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just olya [345]
3 years ago
5

Calculate the distance the gecko covers when sprinting from rest to 2 m/s in a time of 1.5 s.

Physics
1 answer:
KIM [24]3 years ago
8 0

Answer:

1.5 m

Explanation:

a = v/t

a = 2/1,5 = 4/3

x = 1/2 a t² + vt + x

x = 1/2 × 4/3 × 9/4 = 3/2 =1.5 m

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The diagram assigned B

explanation:

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3 years ago
A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu
kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the​ rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the​ cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) =  ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i)  work required to pump the contents to the​ rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the​ cone's rim

Extra work required = 28,914.229 lb ×  5ft = 144571.145 lb.ft

Total work required = (281,913.733 +  144571.145) lb.ft

                                 = 426,484.878 lb.ft

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3 years ago
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2 years ago
Is the magnitude of the acceleration of the center of mass of spool A greater than, less than, or equal to the magnitude of the
Anon25 [30]

Answer:

The acceleration of the centre of mass of spool A is equal to the magnitude of the acceleration of the centre of mass of spool B.

Explanation:

From the image attached, the description from the complete question shows that the two spools are of equal masses (same weight due to same acceleration due to gravity), have the same inextensible wire with negligible mass is attached to both of them over a frictionless pulley; meaning that the tension in the wire is the same on both ends.

And for the acceleration of both spools, we mention the net force.

The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.

For this system, the net force on either spool is exactly the same in magnitude because the net force is a difference between the only two forces acting on the spools; the tension in the wire and their similar respective weights.

With the net force and mass, for each spool equal, from

ΣF = ma, we get that a = ΣF/m

Meaning that the acceleration of the identical spools is equal also.

Hope this Helps!

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