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Molodets [167]
3 years ago
12

A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1)

Calculate the final velocity of the 2.00-kg ball. (Express your answer to three significant figures.)
Physics
1 answer:
loris [4]3 years ago
7 0
Elastic collision is when kinetic energy before = kinetic energy after

Ek= 1/2mv^2

total before
Ek=1/2(2)(2.2^2) = 4.84 J

total after
Ek= 1/2(2+4)(v^2) = 3v^2

Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s
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A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of
posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0
3 years ago
In a two-body collision, if the momentum of the system is conserved, then which of the
lakkis [162]

Answer:

c) may also be conserved

Explanation:

Momentum is conserved  in both elastic and inelastic type of collisions.

But the differences is that:

In an ELASTIC type of collisions, KINETIC ENERGY IS ALSO CONSERVED.

whereas, In an INELASTIC type of collision, KINETIC ENERGY IS NOT CONSERVED.

So unless until type of collision is specified, we can not say anything about the conservation of kinetic energy after collision.

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5 0
3 years ago
1. A car accelerates uniformly from 12 m/s to 39 m/s in 12 seconds. What is the car's average acceleration
Alinara [238K]

Answer:

(1) 2.25m/s^2

(2) 45.6m

Explanation:

(1) A car accelerates uniformly from 12m/s to 39m/s in 12 seconds

Therefore the average acceleration can be calculated as follows

a = 39-12/12

a = 27/12

a= 2.25m/s^2

(2) A butterfly is flying at 4m/s , it accelerates uniformly at 1.2 m/s for 6 seconds

u= 4

a= 1.2

t= 6

Therefore the distance can be calculated as follows

S= ut + 1/2at^2

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= 24 + 1/2 × 1.2 × 36

= 24 + 1/2 × 43.2

= 24 + 21.6

S = 45.6m

Hence the butterfly travels at 45.6m

5 0
2 years ago
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