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antoniya [11.8K]
3 years ago
8

Dimension of specific heat capacity​

Physics
1 answer:
Luden [163]3 years ago
3 0

Answer:

The specific heat of a substance (per unit of mass) has dimension L. T. Θ−1, or (L/T)2/Θ. Therefore, the SI unit J/K/kg is equivalent to metre squared per second squared per kelvin (m2 K−1 s−2).

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A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to
mixer [17]

Given :

A jogger runs 5.0 km on a straight trail at an angle of 60° south of west.

To Find :

The southern component of the run rounded to the nearest tenth of a kilometre.

Solution :

We know, southern component is given by :

d = Dsin \ \theta\\\\d = 5\times sin \ 60^{\circ}\\\\d = -4.33\  km

Here, negative sign indicates south direction.

Therefore, the southern component of the run rounded to the nearest tenth of a kilometre is 4.3 km.

Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
an object has a velocity of 0.12 rad/s. how many revolutions will the object make after traveling for 30 seconds?
Basile [38]

Answer:

We can write   S = v * t    where S is the distance traveled in time t

S = .12 rad/sec * 30 sec = 3.6 rad     distance traveled

Also 2 * pi * N = S   since there are 2 * pi rad / revolution

So 2 * pi * N = 3.6

N = 3.6 / (2 * pi) = .573 revolutions

7 0
3 years ago
A car engine uses the combustion of fuel to move a system of gears that
evablogger [386]
C it is transformed into chemical energy
3 0
3 years ago
5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by v =6t^2 i + 2t j where v is
otez555 [7]

The concept of derivatives and integrals allows to find the results for the questions are the motion of a particle where the speed depends on time are:

       a)the position is:  r = 2 t³ i + t² j

       b) the position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

       c) The acceleration is: a = 12 t i + 2 j

       d) the force is: F = 60 t i + 10 j

       e) the torque is:  τ = 40 t³ k^

       f) tha angular momentum is:  L = 4t³ - 6 t² k^

       g) The kinetic energy is: K = 2 m t² (9t² +1)

       h) The power is:   P = 2m (36 t³ + 2t)

Kinematics studies the movement of bodies, looking for relationships between position, speed and acceleration.

a) They indicate the function of speed.

        v = 6 t² i + 2t j

Ask the function of the position.   The velocity is defined by the variation of the position with respect to time

          v = \frac{dr}{dt}  

          dr = v dt

we substitute and integrate.

        ∫ dr = ∫ (6 t² i + 2t j) dt

        r - 0 = 6 \frac{t^3 }{3} \ \hat i + 2 \frac{t^2}{2 \ \hat j }  

       r = 2 t³ i + t² j

b) The position of the body on the y axis is a parabola and on the x axis it is a cubic function.

c) Acceleration is defined as the change in velocity with time.

           a = \frac{dv}{dt}  

           a = \frac{d}{dt} \ (6t^2 i + 2t j)  

           a = 12 t i + 2 j

d) Newton's second law states that force is proportional to mass times the body's acceleration.

          F = ma

          F = m (12 t i + 2 j)

          F = 5 12 t i + 2 j

          F = 60 t i + 10 j

e) Torque is the vector product of the force and the distance to the origin.

           τ = F x r

The easiest way to write these expressions is to solve for the determinant.

         \tau = \left[\begin{array}{ccc}i&j&k\\F_x&F_y&F_z\\x&y&z\end{array}\right]  

        \tau = \left[\begin{array}{ccc}i&j&k\\60t &10&0\\2t^3 &t^2&0\end{array}\right]  

       τ = (60t t² - 2t³ 10) k

       τ = 40 t³ k ^

f) Angular momentum

        L = r x p

        L =rx (mv)

        L = m (rxv)

The easiest way to write these expressions is to solve for the determinant.

       \left[\begin{array}{ccc}i&j&k\\2t^3 &t^2&0\\6t^2&2t&0\end{array}\right]  

    L = (4t³ - 6 t²) k

 

g) The kinetic energy is

            K = ½ m v²

            K = ½ m (6 t² i + 2t j) ²

            K = m 18 t⁴ + 2t²

            K = 2 m t² (9t² +1)

h) Power is work per unit time

           P = \frac{dW}{dt}dW / dt

The relationship between work and kinetic energy

           W = ΔK

     

          P = 2m \ \frac{d}{dt} ( 9 t^4 + t^2)

          p = 2m (36 t³ + 2t)

In conclusion with the concept of derivatives and integrals we can find the results for the questions are the motion of a particle where the speed depends on time are:

       a) The position is:  r = 2 t³ i + t² j

       b) The position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

       c) The acceleration is: a = 12 t i + 2 j

       d) The force is: F = 60 t i + 10 j

       e) The torque is:  τ = 40 t³ k^

       f) The angular momentum is:  L = 4t³ - 6 t² k^

       g) The kinetic energy is: K = 2 m t² (9t² +1)

       h) The power is:   P = 2m (36 t³ + 2t)

Learn more here:  brainly.com/question/11298125

8 0
2 years ago
Read 2 more answers
A man rides up in an elevator at 12 m. He gains 6500 J of gravitational potential energy. what is the man's mass?
iren2701 [21]

Answer:

We know that potential energy of a body;

= mass(m)× gravitational acceleration(g) × height(h)

Lets find out the mass of the body

P.E. = mgh

=> 6500J = mass × 9.8m/s^2 × 12m

=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)

=> 6500 Nm = m × 117.6 × m^2 / s^2

=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]

=> 55.272 Kg = mass

Therefore the mass of the body = 55.272 kg ~ <em>6</em><em>0</em><em> </em><em>k</em><em>g</em><em> </em>(Ans)

Hope it helps you

6 0
3 years ago
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