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Sergeeva-Olga [200]

3 years ago

9

A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for

this vehicle to maintain a 70 mi/h speed on a 5% upgrade through an air density of 0.002045 slugs/ft3?

1 answer:

emmasim [6.3K]3 years ago

4 0

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

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A semi has a net force of 13,000 N and is accelerating at a rate of 6.5 m/s^2. What is the mass of the semi? *

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2 years ago

Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,

insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction $\mu_{k} = 0.250$

Zak's speed $v = 3 \frac{m}{s}$

Gravitational acceleration $g = 9.8$ $\frac{m}{s^{2} }$

Work done by frictional force is given by,

$W = \Delta K$

$\mu _{k} mg d = \frac{1}{2} m v^{2}$

$d = \frac{v^{2} }{2 g \mu _{k} }$

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2 years ago

A speed-time graph is shown below:

Juliette [100K]

Answer:

It traveled 4 centimeters.

Explanation:

In a speed versus time graph, the distance travelled is given by the area under the graph.

In this graph we have the following:

- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s

- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s

Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:

$d=v\Delta t=(1)(4-0)=4 cm$

4 0

3 years ago