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fredd [130]
2 years ago
9

a 13kg block is at rest on a level floor. A 400 g glob of putty is thrown at the block so that the putty travels horizontally, h

its the block, at sticks to it. The block and putty slide 15cm along the floor. if the coefficient of sliding friction is 0.4, what is the initial speed of the putty
Physics
1 answer:
stich3 [128]2 years ago
7 0

Answer:

36.3m/s

Explanation:

We are given that

Mass of block, m=13 kg

Mass of glob,m'=400g=0.4 kg

1 kg=1000g

Total mass, M=13+0.4=13.4 kg

Distance, s=15 cm=0.15m

1m=100cm

Coefficient of sliding friction, \mu=0.4

We have to find the initial speed of the putty.

Friction force, f=\mu Mg=0.4\times 13.4\times 9.8N

Where g=9.8m/s^2

Work done=Fs

Work done=0.4\times 13.4\times 9.8\times 0.15

Work done=7.8792 J

Work done =K.E=1/2 MV^2

7.8792=\frac{1}{2}(13.4)V^2

V^2=7.8792\times 2/13.4=1.176

V=1.0844m/s

Using conservation of linear momentum

mu+ m'v=MV

13(0)+0.4v=13.4(1.0844)

0.4v=14.53

v=\frac{14.53}{0.4}

v=36.3m/s

Hence, the initial speed of the putty=36.3m/s

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Calcular la longitud del faldón de una Rampa de Acceso , que en planta tiene una longitud de 20 m y la pendiente es 27%.
seraphim [82]

La longitud del faldón de la rampa es de 5.4 m.

 

La pendiente expresada en porcentaje sigue la siguiente ecuación:

m=\frac{y}{x}*100 (1)

Donde:

  • y es la elevacion de la rampa (faldón)
  • x es la longitud de la ramapa (20 m)

Sabemos que la pendiente es de 27%. Por lo tanto, usando la ecuación 1, despejamos y.

27=\frac{y}{20}*100

y=\frac{27*20}{100}

y=5.4\: m        

La longitud del faldón es 5.4 m

Pudes ver más sobre el tema aquí:

brainly.com/question/8906330

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3 years ago
A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br
serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

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3 years ago
If someone is driving 100 miles in 60 minutes then drives 150 miles in 100 minutes west, what is his acceleration rate.
Sliva [168]

Answer:

his acceleration rate is -0.00186 m/s²

Explanation:

Given;

initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)

time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s

final position of the car, x₁ = 150 miles = 241,350 m

time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s

The initial velocity is calculated as;

u = 160, 900 m / 3,600 s

u = 44.694 m/s

The final velocity is calculated as;

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The acceleration is calculated as;

a = \frac{\Delta V}{\Delta t} = \frac{v- u}{t_1 - t_ 0} = \frac{40.225  - 44.694}{6000-3600} = -0.00186 \ m/s^2\\\\

Therefore, his acceleration rate is -0.00186 m/s²

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