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3 years ago

Use the rules for significant figures to find the answer to the following addition problem: "21.4+15+17.17+4.003"

Physics

1 answer:

harina [27]3 years ago

5 0

Answer:

57.6

Explanation:

"Significant figures" refer to figures that have an actual contribution to a number's value.

They refer to all digits except those with "leading zero," which means having a zero digit before a specific number. For example, in the number 0234, the significant figures are 234.

The rule for adding and subtracting significant figures is to round off the answer to the least number of decimal places. However, when it comes to multiplying and dividing significant figures, you have round off the answer to the least number of significant digits.

Let's solve.

21.4

+15

17.17

4.003

57.573

Let's now round off to the least number of decimal places, which is the tenths. Since 0.5 is followed by a number greater than 5, we have to round it off to 6. Therefore, the answer is: 57.6

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PHYSICS 50 POINTS PLEASE HELP

tangare [24]

Answer:

One way to look at Newton’s three laws of motion is this:

The third law states what forces are. That is, all forces are interactions between two different objects. If one object is interacting with another, then equal and opposite forces act on each object. So no force acts alone. When you exert a force on something, it is exerting the identical force back on you.

The first and second laws deal with the consequences of the forces that act on an object. The first law says that in the absence of a net force on an object, it simply continues doing whatever it was already doing. If it is at rest, it will remain at rest. If it is in motion, it will continue with that same motion - at constant speed and in the direction it was already traveling.

The second law says what happens if there is a net force on the object. In that case, the object accelerates - either by changing its speed, its direction, or both - in proportion and in the direction of the net force that acts on it. The amount of acceleration depends the object’s mass. That is, the larger the mass the smaller the acceleration for a given net force. The first and second laws can be summarized in the mathematical expression

F = ma

where F is the vector sum of all the forces that act on the object at any given moment (i.e., the net force), m is the mass of the object, and a is the acceleration of the object due to the net force at that moment - and is always in the same direction of the net force.

And notice that in a way, the first law is then “contained” within the second. That is, if the net force is zero on an object, then so is the acceleration. That is, either the object is (still) at rest or, if already in motion, the velocity didn’t change, in either case, the acceleration was zero.

Explanation:

4 0

3 years ago

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WORTH 50 POINTSSSS!!!!!!!! don't lie either if you do I will report your answer and get my points back idc !!!!!!

BARSIC [14]

Answer:

Explanation:

3 0

3 years ago

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

irina [24]

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

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

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

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

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

To learn more about mass, refer to:

brainly.com/question/3187640

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4 0

2 years ago

Not sure what to put this under

cestrela7 [59]

family 16 cause i said so XD

5 0

3 years ago

In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas

Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

3 0

3 years ago

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