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mr_godi [17]
2 years ago
14

Can some one pls answer them asapppp

Physics
2 answers:
RUDIKE [14]2 years ago
7 0
D is 3G of wood
E is gas
Yuliya22 [10]2 years ago
4 0

A) Oil B) Wood C) 0.02J D) 3g E) Gas

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A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost
densk [106]

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = \frac{1}{2}m₁u₁² +  \frac{1}{2}m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = (\frac{1}{2} x 2.0 x 5.0²) +  (\frac{1}{2} x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = \frac{1}{2}(m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = \frac{1}{2} ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

3 0
3 years ago
I need heelp can anyone heelp me plz
GuDViN [60]

Answer:

e) 1.04

Explanation:

7 0
2 years ago
Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plume
Mamont248 [21]

Answer:

1331.84 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement = 490 km

a = Acceleration

g = Acceleration due to gravity = 1.81 m/s² = a

From equation of linear motion

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -1.81\times 490000}\\\Rightarrow u=1331.84\ m/s

The speed of the material must be 1331.84 m/s in order to reach the height of 490 km

3 0
3 years ago
Read 2 more answers
Two objects of equal mass are a fixed distance apart. If the mass of each object could be tripled, the gravitational force betwe
antiseptic1488 [7]

gravitational force between two objects is given as

F = G m₁ m₂/r²

where m₁ = mass of first object , m₂ = mass of second object , r = distance between the two objects .

Initial case :

m₁ = m₂ = m

gravitational force between the objects is given as

F = G m²/r²

Final Case :

m₁ = m₂ = 3 m

new gravitational force between the objects is given as

F' = G (3m)²/r²

F' = 9 G m²/r²

F' = 9 F

hence the gravitational force between the two objects becomes 9 times.

5 0
3 years ago
Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abund
vova2212 [387]

Answer:

C. 28.09 amu

Explanation:

The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance),  X-29 (28.976 amu, 4.67% abundance) and  X-30 (29.974 amu, 3.10% abundance).

The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.

The atomic weight is computed as follows:

atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of  X-30 × fractional abundance

atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 ×  0.0310

atomic weight = 25.8031871 + 1.3531792 + 0.929194

atomic weight = 28.0855603 amu

To 2 decimal place atomic weight = 28.09 amu

6 0
3 years ago
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