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3 years ago

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A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

. Assuming the acceleration is constant, how far did it travel during those 2.0 s?

1 answer:

Dvinal [7]3 years ago

5 0

Answer:

Explanation:

Given

time taken $t=2\ s$

Speed acquired in 2 sec $v=42\ m/s$

Here initial velocity is zero $u=0$

acceleration is the rate of change of velocity in a given time

$a=\frac{v-u}{t}$

$a=\frac{42-0}{2}=21\ m/s^2$

Distance travel in this time

$s=ut+0.5at^2$

where

s=displacement

u=initial velocity

a=acceleration

t=time

$s=0+\0.5\times 21\times (2)^2$

$s=42\ m$

so Jet Plane travels a distance of 42 m in 2 s

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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

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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 light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = $85\frac{1000}{3600}=23.61\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s$

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s$

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2$

Emergency acceleration is 2.99 m/s² (magnitude)

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What is the kinetic energy of a ball that has a mass of 3kg and Is<br> moving at 5m/s?

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Answer:

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E kin: kinetic energy

m: mass

v: velocity

Explanation:

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An automobile travels on a straight road for 42 km at 45 km/h. it then continues in the same direction for another 42 km at 90 k

Nina [5.8K]

Answer:

a) The average velocity is v = (60 km/h ; 0)

b) The average speed is 60 km/h

Explanation:

The velocity is a vector that has a magnitude and direction. The average speed is the distance traveled over time without taking into account the direction of the motion.

a)The average velocity is calculated as the displacement over time:

v = Δx/Δt

where

v = velocity

Δx = final position - initial position = traveled distance relative to the center of the reference system.

Δt = final time - initial time (initial time is usually = 0)

We know that the displacement is 84 km but we do not know the time. It can be calculated from the two parts of the trip.

In part 1:

v = 45 km/h = 42 km / t

t = 0.93 h

In part 2:

v = 90 km/h = 42 km / t

t = 42 km / 90 km/h

t = 0.47 h

The time of travel is 0.47 h + 0.93 h = 1.4 h

The average velocity will be:

v = 84 km / 1.4 h = 60 km/h

Expressed as a vector in a 2-dimension plane:

v = (60 km/h; 0)

b) The average speed is calculated as the distance traveled over time. Note that in this case, the distance is equal to the displacement since the direction of the motion is always in one direction. But if the direction of the second part of the trip would have been the opposite to the direction of the first part, the displacement would have been 0 (final position - initial position = 0, because final position = initial position), then, the average velocity would have been 0. In change, the average speed would have been the distance traveled (84 km, 42 km in one direction and 42 km in the other) over time.

Then:

average speed = 84 km / 1.4 h = 60 km/h

c) see attached figure.

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3 years ago