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3 years ago

A charge alters the space around it. What is this alteration of space called?

Physics

1 answer:

iogann1982 [59]3 years ago

6 0

Answer: electric field

Explanation: when a charge is placed in space, it alters the space around it by creating an electric field.

This electric field has the ability to exert a force (f) on any test charge(q) placed within this vicinity.

This is the reason why a charge can either attract or repel another charge.

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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.

denpristay [2]

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:

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

8 0

3 years ago

a child in a tree house uses a rope attached to a basket to lift a 26 nn dog upward through a distance of 5.0 mm into the house.

agasfer [191]

Total work done is 0.13 Joules

The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.

According to the given information

We need to find the work done

work done = force × distance

We are given,

force = 26 N

Distance = 0.0005 meter

hence ,

Work done = 26 × 0.005

= 0.13 Joules

Total work done is 0.13 Joules

To know more about Work done

brainly.com/question/13662169

#SPJ4

4 0

1 year ago

A ball is kicked from level ground at an angle 60° with initial velocity 10 m/s. The distance the ball travels, in meters, is:

RUDIKE [14]

Answer:

<em>the ball travels a distance of 8.84 m</em>

Explanation:

Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.

R = (U²sin2∅)/g.............................. Equation 1

Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.

<em>Given: U = 10 m/s, ∅ = 60°</em>

<em>Constant: g = 9.8 m/s²</em>

Substituting these values into equation 1

R = [10²×sin(2×60)]/9.8

R = (100sin120)/9.8

R = 100×0.8660/9.8

R = 86.60/9.8

R = 8.84 m

<em>Therefore the ball travels a distance of 8.84 m</em>

4 0

3 years ago

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan

lianna [129]

Answer: E. 450 K

Explanation:

It is given that the temperature of the gaseous planet is linearly related with height of the atmosphere. we can write this in the mathematical equation:

y = m x +c

where y is the temperature values, x is height, m is the slope and c is the y-intercept. we have been given two sets of value in the image, using which we can find the value of slope in y-intercept.

at x = 18.40 km, y = 147.54 K

⇒147.54 = 18.40 m + c

⇒c = 147.54 - 18.40 m ..(1)

at x =78.11 km, y = 567.00 K

⇒567.00 =78.11 m + c ..(2)

Put equation 1 in 2 and solve:

⇒567.00 =78.11 m + 147.54 - 18.40 m

⇒419.46 = 59.71 m

⇒ m =419.46 ÷59.71 = 7.025 K/km

c = 147.54 - 18.40 × 7.025 = 18.28 K

At height, x = 61.5 km the approximate temperature is :

y = 7.025 K/km × 61.5 km + 18.28 K = 450.3 K

Thus, the approximate temperature at altitude 61.5 km is 450 K.

8 0

4 years ago

Which matter exists in the gaseous state at room temperature?

gulaghasi [49]

I would say (c.oxygen)

6 0

3 years ago