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spin [16.1K]
3 years ago
15

Ian has a mass of 58.0 kg. what i the weight

Physics
1 answer:
nekit [7.7K]3 years ago
6 0

Answer:

568.4N

Explanation:

weight  =58 x gravity

            =58x9.8

             =568.4N

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The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b
taurus [48]

Answer:

I = 0.483 kgm^2

Explanation:

To know what is the moment of inertia I of the boxer's forearm you use the following formula:

\tau=I\alpha  (1)

τ: torque exerted by the forearm

I: moment of inertia

α: angular acceleration = 125 rad/s^2

You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)

\tau=Fr=(1.95*10^3N)(0.031m)=60.45J

Next, you replace this value of τ in the equation (1) and solve for I:

I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2

hence, the moment of inertia of the forearm is 0.483 kgm^2

8 0
3 years ago
a 20 kg sig is pulled by a horizontal force such that the single rope holding the sign make an angle of 21 degree with the verti
True [87]

Answer:

T= 210.15 N

F= 75.31 N

Explanation:

Let the tension in string be T newton.

According to the question

⇒T×cos21°= mg

⇒T= mg/cos21°

⇒T=20×9.81/cos21

⇒T= 210.15 N

now, the magnitude of horizontal force

F= Tsin21°

⇒F= 210.15×sin21°

=75.31 N

6 0
3 years ago
How is high-ordered conditioned performed
Blababa [14]
................................
4 0
3 years ago
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV
denpristay [2]

Answer:

R₁ = 0.126 m

Explanation:

Let's use the definition of intensity which is the power per unit area

          I = P / A

the generated power is constant

          P = I A

power is

         P = E / t

if we perform the calculations for a given time, the wave energy is

        E = q V

we substitute

        P = \frac{q V\ A}{t}

we can write this equation for two points, point 1 the antenna and point 2 the receiver

         V₁A₁ = V₂A₂

          A₁ = \frac{V_2}{V_1} \ A_2

          A₁ = 0.1 10⁻³  5 10⁻⁴ /V₁

          A₁ = 5 10⁻⁸ /V₁

In general, the electric field on the antenna is very small on the order of micro volts, suppose V₁ = 1 10⁻⁶ V

           

let's calculate

          A₁ = 5 10⁻⁸ / 1 10⁻⁶

         A₁ = 5 10⁻² m²

the area of ​​a circle is

          A = π r²

we substitute

         π R1₁²= 5 10⁻²

         R₁ = \sqrt{    \frac{5 \ 10^{-2} }{\pi } }}

         R₁ = 0.126 m

5 0
3 years ago
The latent heat of vaporization for water at room temperature is 2430 J/g.
alukav5142 [94]

Answer:

1)   kinectic energy=7.26*10^-^2^0J

2)  V= 2.0m/s

3)  T=3.5*10^3K

4)  The Molecules do not burn because of the presences of hydrogen bond in place

Explanation:

From the question we are told that

latent heat of vaporization for water at room temperature is 2430 J/g.

1)Generally in determining the molar mass of water evaporated we have that

-One mole (6.02 x 10. 23 molecules)

-Molar mass of water is 18.02 g/mol

Mathematically the mass of water is give as

   

  M=\frac{18.02}{6.02*10^-^2^6}

  M=3*10^-^2^3g

Therefore

  kinectic energy=2430J/g*3*10^-^2^3g

 kinectic energy=7.26*10^-^2^0J

b)Generally the evaporation speed V is given asV= \sqrt{\frac{K.E*2}{m} }

Mathematically derived from the equation

\frac{1}{2} mv^2 =K.E

To Give

V= \sqrt{\frac{K.E*2}{m} }

V= \sqrt{\frac{7.26*10^-^2^0J*2}{3*10^-^2^3g} }

V= 2.0m/s

c)Generally the equation for velocity   Vrms=\sqrt{\frac{3RT}{M} }

Therefore

Effective temperature T is given by

      T=\frac{\sqrt{v}*m}{R}

where

     T=\frac{\sqrt{2.0m/s}*6.02*10^-^2^6}{0.082057 L atm mol-1K-1}

     T=3.5*10^3K

4) The Molecules do not burn because of the presences of hydrogen bond in place

3 0
3 years ago
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