According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.
Answer:
1.28 m
Explanation:
As shown in the diagram attached,
According to the principle of moment,
For a body at equilibrium,
Sum of clockwise moment = sum of anticlockwise moment.
Taking moment about the pivot,
W₁(1.6)+W(0.133) = W₂(x)............... Equation 1
Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.
But,
W = mg
Where g = 9.8 m/s², m = mass of the body
Therefore,
W₁ = 26×9.8 = 254.8 N,
Wₓ = 18×9.8 = 176.4 N
W₂ = 34.4×9.8 = 337.12 N
Substitute these values into equation 1
(254.8×1.6)+(176.4×0.133) = 337.12(x)
407.68+23.4612 = 337.12x
337.12x = 431.1412
x = 431.1412/337.12
x = 1.2789
x ≈ 1.28 m
Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A =
= 6.77 ×
m²
Now Area of cylinder is :
A =
d²
solving for d:
d = 
d = 9.28 cm
Answer:
The new separation distance between adjacent bright fringes will be <u>4 mm</u>
Explanation:
Since, the distance between adjacent bright fringes is given by the formula:
Δx₁ = λL/d = 2 mm -------- eqn (1)
where,
Δx = Distance between adjacent bright fringes
λ = wavelength of light = constant for both cases
L = Distance between the slits and the screen
d = slit separation
Now, for the second case:
Slit Separation = d/2
Therefore,
Δx₂ = λL/(d/2)
Δx₂ = 2(λL/d)
using eqn (1), we get:
Δx₂ = 2 Δx₁
Δx₂ = 2(2 mm)
<u>Δx₂ = 4 mm</u>
Density=mass÷volume
mass=density×volume
mass=2×8=16 g