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Komok [63]
3 years ago
9

When a moving object collides with an object that isn’t moving, what happens to the kinetic energy of each object?

Mathematics
1 answer:
djyliett [7]3 years ago
6 0
The kinetic energy transfers from the moving ball to the non-moving ball.
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Evaluate the expression under the given conditions. tan(2); cos() = 5 13 , in quadrant i
Nataliya [291]

The solution to given expression tan(2θ) is 22.615°

For given question,

We have been given an expression tan(2θ)

Given that cos(θ) = 5/13, and θ is in quadrant 1.

We know that the trigonometric identity

sin²θ + cos²θ = 1

⇒ cos²θ = (5/13)²

⇒ sin²θ = 1 - 25/169

⇒ sin²θ = 169 - (25/169)

⇒ sin²θ = 144/169

⇒ sin(θ) = 12/13

We know that the identity cos(2x) = cos²x - sin²x

⇒ cos(2θ) = cos²θ - sin²θ

⇒ cos(2θ) = 25/169 - 144/169

⇒ cos(2θ) = -119/169

And sin(2x) = 2sin(x)cos(x)

⇒ sin(2θ) = 2sin(θ)cos(θ)

⇒ sin(2θ) = 2 × 12/13 × 5/13

⇒ sin(2θ) = 120/169

We know that, tan(x) = sin(x)/cos(x)

⇒ tan(2θ) = sin(2θ)/cos(2θ)

⇒ tan(2θ) = (120/169) / (-119/169)

⇒ tan(2θ) = 120 / (-119)

⇒ tan(2θ) = -1.008

Since θ is in quadrant 1, tan(2θ) = 1.008

⇒ 2θ = arctan(1.008)

⇒ 2θ = 45.23

⇒ θ = 22.615°

Therefore, the solution to given expression tan(2θ) is 22.615°

Learn more about the expression here:

brainly.com/question/14961928

#SPJ4

7 0
2 years ago
F(b) = 7b^3 +8b and g(b) = b ^2+ b - 10. What is f(b)-g(b)?
Elden [556K]

Answer:

{ \bf{f(b) - g(b) : }} \\ { \tt{ = ( {7b}^{3} + 8b) - ( {b}^{2}   + b - 10)}} \\  = ( {7b}^{3}  -  {b}^{2}  + 7b + 10)

7 0
3 years ago
Read 2 more answers
Is the first side “SW=XW” or “ST=XU”?
olya-2409 [2.1K]

Answer:

SW =XW and ST=XU and UW=TW

Step-by-step explanation:

All are true and correct since the two triangles are congruent

5 0
3 years ago
Read 2 more answers
Point T is at (-3, 8). What are the coordinates of T' after R(y-axis) o R(x-axis)?
Whitepunk [10]

Given:

Point is T(-3,8).

To find:

The coordinates of T' after R(y-axis)\circ R(x-axis).

Solution:

We know that, R(y-axis)\circ R(x-axis) means the figure reflected across the x-axis then reflected across y-axis.

If a figure reflected across x-axis, then

(x,y)\to (x,-y)

T(-3,8)\to T_1(-3,-8)

If a figure reflected across y-axis, then

(x,y)\to (-x,y)

T_1(-3,-8)\to T'(-(-3),-8)

T_1(-3,-8)\to T'(3,-8)

Therefore, the required point is T'(3,-8).

4 0
3 years ago
Look at this expression
vladimir2022 [97]

Answer:

OD22264 is the correct answer for that

4 0
3 years ago
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