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Aliun [14]
3 years ago
10

A person of mass 70kg runs up a flight of stairs with a vertical 5m.If the trip takes 7s to complete,calculate the persons power

.
Physics
1 answer:
strojnjashka [21]3 years ago
8 0

Answer:

power =? , Work done (WD) =? , mass (m) = 70g , Gravity (g) = 10 m/s² , height (h) = 5m , time = 7s

WD=Ep = mgh \\  = 70 \times 10 \times 5 \\  = 3500J \\  \\  \\ power =  \frac{work \: done}{time}  =  \frac{3500}{7}  \\  \\ power = 500 \: w

I hope I helped you^_^

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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
m_a_m_a [10]

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

5 0
3 years ago
Read 2 more answers
In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

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Tina's calculations of the tarantula found that the spider was able to cover 20 centimeters in five seconds what was the average
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4 years ago
A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.
stiks02 [169]

Ignoring air resistance, the bullet's horizontal velocity is constant:

v_x=v_{0x}=100\,\dfrac{\mathrm m}{\mathrm s}

In 1.3 seconds, we can expect it to travel

v_xt=\left(100\,\dfrac{\mathrm m}{\mathrm s}\right)(1.3\,\mathrm s)=130\,\mathrm m

4 0
4 years ago
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