How many laws does newton have?

Where are all stars bornWhat is the brightest star in the night sky? What is its magnitude?

sergij07 [2.7K]

Answer:Stars are born within the clouds of dust and scattered throughout most galaxies.It's the star Sirius in the constellation Canis Major, brightest star in the sky. The bright planet Venus is also up before dawn now. But you'll know Sirius, because Orion's Belt always points to it.With an apparent magnitude of −1.46

Explanation: Brainliest Please!!!!

5 0

3 years ago

A rock at rest has weight 138 n. what is the weight of the rock when it is accelerating downward at 3.5 m/s2?

tamaranim1 [39]

<span>The weight of the rock is 138n when accelerating downward at 3.5m/ s2.Weight is a measurement of the force of gravity on an objects mass. Therefore as long as the force of gravity is the same and the mass of the rock is the same the weight should also be the same.</span>

7 0

3 years ago

An object traveling at constant speed v in a circle of radius R has an acceleration am 5 m/s2. If both R and v are doubled, what

MakcuM [25]

Answer:

im pretty sure its 10 m/s but its kinda hard sorry

Explanation:

7 0

3 years ago

Your friend has decided to make some money during the next State Fair by inventing a game of skill. In the game as she has devel

coldgirl [10]

Answer: from the vertical, one should aim 86.6°

Explanation:

height of the center of object = 7.0 m - 0.05 m = 6.95 m

now let the bullet hits centre at point A height x meters from the ground

also let t be the time taken for the bullet to hit the object

so distance travelled by the target will be

d = h - x = 6.95 - x

now using the equation of motion

d = 1/2gt²

so 1/2gt² = 6.95 - x

x = 6.95 - 1/2gt² .........let this be equ 1

let angle of fire be ∅

so v(cos∅) × t = 100

our velocity v is 1200 ft/sec = 365.76 m/s

365.76(cos∅) × t = 100 ........equ 2

also vertical position of the bullet after t is

y = y₀ + c(sin∅)t - 1/2gt²

y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3

After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;

x = y

we substitute

equ 1 = equ 3

6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²

6.95 - 1 = 365.76(sin∅)t - 1/2gt² + 1/2gt²

5.95 = 365.76(sin∅)t

t = 5.96 / 365.76(sin∅)

now input the above equ into equ 2

365.76(cos∅) × 5.96/365.76(sin∅) = 100

5.95(cos∅)/sin∅ = 100

tan∅ = 5.95/100 = 0.0595

∅ = 3.40°

therefore from the vertical, one should aim (90° - 3.40°) = 86.6°

4 0

4 years ago

g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k

Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

$(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s$

Now, we will apply the law of conservation of momentum:

$m_1v_1 = m_2v_2$

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

$(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s$

Now, we again use the third equation of motion for the upward motion of the ball:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

$(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\$

6 0

3 years ago