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vitfil [10]
2 years ago
9

A pebble rolls off the roof of science hall and falls vertically. just before it reaches the ground, the pebble's speed is 17 m/

s. neglect air resistance and determine the height of science hall.
Physics
1 answer:
nikdorinn [45]2 years ago
3 0
The height of the science hall can be related to final velocity through the equation,

      d = (Vf² - Vi²) / 2a

where d is the distance or height, Vf is the final velocity, Vi is the initial velocity and a is the acceleration due to gravity. Substituting the known values,

  d = ((17 m/s)² - (0 m/s)²) / 2(9.8 m/s²)

Simplifying,
  
   d = 14.74 m

<em>ANSWER: 14.74 m</em>
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A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They
mojhsa [17]

Answer:

the required frequency of waves is 2.066 Hz

Explanation:

Given the data in the question;

μ = 1.50 kg/m

T = 6000 N

Amplitude A = 0.500 m

P = 2.00 kW = 2000 W

we know that, the average power transmit through the rope can be expressed as;

p = \frac{1}{2}vμω²A²

p = \frac{1}{2}√(T/μ)μω²A²

so we solve for ω

ω² = 2P / √(T/μ)μA²

we substitute

ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²

ω² = 4000 / 23.71708

ω² = 168.65

(2πf)² = ω²

so

(2πf)² = 168.65

4π²f² = 168.65

f² = 168.65 / 4π²

f² = 4.27195

f = √4.27195

f = 2.066 Hz

Therefore, the required frequency of waves is 2.066 Hz

3 0
2 years ago
If you were trying to cross a river with the shortest possible time, would you aim your boat slightly upstream, directly across
Maru [420]
Slightly downstream for the shortest possible time
7 0
2 years ago
To measure moderately low pressures, oil with a density of 9.0 x 102 kg/m3 is used in place of mercury in a barometer. A change
Yuki888 [10]

Answer:

Δ P =  13.24 Pa

Explanation:

Given that

Density of oil ,ρ₁ = 9 x 10² kg/m³

We know that density for mercury ,ρ₂  = 13.6 x 10³ kg/m³

The change in the height of column ,Δh = 1.5 mm

The pressure given as

P = ρ g h

Change in the pressure

Δ P =  ρ₁ g Δh

Now by putting the values

Δ P =  9 x 10² x 9.81 x 1.5 x 10⁻³    Pa

Δ P =  13.24 Pa

Therefor the change in the pressure will be 13.24 Pa.  

       

3 0
2 years ago
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2
Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

8 0
2 years ago
A projectile is launched at an angle of 60° from the horizontal and at a velocity of
gayaneshka [121]

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

5 0
2 years ago
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