How are joules measured ?

1. Black Panther ran 567.5 ks to the east in 2.3 hours then tumed around and ran 2218 km to

mario62 [17]

Answer:Average Velocity of the trip=98.77km/h

Explanation:

Given that:

Black Panther started his journey with

Distance ran towards east = 567.5km

Time ran towards east = 2.3 hours

and then turned back,

Distance ran towards west, = 2218 km

Time taken towards west, = 1.2 hours

Let us treat the westward movement as negative movement of the east

So total displacement =567.5 km east + (-2218km east )= 1,650.5km

But it seems the value of the distance ran west is wrong , because i do not think he can run 2218km under 1.2hours considering he ran 567.5m due east at 2.3hours .

So Let we use 221.8 km as distance ran due west.

So that Total Displacement becomes = 567.5 km east + (-221.8km east )= 345.7km

Total time = 2.3 + 1.2 = 3.5hrs.

Average velocity = Total displacement / total time

.345.7km / 3.5hr = 98.77km/h

7 0

2 years ago

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di

vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi (\frac{D}{2})^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= \frac{d}{\pi}$

$r_e = \sqrt{\frac{d}{\pi} }$

replacing d = $\frac{\pi (\frac{D}{2})^2 }{10, 000}$ in the equation above; we have:

$r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }$

$r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}$

$r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

4 0

3 years ago

What is average speed?what is the formula for average speed?what does it mean?

Rama09 [41]

Average speed = total distance traveled/total time taken
it is the total distance traveled in a total time the total distance is attained

4 0

2 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

5 0

1 year ago

A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.

malfutka [58]

The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

time (s) x (m) y(m)

0 0 0

2.0 3.6 1.2

3.0 5.4 0.9

b) The aceleration is g = 0.6 m / s²

c) The launch angle θ = 33.7º

given parameters

the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s

the movement times t = 1.0s, 2.0s and 3.0 s

to find

a) position

b) acceleration

c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

v_y = v_{oy} - g t

where v and v({oy} are the velocities of the body, g the acceleration of the planet's gravity and t the time

0 = v_{oy} - gt

g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

g = 1.2 / 2.0

g = 0.6 m / s²

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

vₓ = x / t

x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

x₀ = 0

t = 2.0 s

x₂ = 1.8 2

x₂ = 3.6 m

t = 3.0 s

x₃ = 1.8 3

x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

y = v_{oy} t - ½ g t²

t = 0.0 s

y₀ = 0

y₀ = 0 m

t = 2.0 s

y₂ = 1.2 2 - ½ 0.6 2²

y₂ = 1.2 m

t = 3.0 s

y₃ = 1.2 3 - ½ 0.6 3²

y₃ = 0.9 m

c) the launch angle use the trigonometry relation

tan θ = $\frac{v_y}{v_x}$

θ = tan⁻¹ $\frac{v_y}{v_x}$

θ = tan⁻¹ $\frac{1.2}{1.8}$

θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

time (s) x (m) y(m)

0 0 0

2.0 3.6 1.2

3.0 5.4 0.9

b) The aceleration is g = 0.6 m / s²

c) The launch angle θ = 33.7ºto)

learn more about projectile launch here:

brainly.com/question/10903823

4 0

2 years ago