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harina [27]
4 years ago
9

Three deer, A, B, and C, are grazing in a field. Deer B is located 62m from deer A at an angle of 51° north of West. Deer C is l

ocated 77° north of east relative to deer A. The distance between deer B and C is 95m. What is the distance between deer A and C?

Physics
1 answer:
Anna71 [15]4 years ago
8 0

Set deer A's position to be the origin. Let c be the distance from deer A to deer C. We're given that deer B is 95 m away from deer C, which means the length of the vector B-C is 95 (or C-B). Then

|B-C|^2=(B-C)\cdot(B-C)=B\cdot B-2B\cdot C+C\cdot C=|B|^2-2B\cdot C+|C|^2

|B-C|^2=|B|^2-2|B||C|\cos(180-77-51)^\circ+c^2

95^2=62^2-2(62)(c)\cos52^\circ+c^2

c^2-124\cos52^\circ c-5181=0\implies c=120\,\mathrm m

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An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp
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Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0
4 years ago
A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after
Aleks04 [339]

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g=9.81ms^{-2}

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

v=u+at

Using the formula and substitute the values

0=24.1-9.81t

Where g is negative because motion of ball is against gravity

24.1=9.81t

t=\frac{24.1}{9.81}=2.46s

3.y=y_0+ut+\frac{1}{2}at^2

Using the formula

y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2

y=30.1 m

4.Time of flight for red ball=3.77-2.9=0.87s

y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2

y'=22.0m

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2

0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2

0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2

0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t

-2.86105=-2.851t

t=\frac{2.86105}{2.851}=1.004 s

Hence,  1.004 s  after the blue ball is thrown  are the two balls in the air at the same height.

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