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bezimeni [28]
3 years ago
14

what is the acceleration of each body of mass 5kg rests on a frictionless table and is connected to a cable that passes over a p

ulley onto a hanging body of mass 10kg​
Physics
1 answer:
Oksanka [162]3 years ago
8 0

Answer:

6.53 m/s²

Explanation:

Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.

Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:

T = m₁a               (1)

m₂g - T = m₂a    (2)

substituting T = m₁a in equation 2:

m₂g - m₁a = m₂a

m₂a + m₁a = m₂g

a(m₁ + m₂) = m₂g

a = m₂g / (m₁ + m₂)

a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²

Both objects have an acceleration of 6.53 m/s²

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3 years ago
An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to ro
Dafna11 [192]

To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,

T_x = nsinA = \frac{mv^2}{r}

T_y = ncosA = mg

Dividing both.

tan A = \frac{v^2}{rg}

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Three systems of two charged particles are shown below. All the particles have the same mass,
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3 years ago
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!!!

4 0
3 years ago
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