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2 years ago

Which of the following equations illustrates the law of conservation of matter?

Physics

1 answer:

Anvisha [2.4K]2 years ago

8 0

Answer:

Explanation:

the balancing is correct in the first one

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Air pressure decreases as

bearhunter [10]

The answer is going to be B: elevation.

8 0

3 years ago

How do mass and force affect the acceleration of an object?

joja [24]

Gravity affects weight, it does not affect mass. Masses always remain the same. Newton's Second Law of Motion: Force = mass x acceleration The acceleration of an object is: a) directly proportional to the net force acting on the object. ... c) inversely proportional to the mass of the object.

3 0

3 years ago

Read 2 more answers

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

inysia [295]

Answer:

$v = 1,582 \ \frac{m}{s}$

Explanation:

We know that for circular motion the centripetal acceleration $a_c$ is:

$a_c = \frac{v^2}{r}$

where v is the speed and r is the radius.

The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

$a_c = 1.27 \frac{m}{s^2}$ .

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

$r = 1.7 * 10^6 \ m + 270 \ km$

$r = 1.7 * 10^6 \ m + 270 * 10^3 \ m$

$r = 1.7 * 10^6 \ m + 0.270 * 10^6 \ m$

$r = 1.97 * 10^6 \ m$

We can obtain the speed as:

$v^2 = a_c r$

$v = \sqrt{a_c r}$

$v = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}$

$v = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}$

$v = 1.582 \ 10^3 \ \frac{m}{s}$

$v = 1,582 \ \frac{m}{s}$

And this is the orbital speed.

7 0

3 years ago

physics a flower pot falls from a windowsill 25.0m above the sidewalk. how much time does a passerby on the sidewalk below have

Usimov [2.4K]

$v_f^2 = v_0^2 + 2*a*d$
Since the flower pot is dropped, it has an initial velocity of zero. Also the flower pot accelerates due to gravity.

so

$v_f = \sqrt{2*g*d}=\sqrt{2*9.81*25.0}$

8 0

3 years ago

A student measured the specific heat of water to be 4.29 J/g.Co. The

leonid [27]

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

$P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%$

So, the student's percent error is 2.63 %.

7 0

3 years ago