A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D
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:
(a). The spring compressed is .
(b). The acceleration is 1.5 g.
Explanation:
Given that,
Acceleration = a
mass = m
spring constant = k
(a). We need to calculate the spring compressed
Using balance equation
....(I)
The spring compressed is .
(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,
The compression is given by
Here, acceleration is zero
So,
We need to calculate the acceleration
Put the value of x in equation (I)
Hence, (a). The spring compressed is .
(b). The acceleration is 1.5 g.