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3 years ago

E) the time it takes the car to reach 20.0ms-1

Physics

1 answer:

Galina-37 [17]3 years ago

8 0

4.2 thats the answer

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Explanation:

I got it off a Quizlet hope this helps!!

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3 years ago

Anyone know this... pls help its for a grade

hodyreva [135]

His average speed is 45 miles an hour

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3 years ago

The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of th

navik [9.2K]

Answer:

(a) the input force is 36.56 N

(b) the input force is 37.49 N

Explanation:

Given;

density of hydraulic oil, ρ = 8.53 x 10² kg/m³

radius of plunger, r₁ = 0.135 m

radius of piston, r₂ = 5.43 x 10⁻³ m

Part (a) The input force needed to support 22600-N weight, when the bottom surfaces of the piston and plunger are at the same level;

$P =\frac{F}{A}$

Where;

P is pressure

F is force

A is circular area = πr²

$\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1* r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N$

Part (b) The input force needed to support 22600-N weight, when the bottom surface of the output plunger is 1.20 m above that of the input plunger

$P_2 = P_1 + \rho gh$

But, F = PA and A = πr²

$F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N$

4 0

3 years ago

Glaciers store about 72% of the world's freshwater false are true

IRISSAK [1]

The answer to this question is True.

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3 years ago

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 * t^2 to s

Mila [183]

Answer:

70560 m

Explanation:

The formula to calculate the distance travelled during a free fall motion is

$d=-\frac{1}{2}gt^2$

where

d is the distance travelled

g = -9.8 m/s^2 is the acceleration due to gravity

t is the time

In this situation,

t = 120 s

Therefore the distance travelled after 120 s is

$d=-\frac{1}{2}(-9.8)(120)^2=70560 m$

6 0

4 years ago

E) the time it takes the car to reach 20.0ms-1​

E) the time it takes the car to reach 20.0ms-1