All categories

1 year ago

Newton's laws of motion are true on earth and space.

Physics

1 answer:

frez [133]1 year ago

8 0

Answer:

False

Explanation:

It is a common misunderstanding that objects in space have no weight. If that were true, they would just float away from the Earth, the Sun and the other planets. Objects in low Earth orbit experience about 90% of the weight that they feel on the surface of the Earth.

You might be interested in

Give two differences between Mass and weight

nydimaria [60]

Answer:

my opi

Explanation:

Mass is like the inside of the weight. weight is just home much it weighs

8 0

3 years ago

Name three elements that are good conductors of heat

Vera_Pavlovna [14]

Copper, gold and silver are three of them.

7 0

3 years ago

Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i

UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

= 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

V = k q/r

Where,

= 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

V₁ = 9 x 10⁹ x 30 / 0.3

= 1.5 x 10¹² J

And,

V₂ = 1.5 x 10¹² J

The energy stored in it is,

W = V₂ - V₁

= 0

Hence, the energy stored in the charge is, W = 0

6 0

3 years ago

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:

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