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kolbaska11 [484]
3 years ago
10

Please Please Please Can Help Me On This Question!!!!! I Give Thanks!!!! Please Do 1-4!!!!

Physics
2 answers:
ololo11 [35]3 years ago
5 0
1.add the amount of the diagram which is M+Y then dived the answer you get.
yuradex [85]3 years ago
4 0

Answer:

add the amount of the diagram which is M+Y then dived the answer

Explanation:

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Which statement best reflects the approach of Gestalt psychology?
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Gestalt psychology is a a theory of mind of the Berlin School of experimental psychology. It tries to understand the laws of our ability to acquire and maintain meaningful perceptions in an apparently chaotic world. It proposes that the whole of an object or scene is more important than its individual parts.
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A 5.5kg mass is pushed with a force of 31N across a table having μk of 0.350. Find how fast it will accelerate, taking friction
Ipatiy [6.2K]

Answer:

3.66m/s^2

Explanation:

First, we need to find the friction which is F=u*N.

After that, find the resultant force by substarcting the friction from the forward force.

Lastly,using the formula F=ma, substitute in the known values of F and m to find a

4 0
3 years ago
The driver of a car slams on the brakes when he sees a tree blocking the road. the car slows uniformly with acceleration of -5.9
Hatshy [7]
Let u =  the speed of the car at the instant when braking begins.

The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.

Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
      u = 27.2546 m/s

Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
   = 2.7696 m/s

Answer: 2.77 m/s (nearest hundredth)

4 0
3 years ago
You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it
VladimirAG [237]

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg  = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd  + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

4 0
3 years ago
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