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3 years ago

12

If the speed of a car is increased by 80%, by what factor will its minimum braking distance be increased, assuming all else is t

he same? Ignore the driver's reaction time.

1 answer:

Vikki [24]3 years ago

6 0

Answer:

$KE_2=3.24 \times KE_1$

Explanation:

given,

Speed of car = v

new speed of car v'= 1.8 v

Kinetic energy of the car

$KE_1 =\dfrac{1}{2}mv^2$

new Kinetic energy of energy

$KE_2 =\dfrac{1}{2}mv'^2$

$KE_2=\dfrac{1}{2}m(1.8v)^2$

$KE_2=3.24 \times \dfrac{1}{2}mv^2$

$KE_2=3.24 \times KE_1$

hence, the kinetic energy is increased by the factor of 3.24.

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3 years ago

Read 2 more answers

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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A cyclist travels 1850 m north and then 1200 m south. Find both the distance it has traveled and the magnitude of its displaceme

finlep [7]

Answer:

The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.

Explanation:

Distance refers to the length between any two points in space, while displacement refers to the distance from a start position to an end position regardless of the path.

In other words, distance refers to how much space an object travels during its movement; is the quantity moved. It is also said to be the sum of the distances traveled. The distance traveled by a mobile is the length of its trajectory and it is a scalar quantity. In this case, the distance is calculated as:

1850 m + 1200 m= 3,050 m

Displacement refers to the distance and direction of the final position from the initial position of an object. The displacement effected is a vector quantity. The vector representing the displacement has its origin in the initial position, its end in the final position, and its module is the distance in a straight line between the initial and final positions. That is, when expressing the displacement it is done in terms of the magnitude with its respective unit of measurement and the direction because the displacement is a vector type quantity. Mathematically, the displacement (Δd) is calculated as:

Δd= df - di

where df is the final position and di is the initial position of the object.

In this case, the displacement is calculated as:

1850 m - 1200 m= 650 m

Since the distance to the north is greater, the direction of travel will be to the north.

<u><em>The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.</em></u>

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2 years ago