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4 years ago

What are the si units for average speed?

1 answer:

8 0

Speed has the dimensions of distance divided by time. The SI unit of speed is the metre per second, but the most common unit of speed in everyday usage is the kilometre per hour or, in the US and the UK, miles per hour.

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A cylindrical container with a cross-sectional area of 60.2 cm^2 holds a fluid of density 826 kg/m^3. At the bottom of the conta

Hatshy [7]

Answer:

The depth of the fluid is: 3.208m

Explanation:

The pressure inside a fluid at rest is obtained with the following equation:

P= fluid density × acceleration of gravity × fluid column height

The pressure at the base of the container only due to the column of the fluid will be the pressure at the base of the container minus the atmospheric pressure:

P= 127KPa - 101KPa= 26KPa= 26000 (kg/(m×s²))

The pressure at the base of the vessel due to the height of the fluid column will be:

P= 26000 (kg/(m×s²)) = 826 (kg/m³) × 9.81 (m/s²) × hight

Therefore, clearing the height of the fluid shaft, the height of the container is obtained which will be:

Fluid column heigh= 26000(kg/(m×s²)) × (1/826) (m³/kg) × (1/9.81) (m/s²)

Fluid column heigh= 3.208m

5 0

4 years ago

A constant force of 12 N acts for 5 s on a 5 kg object. What is the change in object’s velocity?

sattari [20]

Answer:

"solve: given that F -12 N and time 4 seconds and let we have to find out the P.

F = 12 N

t = 4 s

p = ?

F = m×( v - u ) / t

12 = m×v / 4

m×v = 12× 4

p = 48 kg m/s

Linear momentum will be 48 kg m/s.

Explanation:

7 0

3 years ago

Distance Using Hubble's Law II.Find the percent difference (% = |A-O|/A × 100) between the actual (A) distance values and calcul

wlad13 [49]

Answer:

Explanation:

Let us first calculate for Virgo

$A= d_{virgo} = 17;M_{pc}, V_{virgo} = 1200 km/s,
$

Using Hubble's law

$v = H_{0}D$ For Virgo

$V_{virgo} = H_{0}D_{virgo}$

$O = D_{virgo} = \frac{v_{virgo}}{H_{0}} = \frac{1200 km/s}{70 km/s/Mpc}= 17.143 Mpc
$

Percentage difference for the Virgo

$\% = \frac{|A-O|}{A}\times 100 = \frac{|17 Mpc-17.143 Mpc|}{17 Mpc}\times 100 = 0.84 \%
$

Now for calculate for Corona Borealis

$A= d_{Corona Borealis} = 310 Mpc, v_{Corona Borealis} = 22000 km/s,
$

Using Hubble's law

$v = H_{0}D$ For Corona Borealis

$v_{Corona Borealis } = H_{0}D_{Corona Borealis }
\\O = D_{Corona Borealis } = \frac{v_{Corona Borealis }}{H_{0}} = \frac{22000 km/s}{70 km/s/Mpc}= 314.286 Mpc
$