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3 years ago

A force that causes an object to change its motion is called?

1 answer:

8 0

Unbalanced force.

The definition of an unbalanced force is a force that changes the position, speed or direction of the object to which it is applied.

When an object is still it means that is is balanced force. So you would need to add more force to another side to make it move. Which is unbalanced force.

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Light travels at a speed of 3.00 x 10^11 cm/s. What is the speed of light in kilometers/hour?

rewona [7]

As 1 km = 1000 m = 1000,00 cm,

So, 1 cm = (1/1000,00) km

1 hour = 60 × 60 s = 3600 s

So, 1 s = (1 / 3600) hour

The light travels at a speed of $3.00 \times 10^{11} \ cm/s$ .

In kilometer/hour,

$3.00 \times 10^{11} \ cm/s = 3.00 \times 10^{11} \frac{(1/1000,00) km}{ (1 / 3600) hour} = 108 \times 10^8 \ km/hour$

4 0

4 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

4 years ago

Based on the evidence for the impact theory, what was the most probable order of events for the collision that led to the format

anastassius [24]

Answer:

Answered

Explanation:

According to the impact theory, a glancing collision between a Mars-sized object and Earth led to the formation of the Moon. The iron core of Earth had formed before this collision, leading to the similarity between the composition of the Moon and Earth's mantle. After the collision, any iron core of the Mars-sized object would have been left behind on Earth and eventually merged with Earth's core. The Moon then formed out of the debris thrown into space by the collision.

5 0

3 years ago

A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position

irga5000 [103]

(a) The period of the oscillation is 0.8 s.

(b) The frequency of the oscillation is 1.25 Hz.

(d) The amplitude of the oscillation is 3 cm.

(e) The force constant of the spring is 148.1 N/m.

The given parameters:

<em>Mass of the ball, m = 2.4 kg</em>

<em />

From the given graph, we can determine the missing parameters.

The amplitude of the wave is the maximum displacement, A = 3 cm

The period of the oscillation is the time taken to make one complete cycle.

T = 0.8 s

The frequency of the oscillation is calculated as follows;

$f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz$

The angular frequency of the oscillation is calculated as follows;

$\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s$

The force constant of the spring is calculated as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m$

Learn more about general wave equation here: brainly.com/question/25699025

4 0

3 years ago

How does an objects size affect the gravity exerted by that object

egoroff_w [7]

Answer:

The size of an object is directly proportional to the gravity

Explanation:

The size of an object has significant impact on the gravity exerted by such a body.

The more massive a body is, the larger the gravity it exerts.

The reason for this is because of the newton's law of universal gravitation.

It states that "the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
As such, gravity is directly proportional to mass

4 0

3 years ago