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3 years ago

12

in your own words provide two advantages of using meters as a measurement of length rather than old measurements of length such

as hands or steps

1 answer:

Arlecino [84]3 years ago

3 0

Answer:

First, the meters is the unit used in almost all the world today.

Then using meters as your unit will be easier to read and interpret for a larger number of people than if you use an older system, like hands or steps.

And because of the fact that almost all the world uses the meter as the unit for length, almost all the equations and formulas are written in meters.

Precision: The meter is exactly defined as:

"The length of the path traveled by light in a vacuum in 1/299,792,458 of a second"

So the meter is an exact (and accepted) unit.

Then using this unit for measures of length also provides some " legitimacy" for you measure.

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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

The Kinetic Molecular Theory helps explain relationships between:

Aleks [24]

The correct answer is<span> gases, energy, temperature, phases

Gravity and nuclear forces are not encompassed in the kinetic molecular theory as it deals with movement and behavior of gas molecules. It does not include their conversion to other types of energy or anything similar. </span>

8 0

2 years ago

Read 2 more answers

Please do all of i will give you brainlest and thanks to best answer

vlabodo [156]

Answer:the answer is A

Explanation:

3 0

3 years ago

Ohms law is A.(R=E/W). B.(R=E/1). C.(E/Z). D.none of them

nikklg [1K]

Answer:

D. none of them.

Explanation:

This is because Ohm's law is:

Voltage = Current × Resistance

or,

V = IR

6 0

2 years ago

Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f

RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

$p_A = p_B$

Pressure is given by the ratio between force F and area A, so we can write

$\frac{F_A}{A_A}=\frac{F_B}{A_B}$

The force exerted on each piston is just equal to the weight of the corresponding mass: $F=W=mg$ , where m is the mass and g is the gravitational acceleration. So the equation becomes

$\frac{m_A g}{A_A}=\frac{m_B g}{A_B}$

Now we can rewrite the mass as the product of volume, V, times density, d:

$\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}$

We also know that $A_B = 2.0 m^2\\A_A = 1.0 m^2$

So we can further re-arrange the equation (and simplify g as well):

$\frac{V_A d_A}{1}=\frac{V_B d_B}{2}$

$\frac{d_A}{d_B}=\frac{V_B}{2V_A}$

We are also told that block B has bigger volume than block A: $V_B > V_A$ . However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0

3 years ago