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Neporo4naja [7]

3 years ago

15

collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitu

de of their velocity

1 answer:

anzhelika [568]3 years ago

4 0

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

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A runner begins running at the beginning of 7th period and stops running at the end of the period. She runs at a pace of 10 km/h

MrRissso [65]

Answer:

She run for, t = 0.92 s

Explanation:

Given data,

The velocity of the runner, v = 10 km/h

The distance covered by the runner, d = 9.2 km

The relationship between the velocity, displacement and time is given by the formula,

t = d / v

Substituting the given values in the above equation,

t = 9.2 / 10

= 0.92 s

Hence, she ran for, t = 0.92 s

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2 years ago

If i walk to school everyday will i loose weight?

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You would also have to eat right lol

7 0

3 years ago

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

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3 years ago

Physics is explicitly involved in studying which of these activities?

Dmitrij [34]

Hey Friends
The answer to this question would be C
Hope i helped
~Katie

8 0

3 years ago

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In the diagram, ∠D is b

Degger [83]

Answer:

The answer is B

Explanation:

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3 years ago