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olchik [2.2K]
3 years ago
6

Problem 1 An object with m1 = 5kg is attached to a spring of negligible mass. This mass/spring combination is then slid horizont

ally on a frictionless surface with a velocity of 5 m/s towards a stationary object with m2 = 6kg. Upon impact, the spring compresses, then we examine two cases. First, find the velocities of the two objects assuming the spring completely relaxes again after the interaction. Second, assume that m2, after they separate, slides up a frictionless incline. (a)What is the relative speed of the masses when the spring is maximally compressed?

Physics
1 answer:
harkovskaia [24]3 years ago
7 0

Answer and Explanation:

The answer is attached below

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a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by
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The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

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2 years ago
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Answer:

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2 years ago
A sound wave is produced when a medium begins to _________
Natasha2012 [34]
The answer is D) vibrate. vibrations are what cause sound. Consequently, this is why there is no sound in space, because there is no medium for the for which sound to travel 
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A COPPER WIRE OF LENGTH 4M AND AREA OF CROSS-SECTION 1.2CM^2 IS STRECHED WITH A FORCE 4.8×10^3 N . IF YOUNGS MODULAS FOR COPPER
VARVARA [1.3K]

Stress = force / area

= 4.8 x 10³ / 1.2 x 10^-4

= 4 x 10⁷ N /m²

YOUNGS MODULAS = stress / strain

= 4 x 10⁷ / 1.2 x 10^11

= 3.3 x 10^-4

INCREMENT OF LENGTH = longitudinal length x intitial length

= ( 3.3 x 10^-4 ) x 4

= 13.2 x 10^-4 m

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6 0
2 years ago
Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.70 m tall.
cupoosta [38]

Answer:

P_2 - P_1 = 1.8 * 10^4\ Pa

Explanation:

Given

Height (h) = 1.70m

Required

Determine the difference in the blood pressure from feet to top

This is calculated using Pascal's second law.

The second law is represented as:

P_2 = P_1 + pgd

Subtract P1 from both sides

P_2 - P_1 = pgd

Where

p = blood\ density = 1.06 * 10^3kg/m^3

g = acceleration\ of\ gravity = 9.8N/kg

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So, the expression becomes:

P_2 - P_1 = 1.06 * 10^3 * 9.8 * 1.70

P_2 - P_1 = 17659.6Pa

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Hence, the difference in blood pressure is approximately 1.8 * 10^4\ Pa

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3 years ago
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