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klio [65]
3 years ago
5

A jet lands with a speed of 100 m/s and can accelerate uniformly at a rate of -5.0 m/s^2 as it comes to rest. What is the minimu

m length runway to land this plane? A. 1000 m
B. 20 m
C. 675 m
D. 500 m
Physics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

A ) 1000 m.

Explanation:

Here initial velocity u = 100 m /s

Final velocity v = 0

Acceleration a = -5 ms⁻²

Distance travelled = S

v² = u² + 2aS

0 = (100)² -2 x 5 S

S = 10000/ 10

=1000 m.

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Answer:

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Explanation:

4 0
2 years ago
The half life of radium 222 is 38 seconds if you had a 12 gram sample how much would be left after 72 seconds
poizon [28]
<span>3.2 grams The first thing to do is calculate how many half lives have expired. So take the time of 72 seconds and divide by the length of a half life which is 38 seconds. So 72 / 38 = 1.894736842 So we're over 1 half life, but not quite 2 half lives. So you'll have something less than 12/2 = 6 grams, but more than 12/4 = 3 grams. The exact answer is done by dividing 12 by 2 raised to the power of 1.8947. So let's calculate 2^1.8947 power = 12 g / (e ^ ln(2)*1.8947) = 12 g / (e ^ 0.693147181 * 1.8947) = 12 g / (e ^ 1.313305964) = 12 g / 3.718446464 = 3.227154167 g So rounded to 2 significant figures gives 3.2 grams.</span>
5 0
3 years ago
A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an
Mademuasel [1]

Answer:

The angular acceleration is 0.209\ rad/s^2

Explanation:

Given that,

Angular velocity, \omega_{i} = 30.0\ rpm

Angular velocity, \omega_{f} = 50.0\ rpm

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

\alpha=\dfrac{\Delta \omega}{\Delta t}

\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}

\alpha=\dfrac{50.0-30.0}{10.0}

Now, we change the angular velocity in rad/s.

\omega=20\times\dfrac{2\pi}{60}

\omega=2.09\ rad/s

\alpha=\dfrac{2.09}{10.0}

\alpha=0.209\ rad/s^2

Hence, The angular acceleration is 0.209\ rad/s^2

5 0
3 years ago
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What is the weight of a 42 kg object if the object was on the moon?
sveticcg [70]
(1.6 m/s²)(42 Kg)= 80 N
4 0
3 years ago
The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco
Pani-rosa [81]

Answer:

  160 W

Explanation:

Power is the ratio of work to time:

  (1600 J)/(10 s) = 160 J/s = 160 W

7 0
2 years ago
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