Along the frictionless path you have chosen, the main force(s) acting on the puck after receiving the kick is (are):

Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ F (para) = -mg sin(21°) - f = ma

• The net force in the perpendicular direction is

∑ F (perp) = n - mg cos(21°) = 0

Solving the second equation for n gives

n = mg cos(21°)

n = (0.200 kg) (9.80 m/s²) cos(21°)

n ≈ 1.83 N

Then the magnitude of friction is

f = µn

f = 0.25 (1.83 N)

f ≈ 0.457 N

Solve for the acceleration a :

-mg sin(21°) - f = ma

a = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

a ≈ -5.80 m/s²

so the block is decelerating with magnitude

a = 5.80 m/s²

down the ramp.

5 0

3 years ago

A mango fruit drops from the top of it's tree which is 40 m. How long does it takes to reach the ground

IrinaVladis [17]

T=distance over speed
T=40m over 9.8ms
T=answer

4 0

3 years ago

What is your wheel and axle

Crank

Answer:

Pls mark me brainliest

Explanation:

The wheel and axle is a simple machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.

8 0

3 years ago

Read 2 more answers

A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object

Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

$Em_{f}$ = U = m g h

Initial mechanical energy, in the lower part of the track

Em₀ = K = ½ m v²

Em= $Em_{f}$

½ m v² = m g h

v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v

Initial before the crash

p₀ = M v₁₀ + 0

Final after the crash

$p_{f}$ = M v1f + m v

p₀ = $p_{f}$

M v₁₀ = M $v_{1f}$ + m v

As the shock is elastic the kinetic energy is conserved

K₀ = $K_{f}$

½ M v₁₀² = ½ M $v_{1f}$ ² + ½ m v²

Let's write the system of equations

M v₁₀ = M $v_{1f}$ + m v

M v1₁₀² = M $v_{1f}$ ² + m v²

We cleared v1f in the first we replaced in the second

$v_{1f}$ = (M v₁₀ - mv) / M

M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

2 m v₁₀ - v (m + 1) m/ M = 0

v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)

V₁₀ = 7.668 (2.68) / 10.82

v₁₀ = 1.90 m / s

5 0

3 years ago

True/False: The points where the potential is the same (in three-dimensional space) lie on a surface.

nikdorinn [45]

1. True
2. False

Does that answer your question?

5 0

3 years ago