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madam [21]
10 months ago
9

how fast would a(n) 73 kgkg man need to run in order to have the same kinetic energy as an 8.0 gg bullet fired at 430 m/sm/s ?

Physics
1 answer:
Mars2501 [29]10 months ago
8 0

Answer:

an 85 kg person run to equal the kinetic energy of an 8.0 g bullet fired at 410 m/s? The speed is, it might be argued,

v=4.0m/s

Explanation:

Typically, the mathematical formula for kinetic energy is as follows:

KE=\frac{1}{2} *m*v^{2}

K.E= kinetic energy

M=mass

V= speed

so,

KE=1/2*0.008*(420)*420

K.E=672.4J

 Therefore,

v*v=672.4/85

v*v=15.821

v=4.0m/s

Therefore

v=4.0m/s

What is Kinetic Energy?

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

To know more about Kinetic Energy? Visit to: -

brainly.com/question/22174271

#SPJ4

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barxatty [35]

Answer:

NE DIYON INGILIZ MISIN SEN

7 0
2 years ago
How much force acts on one surface of a rectangular bedroom wall that is 2.20 m by 3.20 m when atmosspheric pressure is?
irakobra [83]
<span>We know that pressure is the force applied into a surface, in our case the wall of the room, so then first we will calculate the surface of this wall: S = 2.2 * 3.2 = 7.04 m2 Then we also know the atmospheric pressure in normal conditions is 1 atm. That is the same 1 atm = 101325 Pascals or 101325 N/m2 Now we need to use the formula : P = F/S where P is pressure, F is force and S is surface to calculate the force: F = P * S = 101325 * 7.04 = 713,328 Newtons Conclusion: the force acts on the wall due the air inside the room is 713,328 N</span>
3 0
3 years ago
uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

7 0
3 years ago
When working with a powered instrument, the swirling effect produced by the water stream within the confined space of a periodon
Lina20 [59]

Answer:

Acoustic microstreaming

Explanation:

Acoustic microstreaming is the swirling effect produced by water stream confined in a spaced of a periodontal pocket.

  • It is the movement of water in a particular direction as a result of mechanical pressure within the fluid body.
  • They are often used in dental procedures to remove particulates from the teeth.
  • It mostly relies on the properties of sound waves to achieve this goal
3 0
2 years ago
A six pack of your favorite beverage (12 fl. oz. cans, assumed to have the same characteristics as water) is placed in a cooler
nasty-shy [4]

Answer:

Q = 169 BTU

Explanation:

As we know that volume is given as

V = 12 Fl oz

so it is given in liter as

V = 12 fl oz = 0.355 Ltr

now we have six pack of such volume

so total volume is given as

V = 6\times 0.355 Ltr

V = 2.13 ltr

so its mass is given as

m = 2.13 kg

now the change in temperature is given as

\Delta T = 70 - 34 = 36 ^oF

\Delta T = 20^oC

now the heat given to the liquid is given as

Q = ms\Delta T

Q = 2.13(4186)(20)

Q = 1.78 \times 10^5 J

Q = 169 BTU

4 0
2 years ago
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