According to Newton's law of universal gravitation, which of these objects would have the strongest gravitational force?

A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Fin

Ira Lisetskai [31]

Answer:

a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

v = x / t

x = v t

we substitute

x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

y = v₀ t + ½ a t²

y = ½ at t²

we substitute

y = ½ 1.00 t²

y = 0.500 t²

in vector position is

d = x i ^ + y j ^

d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

vₓ = v₀

vₓ = 6.00 m / s

y Axis

there is an acceleration and the initial speed is zero

$v_{y}$ = v₀ + a t

v_{y} = a t

v_{y} = 1.00 t

the velocity vector is

v = vₓ i ^ + v_{y} j ^

v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

d = (6.00 4 i ^ + 0.50 4 2 j⁾

d = (24.00 i ^ + 8.00 j^ ) m

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

v = (6 i ^ + 1 5 j ^)

v = (6.00 i ^ + 5 j^ ) m/s

7 0

3 years ago

A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars

shtirl [24]

In order to compute the final velocity of the trains, we may apply the principle of conservation of momentum which is:
initial momentum = final momentum
m₁v₁ = m₂v₂

The final mass of the trains will be:
10,000 + 10,000 = 20,000 kg
Substituting the values into the equation:

10,000 * 3 = 20,000 * v
v = 1.5 m/s

The final velocity of the trains will be 1.5 m/s

3 0

3 years ago

What is the order of magnitude of the distance of Sun to nearest star in meters?

neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0

2 years ago

Read 2 more answers

Mike has a mass of 97 kg. He jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when

Ne4ueva [31]

Answer:

a) fr = 224.3 N , b) fr = 224.3 N , c) v = 198.0 m/s

Explanation:

a) For this exercise let's start by calculating the acceleration in the fall

v² = v₀² - 2 a (y-y₀)

When it jumps the initial vertical speed is zero

a = -v² / 2 (y-y₀)

a = -68 2/2 (1000-2000)

a = 2,312 m / s²

Let's use the second net law to enter the average friction force

fr = m a

fr = 97 2,312

fr = 224.3 N

b) let's look for acceleration

v² = v₀² - 2 a y

a = (v² –v₀²) / 2 (y-y₀)

a = (4² - 68²) / 2 (0-1000)

a = 2,304 m / s²

fr = m a

fr = 97 2,304

fr = 223.5 N

c) the speed of the wallet is searched with kinematics

v² = v₀² - 2 g (y-y₀)

v = √ (0-2 9.8 (0-2000))

v = 198.0 m/s

4 0

3 years ago

Which part of the central nervous system controls reflexes?

larisa86 [58]

Answer:

spinal cord becuse it does all of the reflexes

6 0

3 years ago

Read 2 more answers