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?
1. 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
2. Mohon bantu dijawab. 1. ............ in a long queue * being finding having making moving shopping travelling taking waiting 2. ............ stuck in a traffic jam * being finding having making moving shopping travelling taking waiting 3. ........... a parking space * being finding having making moving shopping travelling taking waiting 4. ............ with your partner /family * being finding having making moving shopping travelling taking waiting 5............. house * being finding having making moving shopping travelling taking waiting 6. ............ an interview * being finding having making moving shopping travelling taking waiting 7. ........... a speech * being finding having making moving shopping travelling taking waiting 8. ........... by air * being finding having making moving shopping travelling taking waiting 9. ............ an exam or test * being finding having making moving shopping travelling taking waiting
Jawaban:
1. Waiting
2. Being
3. Finding
4. Traveling
5.
6. Waiting
7. Taking
8. moving
9. Taking
3. 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
4. Kinetic energy is the energy possessed by a moving object KE = 1/2^2m = mass (kg) v = velocity (m/s) KE = Kinetic energy (J) Solve the numerical problem about the kinetic energy by using the formula above.
Jawaban:
1/2 pangkat 2 =1/4ke
Penjelasan:
gitu gee ya
5. 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....
6. A body, initial at rest, explodes into two fragments of masses M and 3M having total kinetic energy E. The kinetic energy of the fragment of mass M after the explosion is........
The explosion from rest is categorised as inelastic collision. Kinetic energy of the fragment of mass M after the explosion is 3E.
Penjelasan dan langkah-langkah:
Assumption:Assume m₁ and m₂ move in opposite direction after the explosion so that v₁ is positive and v₂ is negative. Initial velocity is zero (at rest), final velocity m₁ is v₁ and final velocity of m₂ is -v₂.
Use inelastic collision formula
Inelastic collision formula:
(m₁+m₂)V = m₁v₁ + m₂(-v₂)
(3M+M)(0) = 3Mv₁ + M(-v₂)
3Mv₁ = Mv₂
v₂ = 3v₁ ...................(1)Kinetic energy m₁ from the question
Kinetic energy m₁:
[tex]KE = \frac{1}{2}m_{1} v_{1}^{2}[/tex]
[tex]E = \frac{1}{2}(3M)v_{1}^{2}[/tex]
[tex]E = \frac{3}{2}Mv_{1}^{2}[/tex] ..................(2)Kinetic energy m₂:
[tex]KE = \frac{1}{2}m_{2} v_{2}^{2}[/tex]Subtitute (1)
[tex]KE = \frac{1}{2}(M)(3v_{1})^{2}[/tex]
[tex]KE = \frac{9}{2}Mv_{1}^{2}[/tex]
[tex]KE = 3(\frac{3}{2}Mv_{1}^{2})[/tex]Subtitute (2)
[tex]KE = 3E[/tex]
The kinetic energy of the fragment of mass M after the explosion is 3E.
Pelajari Lebih Lanjut
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7. 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]
8. lengkapi kalimat imperatif ini dengan konjungsi yang benar keep moving for two block........turn rigth to sarinah boulevard
Jawaban:
keep moving for two block and then/then turn rigth to sarinah boulevard
Semoga membantu....
Maaf kalau salah :)
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. 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
12. 3. A block of mass m/ = 4.0 kg is put on top of a block of mass m2 = 5.0 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N must be applied to the top block. The blocks assembly is now placed on a horizontal, frictionless table, as shown in FIGURE 1. Find the magnitudes of a.the maximum horizontal force that can be applied to the lower block so that the blocks will move together b.the resulting acceleration of the blocks.
The maximum horizontal force that can be applied to the lower block is 5 N.The resulting acceleration of the blocks are 21,1 m/s².
Penjelasan dengan langkah-langkah:
Diketahui:
Massa 1 = 4 kg
Berat 1 = 40 N
Massa 2 = 5 kg
Berat 2 = 50 N
F1 = 12 N
Ditanya:
Gaya pada blok paling bawah (F2)?Percepatan (a)?Jawab:
∑Fy1 = 0
N1 - W cos θ = 0
12 - 120 cos θ = 0
12 = 120 c0s θ
cos θ = 0,1
∑Fx1 = m1 x a
120 - T = 4a....(1)
∑Fy2 = 0
N2 - w cos θ = 0
N2 = w cos θ
N2 = 50 (0,1) = 5 N
∑Fx2 = m2 x a
50 sin θ + T = 5a
50 x 1 + T = 5a
50 + T = 5a..........(2)
120 - T = 4a
50 + T = 5a
___________ +
190 = 9a
a = 21,1 m/s²
from that calculation, we know that:
The maximum horizontal force that can be applied to the lower block is 5 N.The resulting acceleration of the blocks are 21,1 m/s².Pelajari lebih lanjut
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13. An object of mass 1000 kg, initially having a velocity of 100 m /s, decelerates to a final velocity of 20 m/s. What is the change in kinetic energy of the object, in kJ?
ΔEk = 1/2.m.Vf² - 1/2.m.Vi²
ΔEk = 1/2.1000.20² - 1/2.1000.100²
ΔEk = 500.400 - 500.10000
ΔEk = 200.000 - 5.000.000
ΔEk = - 4.800.000 J
ΔEk = - 4.800 kJ
14. What is the mass of a body having a weight of 45N ? (4.59 kg)
The mass of a body having a weight of 45N. W = m.g m = W/g m = 45/(9,8) = 4,59 kg
15. Make a congratulations card about:a. Moving to a new homeb. Having a new babyc. Graduation dayd. Wedding daye. Winning a contest
Jawaban:
A. congratulations for your new home my friend, i hope you happy there.
B. congratulations for your baby, i hope you guys Will be the happy family ever.
C. congratulations for your gradutaion day, i hope you Will be the succes man in the future.
D. congratulations for your wedding day brother, i hope you guys Will always together forever.
E. congratulations for winning the contesst sister, i hope you Will be the winner in the all contest you join.
Penjelasan:
sorry of wrong.
16. A block of mass 15 kg executes simple harmonic motion under the restoring force of a spring. The amplitude and the time motion are 0.1 m and 3.14 s respectively. Find the maximum force exerted by the spring on the block !
Jawaban:
30 blok
Penjelasan:
17. A block of monel alloy consist of 70% nickel and 30% copper.if it contains 88.2 g of nickel determine the mass of copper on the block ?
semoga jawabannya bener
18. 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
19. A bullet with mass 10 gram is fired and hits a block which mass is 1.49 kg. The block is haning freely on a piece of rope of length with 0.2 m. If the acceleration due to the earth gravity is 10m/s² and the rope is displaced by 60 cm measured from its initial position/vertical position, the velocity of the bullet when fired is ... m/s
Hk Kekekalan Energi
(m+M).g.h = 1/2.(m+M).v'²
v' = √2gh
v' = √(2. 10. 0,6)
v' = 2√3 m/s
Hk Kekekalan Momentum
m.v + M.V = (m+M).v'
0,01. v + 1,49. 0 = (0,01+1,49). 2√3
0,01. v = 1,5. 2√3
v = 300√3 m/s
20. If an object has a mass of 8 kg moving at a speed of 10 m/s. the kinetic energy ( the object is… J A.400 B.800 C.80 D.40 Dengan cara ya
Jawaban:
400 J
Penjelasan:
The formula for kinetic energy:
Ek = [tex]\frac{1}{2}[/tex]mv²Ek = Kinetic Energy
m = mass
v = velocity
So, if the mass of the object is 8 kg, and it's moving at 10 m/s, then:
Ek = [tex]\frac{1}{2}[/tex] . 8kg . (10m/s)² = [tex]\frac{1}{2}[/tex] . 8kg . 100 m²/s² = 400 kgm²/s² = 400 JHence, the kinetic energy of the object is 400 J
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