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2 years ago

PLEASE HELP.

1 answer:

3 0

The Solar System moves through the galaxy with about a 60° angle between the galactic plane and the planetary orbital plane. The Sun appears to move up-and-down and in-and-out with respect to the rest of the galaxy as it revolves around the Milky Way and
This speed is not unusual for the stars around us and is our “milling around” speed in our suburban part of the Galaxy. In addition to the individual motions of the stars within it, the entire Galaxy is in spinning motion like an enormous pinwheel.

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You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis

tangare [24]

Answer:

56.86153 N

Explanation:

t =Time taken

F = Force

Power

$P=\frac{W}{t}\\\Rightarrow W=P\times t\\\Rightarrow W=77\times 0.96\\\Rightarrow W=73.92\ Joules$

Work done

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.92}{1.3}\\\Rightarrow F=56.86153\ N$

The magnitude of the force that is exerted on the handle is 56.86153 N

5 0

3 years ago

When water in a lake freezes, the ice that forms floats on top of any water that is still liquid. Why does the ice float?

ddd [48]

Answer:

it floats because the lake is cold at that moment so when part of the lake freezes it still remains solid and floats because of the lake and the surrounding of the lake is still cold

5 0

3 years ago

2 Points

Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

$:\implies$ Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

$:\implies$ 50 N = 10 kg × Acceleration

$:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}$

$:\implies$ Acceleration of the Object = 5 m/s²

4 0

3 years ago

The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from

Allushta [10]

Answer:

y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

let's substitute

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

0 = v₀ sin θ - 9.8 t

Let's write our system of equations

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

v₀ sin θ = 9.8 t

we substitute

10 = (9.8 t) t - ½ 9.8 t2

10 = ½ 9.8 t2

10 = 4.9 t2

t = √ (10 / 4.9)

t = 1,429 s

Now let's use the first equation and the last one

45 = v₀ cos θ t

0 = v₀ sin θ - 9.8 t

9.8 t = v₀ sin θ

45 / t = v₀ cos θ

we divide

9.8t / (45 / t) = tan θ

tan θ = 9.8 t² / 45

θ = tan⁻¹ ( 9.8 t² / 45 )

θ = tan⁻¹ (0.4447)

θ = 24º

Now we can calculate the maximum height

v_y² = $v_{oy}^2$ - 2 g y

vy = 0

y = v_{oy}^2 / 2g

y = (20 sin 24)²/2 9.8

y = 3,376 m

the other angle that gives the same result is

θ‘= 90 - θ

θ' = 90 -24

θ'= 66'

for this angle the maximum height is

y = v_{oy}^2 / 2g

y = (20 sin 66)²/2 9.8

y = 17 m

thisis the correct

6 0

3 years ago

Before partacting, if correct marked as brainest

Molodets [167]

Answer:

C. Basic swimming capability.

Explanation:

one of the way before participating in any water-based sport is know how to do the basic swimming way.

7 0

2 years ago