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4 years ago

5

a girl pushes a cart to the left a 100-N force. a boy pushes it to the right with a 50-N force. The net force exerted on the car

t is

1 answer:

AURORKA [14]4 years ago

5 0

Since the net force is the amount of force favoring the side with the most force acted upon it,

100 - 50 = 50

50N to the left

Hope this helps!

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Which type of circuit is becoming more common in residential electrical design and construction?

jekas [21]

Communication circuit <em>(D)</em> is becoming more common in residential electrical design and construction.

LAN Ethernet cables, outlets, and even hubs and bridges, are being built into the walls of new homes, along with the usual electrical outlet wiring, to give the owner the networking infrastructure and internet access that everybody needs now ... without stringing a mess of cables on the floor and through doors all over the house.

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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

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3 years ago

A 1200-kg car initially at rest undergoes constant acceleration for 8.8 s, reaching a speed of 10 m/ s. It then collides with a

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Answer:

The final kinetic energy of the two-car system is 60,000 J.

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Given;

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time of motion, t = 8.8 s

final velocity of the car, v = 10 m/s

Apply the principle of conservation of kinetic energy; the initial kinetic energy is equal final kinetic energy.

$K.E_i = K.E_f\\\\K.E_f = \frac{1}{2}mv^2\\\\K.E_f = \frac{1}{2}(1200)(10)^2\\\\K.E_f = 60,000 \ J$

Therefore, the final kinetic energy of the two-car system is 60,000 J.

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3 years ago

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5207 because I said so

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Obligation of reasearchers to review true or fasle

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