All categories

2 years ago

What is the average velocity of the particle from rest to 15 seconds?

Physics

2 answers:

Hitman42 [59]2 years ago

5 0

Displacement/Time = 21m / 15s .

That's choice-D .

sdas [7]2 years ago

4 0

Answer : The correct option is, (D) 1.4 meters/second

Solution :

Average velocity : It is defined as the displacement per unit time.

Formula used for average velocity :

$v_{av}=\frac{d}{t}$

where,

$v_{av}$ = average velocity

d = displacement of the particle

t = time taken

Now put all the given values in the above formula, we get the average velocity of the particle.

$v_{av}=\frac{(21-0)m}{15s}=1.4m/s$

Therefore, the average velocity of the particle is, 1.4 meter/second.

You might be interested in

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

2 years ago

When an object falls, it trades gravitational potential energy for kinetic energy, accelerating toward the ground. calculate the

liq [111]

The speed of the energy decreases

5 0

3 years ago

Moving a charged object against an electrical force causes it to

Alinara [238K]

Decelerate

Explanation:

If a charged object is moving against an electric force, the electric force would cause the charged object to decelerate. Rate of deceleration would depend on the amount of the charge the object posses and amount of the opposing electric force.

This could be understood by visualising a hypothetical situation where a charged object is moving against an electric force. Since the object is charged, it would exert a force in its direction of motion which would be opposed by the electric force, thus causing it to decelerate

5 0

3 years ago

A young parent is dragging a 65 kg (640 N) sled (this includes the mass of two kids) across some snow on flat ground, by means o

Delicious77 [7]

Answer:

b) N = 560 N, c) fr = 138.56 N, d) μ = 0.247

Explanation:

a) In the attachment we can see the free body diagram of the system

b) Let's write Newton's second law on the y-axis

N + T_y -W = 0

N = W -T_y

let's use trigonometry for tension

sin θ = T_y / T

cos θ = Tₓ / T

T_y = T sin θ

Tₓ = T cos θ

we substitute

N = W - T sin 30

we calculate

N = 640 - 160 sin 30

N = 560 N

c) as the system goes at constant speed the acceleration is zero

X axis

Tₓ - fr = 0

Tₓ = fr

we substitute and calculate

fr = 160 cos 30

fr = 138.56 N

d) the friction force has the formula

fr = μ N

μ = fr / N

we calculate

μ = 138.56 / 560

μ = 0.247

4 0

2 years ago

a telescope is oriting on a spacecraft aroudn the earth. Speed of 3.25 x 10^5 mass 7500 kg. What is the de Broglie wavelength of

Scrat [10]

Answer:

2.72 x 10^-43 m

Explanation:

mass of the telescope = 7500 kg

speed of the telescope = 3.25 x 10^5 m/s

de Broglie's wavelength of the telescope is given as

λ = h/mv

where

λ is the wavelength of the telescope

h is the plank's constant = 6.63 × 10-34 m^2 kg/s

m is the mass of the telescope = 7500 kg

v is speed of the telescope = 3.25 x 10^5 m/s

substituting value, we have

λ = (6.63 × 10-34)/(7500 x 3.25 x 10^5)

λ = 2.72 x 10^-43 m

8 0

3 years ago