What happens when two charged objects are moved closer together

Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

elena-14-01-66 [18.8K]

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, $1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}$

$1\ kcal/h=1.16\ J/sec$

J/sec is nothing but watts.

So, $1\ kcal/h=1.16\ watts$

and $88\ kcal/h=88\times 1.16\ watts = 102.08\ watts$

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

8 0

3 years ago

A box is moving along the x-axis and its position varies in time according to the expression:

Colt1911 [192]

Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. $x = 6 * 3.2^2 = 61.44 m$

b) at t = 3.2 + Δt. $x = 6*(3.2 + \Delta t)^2$

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

$v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}$

$v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}$

$v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}$

$v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}$

$v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}$

$v = 38.4 + \Delta t$

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0

3 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

May some one help a poor MHA fan out with science

guapka [62]

Answer:

nm

Explanation:

4 0

3 years ago

Adita lifts a book from the floor, carries it across the room, and places it on a high shelf. When is Adita doing work on the bo

noname [10]

She does work from the moment she touches the book until she lets it go. Work is anything that requires energy. Therefore, she is working as she picks up the book, carries it, and when she is lifting it onto the shelf.

5 0

3 years ago

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What happens when two charged objects are moved closer together ​

What happens when two charged objects are moved closer together