Which element is malleable and ductile
1 answer:
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Refer to the diagram shown below.
In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.
The second airplane travels
b = 620*2.4 = 1488 mile
The angle between the two airplanes is
163° - 51.3° = 111.7°
Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles
Answer: 2335.4 miles
In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.
The second airplane travels
b = 620*2.4 = 1488 mile
The angle between the two airplanes is
163° - 51.3° = 111.7°
Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles
Answer: 2335.4 miles
Answer:
The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>
Explanation:
Given:
Time taken by the ball to reach maximum height is,
We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is,
Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is,
The negative sign is used as acceleration is a vector and it acts in the downward direction.
Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:
Where, 'u' is the initial velocity.
Plug in the given values and solve for 'u'. This gives,
Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s