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sertanlavr [38]

3 years ago

13

The 94-lb force P is applied to the 220-lb crate, which is stationary before the force is applied. Determine the magnitude and d

irection of the friction force F exerted by the horizontal surface on the crate. The friction force is positive if to the right, negative if to the left.

1 answer:

Makovka662 [10]3 years ago

8 0

Answer

given,

force = 94 lb

weight of crate = 220 lb

Assuming the static friction be equal = 0.47

kinetic friction = 0.36

Maximum force applied to move the object is when object is just start to move.

F = μ N

F = 0.47 x 220

F = 103.4 lb

As the frictional force is more than applied then the object will not move.

so, the friction force will be equal to the force applied on the object that is equal to 94 lb.

hence, the direction of force will left.

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3 years ago

You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba

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Answer:

The ball reaches Barney head in $t = 8 \ s$

Explanation:

From the question we are told that

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The height considered is $h = 196 \ m$

The horizontal velocity of the large object is $v_h = 8.50 \ m/s$

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Here s is the distance of the object from Barney head ,

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So

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So

$196 = -14.7 t + \frac{1}{2} * 9.8 * t^2$

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3 years ago

Which force controls the size of an atomic nucleus?

IrinaVladis [17]

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Explanation:

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2 years ago

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

AlekseyPX

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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:

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

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,

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

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

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3 years ago

Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure =

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3 years ago