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Keith_Richards [23]
2 years ago
12

What is the mass of a dog running at a speed of 5 m/s and a momentum of 120.5 kgm/s?

Physics
1 answer:
viktelen [127]2 years ago
6 0

Given:

Momentum of the dog (p) = 120.5 kg m/s

Speed of the dog (v) = 5 m/s

To Find:

Mass of the dog (m)

Concept/Theory:

\underline{\underline{ \bf{\Large{Momentum}}}}

  • It is defined as the quantity of motion contained in a body.
  • It is measured as the product of mass of the body and it's speed.
  • It is represented by p.
  • It's SI unit is kg m/s
  • Mathematical Representation/Equation of Momentum: \boxed{ \bf{p = mv}}

Answer:

By using equation of momentum, we get:

\rm \longrightarrow m =  \dfrac{p}{v}  \\  \\  \rm \longrightarrow m =  \dfrac{120.5}{5}  \\  \\  \rm \longrightarrow m = 24.1 \: kg

\therefore Mass of the dog (m) = 24.1 kg

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You have a rock sample and analyze it for the presence of radioactive isotopes in order to determine when it was formed. You fin
Firdavs [7]

Answer:

the decay of half of the nuclei only a half-life has passed ,  b) in rock time it is 1 108 years

Explanation:

The radioactive decay is given by

         N = N₀ e^{\lambda t}

If half of the atoms have decayed

       ½ N₀ = N₀ e^{\lambda t}

       ½ = e^{\lambda t} ₀

       Ln 0.5 = - λ t

       t = - ln 0.5 /λ

The definition of average life time is

      T_{1/2}= ln 2 / λ

       λ = ln 2 /  T_{1/2}

       λ = 0.693 / 100 10⁶

       λ = 0.693 10⁻⁸ years

We replace

       t = -ln 0.5 / 0.693 10⁻⁸

       t = 10⁸ years

We see that for the decay of half of the nuclei only a half-life has passed

b) in rock time it is 1 108 years

8 0
3 years ago
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
3 years ago
A block of wood 3.0 cm on each side has a mass of 27g. what is the density of this block?
schepotkina [342]
The block of wood is 3cm on each side so it is a cube. The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3. density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3
8 0
3 years ago
20.0 m [N] - 15 m [S20degreesE]
denis-greek [22]

Answer:

thank for making me give up on life

Explanation:

I thought the stuff I had was hard wth is even that

3 0
3 years ago
A 6.00 g lead bullet traveling at420 m/s is stopped by a large
Westkost [7]

Answer:

The increase in temperature of the bullet is 351.1 kelvin

Explanation:

First, we should find the kinetic energy of the bullet is:

K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}

with m the mass and v the velocity.

K=529.2 J

Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:

Q=\frac{K}{2}= 264.6\,J

To know what the increase in temperature is, we should use specific heat of lead:

c=125.604 \frac{J}{kg\.K}

The equation that relates specific heat, change in temperature and mass is:

Q=cm\varDelta T

solving for \varDelta T:

\varDelta T=\frac{Q}{cm}=\frac{264.6}{(125.604)(6\times10^{-3})}

\varDelta T= 351.10\, K

4 0
3 years ago
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