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2 years ago

8

Air escapes from a balloon at a rate of 2 60 ( ) 1 R t t   3 ft / min , where t is measured in minutes. How much air (in 3 ft

) escapes during the first minute?

1 answer:

kotykmax [81]2 years ago

4 0

Answer:

Given R (t) = 60/(1+t²), ft /min

To find the amount of air in ft³, during the first min,

R (t) = 60/(1+t²

at t=1min, R is the air amount in ft³/min

Take the integral, and evaluate over [0,1]

Integral Of R (t) = 60/(1+t²) = 60 tan⁻¹(t),

60(Tan⁻¹(1) - Tan⁻¹(0)) = 60(pi/2) = 30 π

Therefore, it means that in the first minute, 30π ft³ of air escaped.

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

2 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

2 years ago

What is the approximate pressure of a storage cylinder of recovered r-410a that does not contain any non-condensable impurities

Aliun [14]

Answer:

173psig

Explanation:

The storage cylinder of recovered R-410A is mixture of difluoromethane and pentafluoroethane which is used as a refrigerant in air conditioning application. The refrigeration sector has low temperatures for installation. The pressure of cylinder at 80 F will be 173 psig. The pure refrigerants have inside a container have saturation temperature which is equal to ambient temperature.

8 0

2 years ago

If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they? (

valkas [14]

The gravitational force between two masses m₁ and m₂ is
$F=G \frac{m_{1} m_{2}}{d^{2}}$
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm

8 0

2 years ago

The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi

ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done = mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = $= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})$

kinetic energy = $= \frac{1}{2}*4*(1.5^{2}-11^{2})$ = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0

2 years ago