10. What are the only types of passes allowed in floor hockey?

Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi

Leviafan [203]

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

W = 12.96 J

5 0

3 years ago

A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block

mel-nik [20]

Answer:

v₁ = 0.375 m / s , x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

$p_{f}$ = (m + M) v₁

how the forces are internal

p₀ =p_{f}

m v₀ = (m + M) v₁

v₁ = m /m+M v₀

let's calculate

v₁ = 0.4 /(0.4 + 2.8) 3

v₁ = 0.375 m / s

Let's apply Newton's second law to the Block, to find the friction force

Y axis

N - W = 0

N = W

N = m g

where m is the mass of the block

the friction force has the formula

fr = μ N

fr = μ m g

We apply Newton's second law to slab

X axis

fr = M a

where M is the mass of the slab

μ m g = M a

a = μ g m / M

let's calculate

a = 0.15 9.8 0.4 / 2.8

a = 0.21 m / s²

With kinematics we can find the position

v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

v² = 2 a x

x = v2 / 2a

let's calculate

x = 0.375²/2 0.21

x = 0.335 m

4 0

2 years ago

A bicycle is 10 m from a wall traveling at a rate of 4 m/s. How far from the wall will it be if it accelerates at a rate of 12 (

kompoz [17]

The bicycle is a distance x away from the wall at time t according to

x = 10 m + (4 m/s) t + 1/2 (12 m/s²) t ²

so that after t = 8 s, it will be

x = 10 m + (4 m/s) (8 s) + 1/2 (12 m/s²) (8 s)² = 426 m

away from the wall.

5 0

2 years ago

To pull a 57 kg crate across a horizontal frictionless floor, a worker applies a force of 230 N, directed 34° above the horizont

lisabon 2012 [21]

Answer:

$a. W_w=235\ J\\b. W_g=-343.54\ J\\c. F_N=463.100\ N\\d. W_t=235\ J$

Explanation:

Given: that,

Angle of inclination of the surface, $\theta=34^{\circ}$

mass of the crate, $m=57\ kg$

Force applied along the surface, $F=230\ N$

distance the crate moves after the application of force, $s=1.1\ m$

a) work done = F× s

work done = 230 × 1.1

work done = 253 J

b) Work done by the gravitational force:

$W_g=m.g\times h$

where:

g = acceleration due to gravity

h = the vertically downward displacement

Now, we find the height:

$h=s\times sin\ \thetah=1.1\times sin\ 34^{\circ}h=0.615\ m$

So, the work done by the gravity:

$W_g=57\times 9.8\times (-0.615) \\= - 343.54 J$

∵direction of force and displacement are opposite.

= - 343.54J

c)

The normal reaction force on the crate by the inclined surface:

$F_N=m.g.cos\ \thetaF_N=57\times 9.8\times cos\ 34F_N=463.100\ N$

d)

Total work done on crate is with respect to the worker:

$W_t=235\ J$

5 0

3 years ago

How far would a cantaloupe fall in 5.0 seconds of true free fall?

RUDIKE [14]

Answer:

i think the answer is about 50 m/s

7 0

2 years ago

10. What are the only types of passes allowed in floor hockey?​

10. What are the only types of passes allowed in floor hockey?