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nataly862011 [7]
2 years ago
15

In the sum of 4.34 + 5.66, the number of significant figures is

Physics
1 answer:
kifflom [539]2 years ago
4 0

Answer:

1

Explanation:

4.34 + 5.66 = 10

number of significant figures in 10: 1

(the significant figure here is 1)

here's a clear explanation from Northern Kentucky University (NKU)

[To determine the number of significant figures in a number use the following 3 rules:

Non-zero digits are always significant

Any zeros between two significant digits are significant

A final zero or trailing zeros in the decimal portion ONLY are significant

Example:  .500 or .632000 the zeros are significant

                .006  or .000968 the zeros are NOT significant ]

[For addition and subtraction use the following rules:

Count the number of significant figures in the decimal portion ONLY of each number in the problem

Add or subtract in the normal fashion

Your final answer may have no more significant figures to the right of the decimal than the LEAST number of significant figures in any number in the problem.]

(https://www.nku.edu/~intsci/sci110/worksheets/rules_for_significant_figures.html#:~:text=To%20determine%20the%20number%20of,decimal%20portion%20ONLY%20are%20significant)

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There is no atmspheric pressure

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2 years ago
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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
3 years ago
You are sitting on a deck of your house surrounded by oak trees. You hear the sound of an acorn hitting the deck. You wonder if
Black_prince [1.1K]

Answer: 96N

Explanation:

To calculate the velocity of the impact On the persons head, we have

h = gt²/2

14 = 9.81t²/2

t² = 28/9.8

t² = 2.86

t = 1.69s

V = u + at

V = 0 + 9.81*1.69

V = 16.58m/s

a(average) = (v1² + v2²) /2Δy

a(average) = 16.58² + 0)/2 * 0.005

a(average) = 274.8964/0.01

a(average) = 27489.64m/s²

Using newton's second law of motion,

F(average) = m * a(average)

F(average) = 0.0035 * 27489.64

F(average) = 96.21N

Therefore the force needed by the acorn to do much damage starts from 96N

8 0
3 years ago
A child who is swimming toward shore at 0.78 m/s sees shark and picks up his speed
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Answer:

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Use v²=v0²+2a(d)

solve for a

v²-v0²/2d=a

Plug in givens

1.89²-0.78²/2*17.5=a

Plug into calculator

a=0.085m/s²

3 0
3 years ago
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Artyom0805 [142]

Energy consumed in doing the work = 300 Joules

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Let the distance moved by the object be d.

Work done by the force is determined by the force applied and the displacement happened in the direction of the force applied.

Work done = Force x displacement

300 = 75 x d

d = \frac{300}{75}

d = 4 m

Hence, the maximum distance moved by the object = 4 meters

6 0
3 years ago
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