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3 years ago

While traveling along a highway, a driver slows from 27 m/s to 13 m/s in 11 seconds. What is the automobile’s acceleration?

Physics

1 answer:

Travka [436]3 years ago

4 0

Answer:

$a=-1.27\ m/s^2$

Explanation:

Given that,

The initial speed of a driver, u = 27 m/s

Final speed of the driver, v = 13 m/s

Time, t = 11 s

We need to find the automobile's acceleration. We know that the acceleration of an object is equal to the change in speed divided by time taken.

$a=\dfrac{13-27}{11}\\\\a=-1.27\ m/s^2$

So, the magnitude of automobile's acceleration is $1.27\ m/s^2$ .

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Two 100kg bumper cars are moving towards eachother in oppisite directions. Car A is moving at 8 m/s and Car B at -10 m/s when th

DerKrebs [107]

Answer:

$-10 m/s$

Explanation:

When two cars collide then the momentum of two cars will remains conserved

Mass of two cars = 100 kg

Speed of car A = 8 m/s

Speed of car B = - 10 m/s

After collision the speed of car B = +8 m/s

By momentum conservation equation

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$(100)(8)+(100)(-10)=(100v)+(100)(8)\\ v=-10 m/s$

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3 years ago

In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es

tigry1 [53]

Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

4 0

4 years ago

The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the

irga5000 [103]

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

$P_{abs}=P_{gauge}+P_{atm}$ (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

$P_{gauge}=\rho gh$ (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

$P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh$

$P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)$

$P_{atm}=95950Pa$

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

$S.G=\frac{\rho_{fluid}}{\rho_{water}}$

$\rho_{liquid}=S.G*\rho_{water}$

$\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}$

so:

$P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa$

$P_{abs}=137642.5 Pa$

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pashok25 [27]

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3 years ago

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pav-90 [236]

Answer:

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3 years ago