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3 years ago

11

Once broken into parts curved motion can be worked as ________________ problems along both axes.

1 answer:

vovikov84 [41]3 years ago

8 0

Answer:

projectile motion

Explanation:

i am not sure sorry

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Velocity of B rays ?<br>

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is from 9 x 107 m/sec to 27 x 107 m/s

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(FIRST TO ANSWER CORRECTLY WILL BE THE BRAINIEST!!!)How will the temperatures of the water in the beakers compare if an equal am

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the answer is the temperatures of both beakers' water will increase by the same amount...

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Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

2)

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

#LearnwithBrainly

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3 years ago

Blaine steps onto a ski slope with an angle of 25°. There is a coefficient of kinetic friction of 0.15 between him and the groun

Ymorist [56]

θ = angle of the incline surface from the horizontal surface = 25⁰

μ = Coefficient of friction = 0.15

m = mass of the person = 65 kg

$f_{k}$ = kinetic frictional force acting on the person as he slides down

mg = weight of the person acting in down direction

$F_{n}$ = normal force by the incline surface on the person

the free body diagram showing the forces acting on the person is given as

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