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scZoUnD [109]
2 years ago
13

An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire

ction of its acceleration?
Physics
1 answer:
dem82 [27]2 years ago
7 0

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

Acceleration

Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

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3 0
3 years ago
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Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a
skelet666 [1.2K]

Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

F = Force = 151 N

d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

<u>W = 1418.9 J = 1.418 KJ</u>

6 0
3 years ago
How do you know when water is boiling?
vladimir1956 [14]

Answer:

when the steam starts coming out

Explanation:

5 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
3 years ago
In a two-slit experiment using coherent light, the distance between the slits and the screen is 1.10 m, and the distance between
Paul [167]

Answer:

D) 763 nm

Explanation:

Calculation for the wavelength of light

Using this formula

Wavelength of light=Delta Y*Distance / Length

Where,

Delta Y represent the 2nd order bright fringe

Length represent the distance between both the slits and the screen

Distance represent the Distance between the slits

Let note that cm to m = (4.2) x 10^-2 and mm to m= ( 0.0400x 10^-3)

Now Let plug in the formula

Wavelength of light=[(4.2 x 10^-2m)(0.0400 x 10^-3m) / 2(1.1m)]*10^-7 meters

Wavelength of light=[(0.042m) (0.0004m)/2.2m]*10^-7 meters

Wavelength of light =(0.0000168m/2.2m)*10^-7 meters

Wavelength of light =7.63 *10^-7 meters

Wavelength of light =763 nm

Therefore the Wavelength of light will be 763 nm

3 0
3 years ago
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