}[/tex] Gaya, Gerak, Jarak, Percepatan
15. The mass of a dictionary is 1.28 kg. The mass of the dictionary is twice the mass of a storybook. The mass of an encyclopaedia is 1.4 kg more than storybook. What is the mass of the encyclopaedia?
Jawaban:
2.04 kg
Penjelasan dengan langkah-langkah:
Jika berat sebuah kamus 2 kali berat buku cerita, maka berat buku cerita adalah setengah dari berat kamus. Sehingga berat ensiklopedia adalah
1.4 kg + (1.28 ÷ 2) kg
= 1.4 kg + 0.64kg
= 2.04 kg
16. The total mass of a block of butter and a box of cherries is 384 g. The mass of theblock of butter is twice the mass of the box of cherries. Find the mass of the blockof butter.
draw model: ∆∆∆
384÷ 3 = 128
cherries = 128 g
butter = 2x 128
butter = 256 g_____
semangat belajar....
17. A bag of grapes has a mass of 4.55 kg. A bag of mangoes has a mass of16.90 kg. What is the total mass of the two bags of fruits?
Jawaban:
total mass = grapes mass + mangoes mass
total mass = 4.55 kg + 6.90 kg
total mass = 11.45 kg
18. A 5g bullet moving at 100 m/s collides a wooden target, initially at rest. The mass of wooden target is 1kg. If after the collision the bullet is embed in the wooden block, what is the final speed of the wooden block?
This is innelastic collision. so we can use formula:
m1.v1+m2v2=(m1+m2)vf
5x100+1kgx0=(5+1000)vf
500+0=1005Vf
Vf=500/1005=0.49 m/s
19. kinetic energy possessed by a 25g bullet is 2000J. The velocity of the bullet is?
Jawaban:
282,84 m/s
Penjelasan dengan langkah-langkah:
[tex]ke = m {v}^{2} \\ 2000 = 0.025 \times {v}^{2} \\ {v}^{2} = \frac{2000}{0.025} \\ {v}^{2} = 80000 \\ v = \sqrt{80000} = 282.84[/tex]
20. Trini add 10g of bakingsoda to 100g of vinegar. The mixture begins to bubble. When the bubbling stop,trini find the mass of resulting mixture. She determine its mass is 105 g. Why has the mass change?
because when baking soda react with vinegar It will produce CO2 which is phase gas ..And this CO2 will evaporate
NaHCO3(s) + CH3COOH(aq) → CH3COONA(aq) + CO2(g) + H2O(l)
so the mass wil decrease because CO2 will evaporate
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