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Liono4ka [1.6K]
3 years ago
13

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

Physics
1 answer:
denpristay [2]3 years ago
8 0

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

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A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile
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Answer:

θ = 62.72°

Explanation:

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x = x₀+ vx*t   Formula (1)

vx = v₀x

Where:  

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s

v₀x: Initial speed in x  in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

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y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

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g: acceleration due to gravity in m/s²

Data

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v₀x= 55 m/s

x-y components of the initial  velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ   Equation (2)

Calculating of the angle θ

We replace data in the  Equation (1)

55 =  120*cosθ

cosθ = 55 / 120

\theta = cos^{-1}(  \frac{55}{120} )

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3 0
3 years ago
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Answer:

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Explanation:

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Use formula:

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Replace:

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Solve the subtraction of the numerator:

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It divides:

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How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

8 0
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