In which wave do the particles of the medium vibrate farthest from their rest

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

3 years ago

A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc

Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

Given the following data;

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

Angle = 41.7° Northeast.

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

Vr = 7.08m/s

6 0

3 years ago

A block is at rest. The coefficients of static and kinetic friction are s = 0.81 and k = 0.69, respectively. The acceleration of

lys-0071 [83]

Answer:

Explanation:

a ) The angle required = angle of repose = θ

Tanθ = .81

θ = 39⁰

b ) when angle of incline θ = 44

Net force on the block = mg sinθ - μ mg cosθ where μ is coefficient of kinetic friction

acceleration = gsinθ - μ g cosθ

= 9.8 ( sin44 - μ cos44 )

= 9.8 ( .695 - .69 x .72 )

= 9.8 ( .695 - .497 )

= 1.94 m /s²

7 0

3 years ago

A shopper weighs 4.00 kg of apples on a supermarket scale whose spring obeys Hooke's law and notes that the spring stretches a d

Irina-Kira [14]

Answer:

a) 0.040625 m

b) 5.02272 J

Explanation:

k = Spring constant

x = Stretched length

F = Force

a)

$F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{4\times 9.81}{0.025}\\\Rightarrow k=1569.6\ N/m$

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{6.5\times 9.81}{1569.6}\\\Rightarrow x=0.040625\ m$

Extension of the spring would be 0.040625 m

b) Work done in a spring

$W=\frac{1}{2}kx^2\\\Rightarrow W=\frac{1}{2}\times 1569.6\times 0.08^2\\\Rightarrow W=5.02272\ J$

The work done by the shopper to stretch this spring a total distance of 8.00 cm is 5.02272 J

4 0

3 years ago

Help me please........................

stepan [7]

Answer:

Left

Explanation:

3 0

3 years ago