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Blizzard [7]
3 years ago
13

Acceleration involves a change in __________.

Physics
2 answers:
uysha [10]3 years ago
7 0

"Acceleration" is any change in speed or direction of motion ...
slowing down, speeding up, or curving away from a straight line.

You and I, and all other students of Physics, must re-educate
people.  We have to tell everyone that 'acceleration' does NOT
mean 'speeding up'.  AND, it doesn't even necessarily mean
any change in speed.

sweet-ann [11.9K]3 years ago
7 0
Speed. It is used in the equation force = mass X acceleration
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For a star with a parallax angle of 1/2 of an at arcsecond, what will be its distance in parsec?
avanturin [10]

(1 parsec) is the distance at which an object has a parallax of 1 arcsecond. The distance is about 3.26 light years.

Another way to understand it is:  The distance from which the Earth's orbit appears 1 arcsecond across.

For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).

1 arcsecond is 1/3600 of a degree, 0.00028 degree.  

8 0
3 years ago
A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of
posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0
3 years ago
A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

<u>Solution</u>:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

5 0
3 years ago
En la siguiente expresión matemáticas w=mg el peso w con relación a se relaciona con la masa m en una proporción
s2008m [1.1K]

Answer:

a) Directamente proporcional

Explanation:

El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.

Matemáticamente, el peso de un objeto viene dado por la fórmula;

Peso = mg

Donde;

m es la masa del objeto.

g es la aceleración debida a la gravedad.

De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.

Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.

4 0
3 years ago
Two positive charges of same magnitude are separated by some distance if we bring a unit positive charge from one charge to anot
Studentka2010 [4]

Answer:

Increases.

Explanation:

The electric potential increases when the two positive charges of same magnitude bring close to one charge to another because there is repulsive force between them due to same charge and when the two opposite charges move away from each other, the potential energy decreases. When two opposite charges are brought closer together, electric potential energy decreases while on the other hand, when we move opposite charges apart from each other than the work done against the attractive force that leads to an increase in electric potential energy.

3 0
3 years ago
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