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4 years ago

what will be temperature on the kelvin scale if the indicated temperature on the Fahrenheit scale is 200degree Fahrenheit

Physics

2 answers:

Wittaler [7]4 years ago

7 0

Kelvin=366.38(approximately)

Mnenie [13.5K]4 years ago

3 0

We can use Kelvin = 5/9 (Fahrenheit - 32) + 273.15

K = 5/9(168) + 273.15

K = (93 and 1/3) + 273.15

K = <u>366.48</u> (rounded)

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pav-90 [236]

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4 years ago

a force vector f has a magnitude of 12.0 n. it is oriented 60° to the left of the y ax what are its x and y components?

Arte-miy333 [17]

The force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:

v(x) =6 N
v(y) = 10.39 N

To solve this exercise the formulas and procedures we will use are:

v(x) = v * cosine (angle)
v(y) = v * sine (angle).

Where:

v= magnitude of the vector
v(x) = component of the vector on the (x) axis
v(y) = component of the vector on the (y) axis
angle = angle

Information about the problem:

angle = 60º
v = 12.0 N
v(x)= ?
v(y)= ?

Applying the formula of the component of the vector in the (x) axis we have:

v(x) = v * cosine (angle).

v(x) = 12.0 N * cosine (60º)

v(x) =6 N

Applying the formula of the component of the vector in the (y) axis we have:

v(y) = v * sine (angle)

v(y) = 12.0 N * sine (60º)

v(y) = 10.39 N

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2 years ago

Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them

denpristay [2]

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

3 0

4 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

7 0

4 years ago