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3 years ago

3.25 kcal is the same amount of energy as

Physics

2 answers:

PIT_PIT [208]3 years ago

8 0

Phantasy [73]3 years ago

5 0

Answer:

$E = 13.6 kJ$

Explanation:

As we know that

1 cal = 4.18 J

$1 kcal = 10^3 cal$

so we have

$1 kcal = 10^3 \times 4.18 J$

now we have total energy in k cal unit given as

$E = 3.25 kcal$

$E = 3.25 \times 4.18 \times 10^3$

$E = 13585 J$

$E = 13.6 kJ$

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S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

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To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .

We know that the surface area and volume of the sphere is given by:

$A=4 \pi r^{2}\\V=\frac{4}{3} \pi r^{3}$

Therefore, the ratio between the surface area and the volume for the sphere will be:

$\frac{A}{V}=\frac{4 \pi r^{2}\\}{\frac{4}{3} \pi r^{3}}=\frac{3}{r}$

Equating the volume to the constant c, we will find the value of $r$ .

$V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }$

Substituting the value of r in the ration between surface area and volume, we get:

$\frac{A}{V}=\frac{3}{ (\frac{3c}{4\pi} )^{\frac{1}{3} }}$

Calculating the constants, we get:

$\frac{4.83598}{c^{\frac{1}{3} } }$

Hence, the ration between surface area and volume is $\frac{4.83598}{c^{\frac{1}{3} } }$

To learn more about surface area and volume of sphere, refer to:

brainly.com/question/4387241

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1 year ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

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