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iren [92.7K]
3 years ago
13

How do u convert 7.68 cal/sec to kcal/min

Physics
2 answers:
stich3 [128]3 years ago
5 0
<span>First you have to convert 7.68 cal/sec to cal/min to do that, just multiply by 60 So it will be 460.8 cal/min now convert that to kcal/min kcal = kilocalories = 1000 calories. so just divide by 1000 and youll have your answer in kcal/min </span>
Lady bird [3.3K]3 years ago
3 0

Answer:

7.68 cal/sec = 0.4608 kcal/min

Step-by-step explanation:

Here, we need to convert  7.68 cal/sec to kcal/min. We know that,

1\ cal=10^{-3}\ kcal, kcal is kilo calorie

Also, 1 minutes = 60 seconds

Firstly we will convert calorie to kilo calorie as,

7.68\ cal/s=\dfrac{7.68\ cal\times 10^{-3}\ kcal}{1\ cal}\times \dfrac{1}{sec}=7.68\times 10^{-3}\ kcal/sec

Now converting seconds to minutes as,

7.68\times 10^{-3}\ kcal/sec=\dfrac{7.68\times 10^{-3}\ kcal}{1\ sec}\times \dfrac{60\ sec}{1\ min}

So, 7.68 cal/sec = 0.4608 kcal/min

Hence, this is the required solution.

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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

5 0
3 years ago
If Calcium has 2 valence electrons and Chlorine has 7 valence electrons, how many Cl-1 ions would need to combine with 1 Ca+2 io
Tanya [424]

Answer:

CALCIUM IDENINE ADEININE AND PHOSPHATE

Explanation:

6 0
3 years ago
3. A dog walks a distance of 100 ft. in 20 s of time. What is the dog's
natta225 [31]
The answer is A.
100/20=5
8 0
3 years ago
Read 2 more answers
A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
A car moved 60 meters west in 2 hours. What is its average velocity?
Fofino [41]

Explanation: Velocity is the displacement of an object during a specific unit of time. Two measurements are needed to determine velocity. Displacement and time. Displacement includes a direction, so velocity also includes a direction. Speed with direction. Velocity can be an average velocity or an instantaneous velocity. Units for velocity are the same as for speed: m/s, km/h, and mph. Delta x(Δx) is the symbol used for displacement. Delta (Δ) means to "change in." Δx means to "change in position." Δx is calculated by final position minus initial position. Velocity formula: → v=Δx/t as a fraction.

v=Δx/t

v=\frac{xf-xi}{t}= \frac{60m}{2}=30m

<em><u>Final answer is 30.</u></em>

Hope this helps!

Thanks!

Have a great day!

-Charlie

5 0
3 years ago
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