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BlackZzzverrR [31]

3 years ago

7

a trolley of mass 20kg was originally at rest on a smoth horizontal surface. bu how much will it accelerate if a pulling force o

f 22N is applied on it horizontally?

1 answer:

DochEvi [55]3 years ago

6 0

Answer:

just divide 22 N by 20 kg to get the acceleration in m/s2

Explanation:

I hope this is right-

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MA_775_DIABLO [31]

Answer: C) Label 1: X

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Label 1: Lacks free electrons must be placed in the area marked X as they correspond to insulators.

Label 2: Aluminum is an example of conductor and thus must be labelled in area marked by Z.

Area Y corresponds to semiconductors which have properties intermediate to conductors and insulators.

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A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and

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The eccentricity of its orbit is $U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

<h3>What is mass?</h3>

Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where $, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$$

$M=1.99 \times 10^{30} \mathrm{~kg}=$ mass of sur

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$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

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1 year ago