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Natalka [10]
3 years ago
8

A toy car runs off the edge of a table that is 2.225 m high. If the car lands 0.400 m away from the base of the table, how fast

was the car going when it fell off?
Physics
1 answer:
ruslelena [56]3 years ago
8 0

Answer:

0.594 m/s

Explanation:

First, find the time it takes to land.

Given, in the y direction:

Δy = 2.225 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.225 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.674 s

Next, find the horizontal velocity.

Given, in the x direction:

Δx = 0.400 m

a = 0 m/s²

t = 0.674 s

Find: v₀

Δx = v₀ t + ½ at²

(0.400 m) = v₀ (0.674 s) + ½ (0 m/s²) (0.674 s)²

v₀ = 0.594 m/s

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Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
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7800kg/m³

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Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

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A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

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