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3 years ago

A 2000 kg car experiences a constant braking force

1 answer:

4 0

In that case, there are three possible scenarios:

-- If the braking force is less than the force delivered by the engine,
then the car will continue to accelerate, and the brakes will eventually
overheat and erupt in flame.

-- If the braking force is exactly equal to the force delivered by the engine,
then the car will continue moving at a constant speed, and the brakes will
eventually overheat and erupt in flame.

-- If the braking force is greater than the force delivered by the engine,
then the car will slow down and eventually stop. If it stops soon enough,
then the absorption of kinetic energy by the brakes will end before the
brakes overheat and erupt in flame. Even if the engine is still delivering
force, the brakes can be kept locked in order to keep the car stopped ...
They do not absorb and dissipate any energy when the car is motionless.

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A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal

melamori03 [73]

Answer:

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

$\sum M__A }=0$

- 40 (180) - (80) $F_E$ Cos 37° + 55

-7200 + $F_E$ (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79 $F_E$

- 30.79 $F_E$ = 7200

$F_E$ = $-\frac{7200}{30.79}$

$F_E$ = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

$\sum f_y = 0$

$F_E$ sin 37 + $F_A$ cos 20 - $F_G$ = 0

- 233.84 sin 37 + $F_A$ cos 20 - 180 = 0

$F_A$ cos 20 = 320.73

$F_A$ = $\frac{320.73}{cos 20}$

$F_A$ = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

$\sum f_y = 0$

$F_E$ cos 37 - $F_{CB} + F_A$ sin 20° = 0

-233.84 cos 37 - $F_{CB}$ + 341.31 sin 20° = 0

-186.75 - $F_{CB}$ + 116.74 = 0

- $F_{CB}$ -70.01 = 0

- $F_{CB}$ = 70.01

$F_{CB}$ = - 70.01 lb

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

7 0

4 years ago

How fast is sound when traveling through water?

IgorLugansk [536]

Sound travels 1,480 meters per second, which is about 4.3 times as fast as air. Sound travels much slowly in air. This has to do with the frequency, intensity and amplitude of waves, which are affected differently in water and air.

4 0

3 years ago

When enough energy is produced to a gas the electrons on the gaseous atoms being to detach themselves leaving ions of the typica

Luba_88 [7]

Answer:

it's called ionization

4 0

3 years ago

What is the function of the mitochondria?

Aliun [14]

The most prominent roles of mitochondria<span> are to produce the energy currency of the cell, ATP </span>

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3 years ago

Read 2 more answers

The isotope 23893Np has a half-life of 2.0 days. If 4.00 grams are produced at noon on Monday, what will be the mass of neptuniu

viva [34]

Answer:

0.25 gram of neptunium is remaining

Explanation:

First we calculate the no. of half lives passed. For that we have formula:

n = t/T

where,

n = no. of half lives passed = ?

t = total time passed = 8 days (From Monday noon to Tuesday noon of following week)

T = Half Life Period = 2 days

Therefore,

n = 8 days/2 days

n = 4

Now, for the remaining mass of neptunium, we use the formula:

m = (mi)/(2)^n

where,

mi = initial mass of neptunium = 4.00 grams

m = remaining mass of neptunium = ?

Therefore,

m = 4 grams/2⁴

3 0

3 years ago