An elevator moves from rest to 2 m/s over 8 seconds. What is the elevator's acceleration?

A 46 g domino slides down a 30 degrees incline at a constant speed. What is the coefficient of friction?

blondinia [14]

Answer:

40

Explanation:

30

6 0

3 years ago

A light-year is a unit of a. time. c. mass. b. distance. d. density. please select the best answer from the choices provided a b

11111nata11111 [884]

Answer:

distance

Explanation:

it is the distance traveled by light in one year

3 0

1 year ago

Read 2 more answers

A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the

const2013 [10]

Answer:

The equation for the object's displacement is $u(t)=0.583cos11.35t$

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

$k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in$

The angular speed is:

$w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s$

The damping force is:

$F_{D} =cu$

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

$c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in$

The critical damping is equal:

$c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in$

Like cc>c the system is undamped

The equilibrium expression is:

$u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t$

3 0

3 years ago

Explain why a moving object cannot come to a stop instantaneously (in zero seconds). Hint: Think about the acceleration that wou

gizmo_the_mogwai [7]

To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation:F=maSo when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, mv⃗ mv→, we can use the following equation to calculate force:F=p/hHowever, because the force is nowhere close to infinity, time will never tend to zero either, which means that you cannot come to an instantaneous stop.

7 0

3 years ago

A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc

Ray Of Light [21]

Answer:

A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.

Explanation:

A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.

If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

vₓ = 16.6 m/s i

In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s

Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:

$v =\sqrt{(16.6m/s)^{2}+ (-17.2m/s)^{2}} = 23.9 m/s$

v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ = $\frac{-17.2}{16.6} =-1.036$

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0

3 years ago