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podryga [215]
3 years ago
13

Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest p

oint. We want to find how long it takes for the ball to reach this point, and how high the ball goes. (e) What is the component of the ball’s velocity at the instant when the ball reaches its highest point (the end of this time interval)?
Physics
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:A ball is kicked from a location < 9, 0, -8 > (on the ground) with initial velocity < -11, 18, -5 > m/s. The ball's speed is low enough that air resistance is negligible. What is the velocity of the ball 0.5 seconds after being kicked? (Use the Momentum Principle!) = m/s In this situation (constant force), which velocity will give the most accurate value for the location of the ball 0.5 seconds after it is kicked? The arithmetic average of the initial and final velocities. The final velocity of the ball. The initial velocity of the ball. What is the average velocity of the ball over this time interval? avg = Use the average velocity to find the location of the ball 0.5 seconds after being kicked: = m Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest point. We want to find how long it takes for the ball to reach this point, and how high the ball goes. What is the y-component of the ball's velocity at the instant when the ball reaches its highest point (the end of this time interval)? vyf = m/s. Fill in the missing numbers in the equation below (update form of the Momentum Principle): mvyf = mvyi + Fnet,y?t m = m + ?mg?t How long does it take for the ball to reach its highest point? ?t = s. Knowing this time, first find the y-component of the average velocity during this time interval, then use it to find the maximum height attained by the ball: ymax = m. Now take a moment to reflect on the reasoning used to solve this problem. You should be able to do a similar problem on your own, without prompting. Note that the only equations needed were the Momentum Principle and the expression for the arithmetic average velocity.

Explanation:

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A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro
NARA [144]

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width  = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

5 0
3 years ago
Some unpolarized light has an intensity of 1375 W/m2 before passing through three polarizing filters. The transmission axis of t
chubhunter [2.5K]

Answer:

intensity of the light that emerges from the three filters is 560.80 W/m²

Explanation:

Given data

intensity  I = 1375 W/m2

angle 1 = 31.0°

angle 2  = 41.0°

to find out

intensity of the light that emerges from the three filters

solution

we know intensity of light pass 1st polarize = I/2 = 1375 / 2 = 687.5  W/m2

so intensity after 2nd polarize pass = I 1st cos²(θ)

I 2nd = 687.5 cos²(31) = 687.5 ( 0.836754)  = 575.27 W/m2

and

intensity after 3rd polarize pass = I 2nd cos²(θ)

I 3rd = 575.27 cos²(41) = 575.27 (0.974839) = 560.80 W/m2

so that intensity of the light that emerges from the three filters is 560.80 W/m²

4 0
4 years ago
A car traveling 75 km/h slows down at a constant 0.50 m/s2 just by "letting up on the gas." calculate (a) the distance the car c
Serjik [45]

Solution:

At 1st convert km/h to m/s. 1 km = 1000 m, 1 h = 3600 s, 1 km/m = 1000/3600 = 5/18 m/s  

Initial velocity = 75 * 5/18 = 20.8 m/s  

The car’s velocity decreases from 20.8 m/s to 0 m/s at the rate of 0.5 m/s each second. We have the final velocity, initial velocity, and the acceleration.  

Now according to the equation determine the distance.  

vf^2 = vi^2 + 2 * a * d  

a = -0.5 m/s^2  

0 = 20.8^2 + 2 * -0.5 * d  

so d = 431.64 m  

since we have the final velocity, initial velocity, and the acceleration. Use the following equation to determine time.  

vf = vi + a * t  

0 = 20.8 – 0.5 * t  

Solve for t = 41 seconds  

(c) the distance travels by it during the first and fifth second are.  

d = vi * t + ½ * a * t^2  

d1 = 20.8 * 1 – ½ * 0.5 * 1^2 = 20.55 m  

The easiest way to the distance for the 5th second is:

d = vi * t + ½ * a * t^2, a = -0.5  

d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m  

d6 = 20.8 * 6 – ½ * 0.5 * 6^2 =  106.8m

d6 – d5 = 15.3 m  

this is the required solution.


3 0
3 years ago
What secondary color is made by mixing red and green in the rgb additive-color scheme for light?.
neonofarm [45]

Explanation:

Looking at the placement of primary colors in the RGB color wheel, the mixing of the red and green colors will create Yellow secondary color.

7 0
2 years ago
Can yall Please fill out them blanks i need help it's due Today.
hjlf

Answer:

tech ed class

Explanation:

4 0
3 years ago
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