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Elanso [62]
3 years ago
13

An object weighs 10N in air and 70N in water .What is it's weight when immersed in a liquid of relative density 1.5?​

Physics
1 answer:
Morgarella [4.7K]3 years ago
5 0
<h3>Answer:</h3>

5.5 N

<h3>Explanation:</h3>

<u>We are given;</u>

  • Real weight of an object in air as 10 N
  • Apparent weight in water as 7 N
  • Relative density of a liquid is 1.5

We need to calculate the apparent weight when in liquid .

  • First we calculate upthrust

Upthrust = Real weight - Apparent weight

              = 10 N - 7 N

               = 3 N

  • Then calculate the upthrust in the liquid.

we need to know that;

Relative density of a liquid  = Upthrust in liquid/Upthrust in water

Therefore;

1.5 = U ÷ 3 N

Upthrust in liquid = 3 N × 1.5

                             = 4.5 N

  • Therefore, the upthrust of the object in the liquid is 4.5 N

But, Upthrust = Real weight - Apparent weight

Therefore;

Apparent weight in Liquid = Real weight - Upthrust

                                           = 10 N - 4.5 N

                                           = 5.5 N

Thus, the weight when the object is immersed in the liquid is 5.5 N

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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit
sveticcg [70]

Answer:

a)   I = 3.63 W / m² , b)   I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

         I = P / A = ½ ρ v (2π f s_{max})²

         I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

        cte = (½ ρ v 4π² s_{max}²)

         I₀ = cte s² f₀²

        I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

         I = constant (2.20 10³) 2

         I = cte 4.84 10⁶

let's find the relationship of the two quantities

        I / Io = 4.84

        I = 4.84 Io

        I = 4.84 0.750

        I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

        I = cte (f s)²

        I = constant (0.250 10³ 4)²

 

        I = cte 1 10⁶

         

the relationship

        I / Io = 1

        I = Io

        I = 0.750 W / m²

6 0
2 years ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
Someone help me, I'm stuck
Bas_tet [7]
C is the answer hope that helps you
3 0
3 years ago
Shawn uses 45 N of force to stop the cart 27 meter from running his foot over. How much work does he do?
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Answer:

W=f×d

w=45×27

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3 0
3 years ago
ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo
vaieri [72.5K]
Distance is the total length covered = 2m + 3m = 5m

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Displacement =  2m + (-3)m.               Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement  is 1m behind his original starting point.
4 0
3 years ago
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