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3 years ago

15

A jet starts from rest and accelerates to a speed of 100 m/s in 10 seconds before taking off. how far does it travel before taki

ng off?

2 answers:

Sergio [31]3 years ago

8 0

Answer:

$\boxed {\tt 1,000 \ meters}$

Explanation:

We want to find the distance the jet travels. Distance can be found by multiplying the speed by the time.

$d=s*t$

The speed is 100 meters per second.

The time is 10 seconds.

$s= 100 \ m/s\\t= 10 \ s$

Substitute the values into the formula.

$d=100 \ m/s * 10 \ s$

Multiply. Note that the seconds, or "s" will cancel each other out.

$d= 100 \ m * 10$

$d= 1,000 \ m$

The jet travels <u>1,000 meters</u> before taking off.

dlinn [17]3 years ago

7 0

Answer:

1000 m

Explanation:

If the jet accelerates to a speed of 100m/s in 10 sec then till 10 sec it traveled 1000 m.

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7 0

1 year ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago