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Ray Of Light [21]
3 years ago
8

Referring to the above diagram, how high will the ball rise on the right-hand incline?

Physics
1 answer:
Marina CMI [18]3 years ago
3 0
I think it’s 15cm
Might be 7cm
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A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
What do we mean by a penetrative pass in the game of football?​
Ivanshal [37]

Answer:

Penetration means forward passes can go through the opposition lines. Once these penetrative passes get through each line, it eliminates the line of players it broke through and leaves the player in possession closer to the opposition goal.

Explanation:

6 0
2 years ago
A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres
Drupady [299]

Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

3 0
2 years ago
What is the Force in Newtons exerted by an object with a mass of 15 kg and an Acceleration of 12 m/s2?
ZanzabumX [31]
Force = mass * acceleration
F = ma
F = (15)(12)
F = 180N
5 0
3 years ago
Light travels at 300,000,000 m/s. This is an example
mafiozo [28]

Answer:

ook soooooo

Explanation:

3 0
2 years ago
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