All categories

3 years ago

~!Here's the question!~

Physics

2 answers:

vazorg [7]3 years ago

8 0

The correct answer is A) Gravity. Hope this helps.

gtnhenbr [62]3 years ago

6 0

The answer would most likely be A since obviously gravity weighs things down which helps the every other masses stay settled in place

You might be interested in

A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its moti

Crank

Answer:

a) μ = 0.475 , b) μ = 0.433

Explanation:

a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

X axis

Wₓ - fr = m a

the friction force has the expression

fr = μ N

y Axis

N - $W_{y}$ = 0

let's use trigonometry for the components the weight

sin 27 = Wₓ / W

Wₓ = W sin 27

cos 27 = W_{y} / W

W_{y} = W cos 27

N = W cos 27

W sin 27 - μ W cos 27 = m a

mg sin 27 - μ mg cos 27 = m a

μ = (g sin 27 - a) / (g cos 27)

very = tan 27 - a / g sec 27

μ = 0.510 - 0.0344

μ = 0.475

b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

μ = tan 25 - 0.3 / 9.8 sec 25

μ = 0.466 -0.03378

μ = 0.433

3 0

3 years ago

Helpon

statuscvo [17]

Answer:

huh ?ion understand

Explanation:

have a bad day <3

7 0

2 years ago

A small current element at the origin has a length of and carries a current of in the direction. Find the magnetic field due to

Harlamova29_29 [7]

Answer:

Answer and Explanation in Attached document

Explanation:

Download docx

7 0

3 years ago

If a ball of mass 0.30kg is dropped onto a flat surface, which it hits at 9m/s and rebounds to 3m. Find the rebound speed just a

VikaD [51]

Answer:

vinegar without v and i

Explanation:

4 0

2 years ago

A body with mass m slides down a frictionless ramp inclined at 600, with an initial speed v1 = 3 m/s,

Brut [27]

Going down the first ramp:

• net force parallel to the ramp:

∑ F = W sin(60°) = m a₁

(W for weight)

• net force perpendicular to the ramp:

∑ F = N + W cos(60°) = 0

(N for normal force)

The body has mass 0.1 kg, and with g = 10 m/s², its weight is W = 1 N. So in the first equation, we get

(1 N) sin(60°) = (0.1 kg) a₁ → a₁ ≈ 8.7 m/s²

Let d₁ be the distance the body moves down the ramp, i.e. the distance along the ramp between the starting point and the point O. Then

sin(60°) = h / d₁ → d₁ = 2h/√(3) ≈ 1.15h

Given an initial speed v₁ = 3 m/s, we find the speed v₂ at point O to be

v₂² - (3 m/s)² = 2 (8.7 m/s²) d₁

v₂ ≈ √(9 m²/s² + (17.4 m/s²) (1.15h))

v₂ ≈ √(9 m²/s² + (20 m/s²) h)

Going up the second ramp:

• net parallel force:

∑ F = -Fr - W sin(30°) = m a₂

(Fr for friction)

• net perpendicular force:

∑ F = N - W cos(30°) = 0

sin(30°) = cos(60°), and cos(30°) = sin(60°), so the second equation gives N = 0.87 N. Then with µ = 0.1, we have Fr = µ N = 0.087 N. The first equation then gives

-0.087 N - 0.5 N = (0.1 kg) a₂ → a₂ ≈ -5.9 m/s²

We now have

tan(30°) = h/R → h = (2.5 m)/√3 ≈ 1.4 m

(which, by the way, tells us that v₂ ≈ 6.2 m/s)

Then the distance traveled up the ramp is

d₂ = √(h² + R ²) ≈ 2.9 m

Use this to solve for the speed at the top of the ramp:

v₃² - v₂² = 2 (-5.9 m/s²) d₂

v₃ = √((6.2 m/s)² - (11.8 m/s²) (2.9 m)) ≈ 2.0 m/s

At the top of the second ramp:

• net parallel force:

∑ F = -Fsp - W sin(30°) = m a₂

(Fsp for spring)

• net perpendicular force:

∑ F = N - W cos(30°) = 0

By Hooke's law, Fsp = kx, so in the first equation we get

-k (0.10 m) - (1 N) cos(60°) = (0.1 kg) (-5.9 m/s²)

→ k ≈ 0.87 N/m

8 0

3 years ago

~~~!Here's the question!~~~

~!Here's the question!~