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Shkiper50 [21]
3 years ago
10

The total electric field consists of the vector sum of two parts. One part has a magnitude of E1 = 1300 N/C and points at an ang

le θ1 = 35o above the +x axis. The other part has a magnitude of E2 = 1700 N/C and points at an angle θ2 = 55o above the +x axis. Find the magnitude and direction of the total field. Specify the directional angle relative to the x axis.
Physics
1 answer:
V125BC [204]3 years ago
5 0

Answer:

2954.6 N/C, 46.36 degree from positive  axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E =  1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

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Yuki888 [10]

Answer:

A) 3.13 m/s

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We are told that the position as a function of time is given by;

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v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

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6 0
3 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
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The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
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r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
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And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

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Many of water's emergent properties, such as its cohesion, its high specific heat, and its high heat of vaporization, result fro
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Answer:

Option 5.

Explanation:

Many of the properties of water like high specific heat, cohesion, high vaporization heat, etc can be contributed to the polar nature of water molecule.

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We have to find the momentum after collision

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<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased. 
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