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STatiana [176]
3 years ago
15

newton's 3rd law is known as the law of______? is it different objects or the same objects or objects in motion?

Physics
1 answer:
Levart [38]3 years ago
5 0

Answer:

th law of action and reaction

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8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.
mamaluj [8]

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle (p), measured in centimeters, is represented by the following formula:

p = 2\cdot (w+l) (1)

Where:

w - Width, measured in centimeters.

l - Length, measured in centimeters.

If we know that w = 6.4\,cm and l = 8.3\,cm, then the perimeter of the rectangle is:

p = 2\cdot (6.4\,cm+8.3\,cm)

p = 29.4\,cm

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter (\Delta p), measured in centimeters, is estimated by differences. That is:

\Delta p = 2\cdot (\Delta w + \Delta l)  (2)

Where:

\Delta w - Uncertainty in width, measured in centimeters.

\Delta l - Uncertainty in length, measured in centimeters.

If we know that \Delta w = 0.2\,cm and \Delta l = 0.2\,cm, then the uncertainty in perimeter is:

\Delta p = 2\cdot (0.2\,cm+0.2\,cm)

\Delta p = 0.8\,cm

The uncertainty in its perimeter is 0.8 centimeters.

5 0
3 years ago
PLEASE HELP WILL GIVE BRANLIEST IF CORRECT
coldgirl [10]

Answer:

Explanation:

  • z: commutator
  • y: brush
  • x: armature

3 0
3 years ago
Read 2 more answers
HELP I WILL GIVE BRAINLIEST!
Ann [662]

Answer:

Answer:B

Explanation:

Because it all stayed consistant

4 0
3 years ago
Two identical billiard balls are rolling toward each other at the same speed. What will be true after they collide head-on?
tino4ka555 [31]

Answer:

They will both bounce back at the same speed they had before the collision

Explanation:

Assuming an elastic collision, kinetic energy will be conserved.  Therefore, the billiard balls will have the same speed after the collision as before the collision.

7 0
3 years ago
n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

3 0
3 years ago
Read 2 more answers
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