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vazorg [7]
3 years ago
9

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti

ve in pulling the sled horizontally along the ground?
Physics
2 answers:
Andreas93 [3]3 years ago
7 0
Answer: 1.88 N

Explanation:

Data:

Force = 4.00N
angle = 62°
horizontal force = ?

Solution:

The trigonometric ratio that relates horizontal - leg to hypotenuse is the cosine.

That ratio is:

                        horizontal - leg
cos(angle) = -------------------------
                          hypotenuse

So, applied to the force, that is:

                             horizontal force
cos (angle) = -----------------------------------
                                 total force

So, clearing the horizontal component you get:
                                         
horizontal force = force * cos (angle)

Substitute the data given:

horizontal force = 4.00N * cos(62°) = 4.00N * 0.4695 = 1.88 N

Answer: 1.88N


Alex73 [517]3 years ago
4 0
The horizontal force will be equivalent to the total force multiplied by the cosine of the angle between them, that is, since the angle is 62.0 degrees from the ground, we multiply the force by cos 62.
So the total force will be 4cos62 = 1.88 Newtons of force.
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The sputnik one satellite orbiting earth mass equals 5.98×10 to the 24th power kilograms in a circle of radius 6.96×10 to the si
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7572 m/s

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The mass of the earth (m_1) is far greater than the mass of the sputnik (m_2). Therefore m_1m_2=m_1. The mass of the sputnik is neglected, therefore:

F=G\frac{m_1}{r^2}=\frac{(6.674*10^{-11})(5.98*10^{24})}{(6.96*10^6)^2} = 8.2389N

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8.2389 = v^2 / 6.96*10^6\\v=7572m/s

6 0
3 years ago
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An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What
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Answer:

a.  9.99625 cm b. 68 °C

Explanation:

(a) What is the length of the rod at the freezing point of water (0 0C)?

Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

L₁ = L₀(1 + α(T₁ - T₀))

So, substituting the values of the variables into the equation, we have

L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 +  0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

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