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Tatiana [17]
3 years ago
5

Convert 41.3 kilocalories into joules.

Physics
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:172799.2

Explanation:

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What is the measured value always positive from one point to another
LekaFEV [45]
It is called "distance". In fact, distance measures how far apart two points are from each other, and it is always a positive number.
In mathematics, for instance, the distance between two points of coordinates (x1,y1) and (x2,y2) on a xy-plane is defined as
d= \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}
and by definition, this number is always positive.
4 0
3 years ago
The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but
Usimov [2.4K]

Answer:

a) 4.33 pC  b) 5.44*10² N/C

Explanation:

a) The vertical deflecting plates of an oscilloscope form a parallel-plate capacitor.

The value of the capacitance, for a parallel-plate capacitor with air dielectric, can be found to be as follows, applying Gauss' law to the surface of one of the plates, and assuming a uniform surface charge density:

C = ε₀*A / d

where ε₀ = 8.85*10⁻¹² F/m, A = (0.03m)², and d = 0.046 m (we assume that the informed value of 4.6 m is a typo, as no oscilloscope exists with this separation between plates).

Replacing by these values, we find the equivalent capacitance of the plates, as follows:

C = \frac{8.85e-12F/m*(0.03m)^{2} }{0.046m} =1.73e-13 F = 0.173 pF

By definition, the capacitance of any capacitor can be expressed as follows:

C =\frac{Q}{V}

where Q= charge on any of the plates, and V= potential difference between them.

As we know C and V, we can find Q as follows:

Q = C*V = 0.173*10⁻¹² F * 25.0 V = 4.33*10⁻¹² C = 4.33 pC

b) We can find the electric field in several ways, but one very easy is applying Gauss' Law to a pillbox with a face outside one of the plates (paralllel to it) and the other inside the surface.

The total electric flux through the surface must be equal to the enclosed charge, divided by ε₀.

If we look to the flux crossin any face, we find that the only one that has a non-zero flux, is the one outside the surface.

As the electric crossing the boundary must be normal to the surface (in electrostatic conditions,  no tangential field can exist on the surface) , and we assume that the surface charge density that creates it is constant across the surface, we can write the Gauss ' Law as follows:

E*A = Q / ε₀

where A = area of the plate = (.03m)² = 9*10⁻⁴ m², Q= charge on one of the plates = 4.33*10⁻¹² C (as we found in a)) and ε₀ = 8.85*10⁻¹² N/C.

Replacing by these values, and solving for E, we have:

E = \frac{4.33e-12C}{(0.03m)^{2} 8.85e-12F/m} =5.44e2 N/C

⇒ E = 5.44*10² N/C

5 0
4 years ago
What could be done to change this carbide ion to a neutral carbon atom? remove 2 electrons add 2 electrons remove 4 electrons ad
ollegr [7]
A carbide ion is an anion of carbon atom with two carbon atoms with a triple bond connected to each other. Since this carbide ion is a negatively charged ion, in order to convert carbide ion to its neutral carbon atom, it has to give up/remove four electrons. 
6 0
3 years ago
Read 2 more answers
A woman performs 2000 J of work in order to push a cart full of groceries 50 meters. How much force did she apply to the cart? *
ad-work [718]
They will give you 4000
4 0
3 years ago
Read 2 more answers
Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble C
Serhud [2]

Answer:

The current value of the Hubble's constant = 73 km/sec/Mpc.

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Explanation:

The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.

Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.

Hence,

v is directly proportional to d

where, v = apparent velocity

d = distance

if we equate velocity and distance then there comes Hubble constant.

v = H_{0} x d

 H_{0} = 73 km/sec/Mpc

where, Mpc = Mega Parsec = 1 Mpc = 3.086 x 10^{19} km      

We can use Hubble constant to tell the age of universe.

t = d/v

t = d/( H_{0} xd)

t = 1/H_{0}

Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.

Now, if we hypothetically change the value of Hubble constant,

from H_{0} = 73 km/sec/Mpc to H_{0} = 700 km/sec/Mpc

then the age of universe will be:

t = 1/H_{0}

first convert the units of new H_{0} into 1/s

H_{0} = (700) x (/3.08 x 10^{19} )

H_{0} = 227.27 x10^{-19}  = 2.27 x 10^{-21} 1/s

So,

Age of universe will be:

t = 1/H_{0} = 1/2.27x10^{-21} 1/s

t = 2.27 x 10^{21} s

t = 71.9 trillion years

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

       

6 0
3 years ago
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