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Sliva [168]
3 years ago
12

Calculate the force needed to bring a 1,019-kg car to rest from a speed of 29 m/s in a distance of 118 m

Physics
1 answer:
son4ous [18]3 years ago
8 0
Acceleration = ▵v/▵t
Time = d/v
Fisrt calculate time : ( 118/29 ) = 4 seconds
Then calculate acceleration
A = 29/4 = 7.25 m/s²
Now the force.
Force = mass * acceleration.
F= 1,019 * 7.25
F= 7,387 N
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AveGali [126]

Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

Hypotenuse^2 = opposite^2 + adjacent^2

The hypotenuse is the length of the ladder.

Hypotenuse = \sqrt{opposite^2 + adjacent^2}

L^2 = BC^2 + (24 + AE)^2........1

Triangle ABC is similar to triangle AEF, hence:

\frac{BC}{8} = \frac{AE + 24}{AE}

BC = \frac{8(AE + 24)}{AE}.................2

Substitute 2 into 1

L^2 = (\frac{8(AE + 24)}{AE})^2 + (24 + AE)^2

Let AE = x

L^2 = (\frac{8x + 192}{x})^2 + (24 + x)^2

    = (8 + \frac{192}{x})^2 + (24 + x)^2

Minimize L with respect to x.

2\frac{dL}{dx} = 2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)

       =

5 0
3 years ago
the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of spar
adell [148]

Answer:

19.3

Explanation:

Assuming we have to find Specific gravity of gold.

As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object

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now we have

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now weight of the liquid displaced is given by

weight of water displaced = (1 g/cm^3)(0.55ml)g

now we have

specific gravity = (19.3×0.55)/(1×0.55)

specific gravity= 19.3

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