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Otrada [13]
3 years ago
13

5. If Brook has a mass of 87 kg and applies a pressure of 9800 N/m2 when in bed. What is I

Physics
1 answer:
Makovka662 [10]3 years ago
3 0

We have,

  • Mass of brook = 87 kg
  • Pressure applied by it on the ground = 9800 N/m²

Now, we know that,

  • Pressure = Force/Area

We are provided with pressure applied butbwe need to calculate force;

  • F = ma
  • F = 87 × 9.8 { Accleration due to gravity }

Finally, just put all these values in pressure's formula:

  • P = F/A
  • 9800 = 87 × 9.8 / A
  • 98000/87 × 98 = A
  • 1000/87 = A
  • 11.49 = A
  • 11.5 m²(Approx) = A

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Explanation:

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Therefore, we can conclude that if Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done.

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Zanzabum

The water tank waves are transverse waves while the sound waves are longitudinal waves.

<h3>What is a wave?</h3>

A wave is a disturbance along a medium which transfers energy. We know that the water tank waves are transverse waves while the sound waves are longitudinal waves.

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A circular loop of wire with 10 turns, radius 0.241 m and resistance 0.235 Ohms is connected to a 13.1 V power supply. The magne
SVETLANKA909090 [29]

Given Information:  

Resistance of circular loop = R = 0.235 Ω 

Radius of circular loop = r = 0.241 m

Number of turns = n = 10

Voltage = V = 13.1 V

Required Information:  

Magnetic field = B = ?  

Answer:  

Magnetic field = 0.00145 T

Explanation:  

In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B

B = μ₀nI/2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space  and current in the loop is given by

I = V/R

I = 13.1/0.235

I = 55.74 A

B = 4πx10⁻⁷*10*55.74/2*0.241

B = 0.00145 T

Therefore, the magnetic field at the center of this circular loop is 0.00145 T

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Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the
joja [24]

As it is given that the air bag deploy in time

t = 10 ms = 0.010 s

total distance moved by the front face of the bag

d = 20 cm = 0.20 m

Now we will use kinematics to find the acceleration

d = v_i*t + \frac{1}{2}at^2

0.20 = 0 + \frac{1}{2}a*0.010^2

0.20 = 5 * 10^{-5}* a

a = 4000 m/s^2

now as we know that

g = 10 m/s^2

so we have

a = 400g

so the acceleration is 400g for the front surface of balloon

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