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2 years ago

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

2 answers:

mihalych1998 [28]2 years ago

6 0

Newton's Second law of motion:

Force = (mass) x (acceleration)

Force = (15kg) x (8m/s²) = 120 kg-m/s² = 120 newtons

USPshnik [31]2 years ago

4 0

The equation for force is force is equal to mass times acceleration or $F = ma$ where:

- Force or F is measured in Newtons or N (which is also kilogram-meter per second squared),
- Mass or m is measured in kilograms or kg, and acceleration is measured in meters per second or m/s2.

This leads to an answer in kilogram-meter per second squared or kg- m/s2So if we multiply a body with a mass of 15 kilograms to a rate of 8 meters per second, we get 120 kilogram-meter per second squared.

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Are the days with the longest daylight always in the summer? Are the days with the shortest daylight always in the winter? Expla

jok3333 [9.3K]

The day with the longest daylight is the beginning of Summer.
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The day with the shortest daylight is the beginning of Winter.
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6 0

3 years ago

Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of

Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) High and low pressures in kilopascals:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{high} = 15.999\,kPa$

$p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{low} = 10.666\,kPa$

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures in pounds per square inch:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}$

$p_{high} = 2.320\,psi$

$p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}$

$p_{low} = 1.547\,psi$

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures in meter water column in meters water column:

We can calculate the equivalent water column of a mercury column by the following relation:

$\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}$

$h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}$ (Eq. 1)

Where:

$\rho_{w}$ , $\rho_{Hg}$ - Densities of water and mercury, measured in kilograms per cubic meter.

$h_{w}$ , $h_{Hg}$ - Heights of water and mercury columns, measured in meters.

If we know that $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , $h_{Hg, high} = 0.120\,m$ and $h_{Hg, low} = 0.080\,m$ , then we get that: