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nasty-shy [4]
3 years ago
13

What is the average acceleration given by this graph:

Physics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

Explanation:

There is no graph.

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All digits shown on measuring device, plus one estimated digit, are considered_______.
vichka [17]

Significant

Explanation:

All the digits shown on a measuring device and one estimated digit are all considered to be significant.

A significant digit is a set of values that shows how precise a measurement is reported.

Measuring devices such as calculators gives their values in significant digits.

  • Non- zero digits in a measurement are always significant.
  • Zero's before a decimal are not significant. Those after a number in a decimal are significant.
  • Zero's between digits are significant.

learn more:

Significant digits brainly.com/question/2743055

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3 0
3 years ago
A 40 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 3.7 m/s just before hitting the ground.
antiseptic1488 [7]

Answer:

A. -5488J

B. 273.8J

C. 372.44N

Explanation:

Given:

m = 40kg

h = 14 m

v= 3.7 m/s

Part(a)

The change in the potential energy of the bear Earth system during the slide

AU = -mgh = -40(9.8) (14) = -5488 J

Part(b)

The kinetic energy of the bear just before hitting the ground is

Ks 1/2 mV^2= (40)(3.7)2 = 547.6 /2 = 273.8J

Part(c)

The change in the thermal energy of the system due to friction is

AEth = fxh=-(AK +AU) = 5488– 273.8 = 5214.2 J

The average frictional force that acts on the sliding bear is

F = Eth / 14= 5214.2/14 =372.44N

4 0
3 years ago
Read 2 more answers
A spring with k = 53 N/m hangs vertically next to a ruler. The end of the spring is next to the 18 cm mark on the ruler. If a 2.
Anarel [89]

Answer:

1.07 m

Explanation:

x = Compression of the spring

k = Spring constant = 53 N/m

Initial length = 18 cm

P = Kinetic energy

K = Kinetic energy

At the lowest point of the mass the energy conservation is as follows

P_{ig}+P_{is}+K_i=P_{fg}+P_{fs}+K_f\\\Rightarrow mgx+0+0=mgx+\frac{1}{2}kx^2\\\Rightarrow x=\frac{2mg}{k}\\\Rightarrow x=\frac{2\times 2.4\times 9.81}{53}\\\Rightarrow x=0.89\ m

At its lowest position the mark on the ruler will be

x_f=0.18+0.89\\\Rightarrow x_f=1.07\ m

The spring line will end up at 1.07 m

4 0
3 years ago
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