Ask question

Ask question

All categories

3 years ago

9

Which of the following would have the highest power rating? 100 joules of energy used in 0.5 seconds 500 joules of energy used i

n 0.5 seconds 100 joules of energy used in 0.25 seconds 500 joules of energy used in 0.25 seconds

2 answers:

cluponka [151]3 years ago

8 0

Answer:

option D

Explanation:

Maksim231197 [3]3 years ago

7 0

A) P = 100/0.5 = 200 Watts
b) P = 500/0.5 = 1000 watts
c) P = 100/0.25 = 400 watts
d) P = 500/0.25 = 2000 watts [ highest ]

In short, Your Answer would be Option D

Hope this helps!

You might be interested in

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

An echo is a reflection of sound against another surface. Suppose you are standing on one side of a canyon and you

Morgarella [4.7K]

Answer:

10-1 Temperature and Expansion. 135 ... text, the laboratory work that you do, or your physics teacher. ... Assume that the speed of sound in air is 340 m/s. How.

Explanation:

hope that helps

4 0

3 years ago

Help. Please! I really need help. It’s timed, and I’m loosing points.

Archy [21]

Answer:

I think balanced

Explanation:

because there is a 2 on each arrow

5 0

3 years ago

Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black

oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

$Q = FA\sigma\Delta T^4$

Where,

F =View Factor

A = Cross sectional Area

$\sigma =$ Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

$L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K$

The view factor between two coaxial parallel disks would be

$\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33$

$\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75$

Then the view factor between base to top surface of the cylinder becomes $F_{12} = 0.26$ . From the summation rule

$F_{13} = 1-0.26$

$F_{13} = 0.74$

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

$\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}$

$\dot{Q_3} = 2\dot{Q_{13}}$

$\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)$

$\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)$

$\dot{Q_3} = 780.76W$

Therefore the rate heat radiation is 780.76W

5 0

2 years ago