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GuDViN [60]
2 years ago
11

A 5.2 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. How far to the left of the pivot must

a 4.0 kg cat stand to keep the seesaw balanced
Physics
1 answer:
algol [13]2 years ago
6 0

Answer:

The cat must stand 0.96 m from the left side of the pivot to keep the seesaw balanced.

Explanation:

Given;

mass of the cat, m₁ = 5.2 kg

mass of the tuna, m₂ = 2.5 kg

length of the seesaw, L = 4.0 m

Apply the following approach;

                                                                   

                  x m                      <-------------2m ---------->              

        0------------------------------Δ------------------------------4m

                  ↓                         2m                                 ↓

                 5.2 kg                                                      2.5 kg

Apply the principle of moment; anticlockwise moment = clockwise moment.

5.2(x ) = 2.5(2)

5.2x = 5

x = 5/5.2

x = 0.96 m

Therefore, the cat must stand 0.96 m from the left side of the pivot to keep the seesaw balanced.

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vampirchik [111]

Answer:

You pull on the oars. By the third law, the oars push back on your hands, but that’s irrelevant to the motion of the boat. The other end of each oar (the blade) pushes against the water. By the third law, the water pushes back on the oars, pushing the boat forward.

8 0
2 years ago
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a. What proportion of resistors have resistances less than 90 Ω? b. Find the mean resistance. c. Find the standard deviation of
Anettt [7]

Answer:

a) 0.0625 = 6.25%

b) 106.67 Ω

c) 9.43 Ω

d) 1

Explanation:

The probability distribution is given as

f(x) = (x - 80)/800 for 80 < x < 120

f(x) = 0 otherwise.

f(x) = (x/800) - (0.1)

a) Proportion of resistors with resistance less than 90 Ω

P(X < 90) = ∫⁹⁰₈₀ f(x) dx

∫⁹⁰₈₀ f(x) dx = ∫⁹⁰₈₀ [(x/800) - (0.1)]

= [(x²/1600) - 0.1x]⁹⁰₈₀

= [(90²/1600) - 0.1(90)] - [(80²/1600) - 0.1(80)]

= (5.0625 - 9) - [4 - 8]

= -3.9375 + 4 = 0.0625 = 6.25%

b) The mean is given by the expected value expression E(X) = = Σ xᵢpᵢ (with the sum done all over the data set for each variable and its corresponding probability)

It can be written in integral form as

Mean = ∫¹²⁰₈₀ xf(x) dx (with the integral done all over the probability function, i.e. from, 80 to 120)

Mean = ∫¹²⁰₈₀ x[(x/800) - (0.1)] dx

= ∫¹²⁰₈₀ [(x²/800) - (0.1x)] dx

= [(x³/2400) - (0.05x²)]¹²⁰₈₀

= [(120³/2400) - (0.05(120²)] - [(80³/2400) - (0.05(80²)]

= [720 - 720] - [213.33 - 320] = 106.67 Ω

c) Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = mean = expected value = 106.67 Ω

Σx²p = ∫¹²⁰₈₀ x²f(x) dx = ∫¹²⁰₈₀ x² [(x/800) - (0.1)] dx = ∫¹²⁰₈₀ [(x³/800) - (0.1x²)] dx

= [(x⁴/3200) - (0.0333x³)]¹²⁰₈₀

= [(120⁴/3200) - (0.0333(120³)] - [(80⁴/3200) - (0.0333(80)³)]

= (64800 - 57600) - (12800 - 17066.667)

= 11466.667

Variance = 11466.667 - 106.67² = 88.85

Standard deviation = √88.85 = 9.43 Ω

d) Cdf = sum of probabilities over the entire probability function

Cdf = ∫¹²⁰₈₀ f(x) dx = ∫¹²⁰₈₀ [(x/800) - (0.1)] dx

= [(x²/1600) - 0.1x]¹²⁰₈₀ = [(120²/1600) - 0.1(120)] - [(80²/1600) - 0.1(80)] = (9 - 12) - (4 - 8) = -3+4 = 1 as it should be!!!

Hope this Helps!!!

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VARVARA [1.3K]
Good morning.

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\mathsf{V = \dfrac{F}{m} \cdot t}

We want the force, so, let's isolate F:

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To find the total number of miles traveled by a person, we add the distance that he has traveled: 3.0 + 5.00 + 4.000. 

Now, to find the accurate number of significant figures when adding measurements, the basic rule for addition is to use the least number of decimal places when reporting the result.

Now, since 3.0 has the least number of decimal places, we report the sum with 1 decimal place and have 12.0 miles as the total distance traveled by the person to reach his destination.

Answer: 12.0 miles
8 0
3 years ago
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