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3 years ago

How does a jack spread out the work over a large distance?

1 answer:

6 0

Each complete rotation of a jack handle applies a small force over a large distance. A small force exerted over a large distance becomes a large force exerted over a short distance. Each rotation lifts the car only a very short distance.

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Robert is accelerating a cart at the rate of 2.2 meters/second2. If the mass in the cart is doubled, and the net force is also d

skad [1K]

F = ma
double f and m gives
(2f) = (2m)a
notice a is unchanged = 2.2
m/s^2

4 0

4 years ago

Read 2 more answers

You measure a watch's hour and minute hands to be 7.4 mm and 12.1 mm long, respectively. Part A In one day, by how much does the

pychu [463]

Answer:

109.385m

Explanation:

In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius

7.4 * 4π = 93 mm

In 1 day, the minute hand travels 24*60 = 1440 circles, or 1440 * 2π = 2880π rad in angular. The distance it travels is

12.1 * 2880π = 109478 mm

So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is

109478 - 93 = 109385 mm or 109.385 m

3 0

4 years ago

Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm.

jasenka [17]

Answer:

$P_{1}-P_{2}=12675Pa$

Explanation:

From the equation of continuity we know that:

$v_{1}A_{1}=v_{2}A_{2}\\Given \\r_{1}=2.0cm\\r_{2}=1.0cm\\v_{1}=1.3m/s\\Density of water p=1000kg/m^{3}\\Now\\A_{1}=\pi r_{1}^{2} \\A_{2}=\pi r_{2}^{2}\\ Therefore\\v_{2}=v_{1}\frac{A_{1}}{A_{2}}\\v_{2}=v_{1}\frac{\pi r_{1}^{2} }{\pi r_{2}^{2}}\\v_{2}=v_{1}\frac{r_{1}^{2} }{r_{2}^{2}}\\v_{2}=1.3*\frac{2.0^{2} }{1.0^{2} } \\v_{2}=5.2m/s\\$

From Bernoulli equation we know that:

$P_{1}+1/2pv_{1}^{2}+pgy_{1}=P_{2}+1/2pv_{2}^{2}+pgy_{2}\\$

Now assuming $y_{1}=y_{2}$

$P_{1}+(1/2)pv_{1}^{2}=P_{2}+(1/2)pv_{2}^{2}\\ P_{1}-P_{2}=1/2pv_{2}^{2} -1/2pv_{1}^{2}\\ P_{1}-P_{2}=1/2p(v_{2}^{2}-v_{1}^{2} )\\ P_{1}-P_{2}=1/2*1000(5.2^{2}-1.3^{2} )\\ P_{1}-P_{2}=12675Pa$

5 0

3 years ago

A woman carries a 10kg box up a set of 5m high stairs and then down a 12m long hallway. How much work does she do on the box?

nika2105 [10]

Answer: 1666J

Explanation:

Given that,

Mass of box (m) = 10kg

Total distance covered by box (h)

= (5m + 12m)

= 17m

work done on the box = ?

Work is done when force is applied on an object over a distance. Hence, the magnitude of work done on the box depends on its mass (m), distance covered (h), and acceleration due to gravity (g)

(g has a value of 9.8m/s²

i.e Work = mgh

Work = 10kg x 9.8m/s² x 17m

Work = 1666J

Thus, 1666 joules of work was done by the woman on the box.

7 0

3 years ago

In a machine, work output is less than work input because some energy is converted into thermal energy. true or false.

tamaranim1 [39]

True ..........................

7 0

4 years ago

Read 2 more answers