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3 years ago

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Particles (mass of each = 0.40 kg) are placed at the 60-cm and 100-cm marks of a meter stick of negligible mass. This rigid body

is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the magnitude of the initial linear acceleration of the end of the body opposite the pivot?

1 answer:

hichkok12 [17]3 years ago

7 0

Answer:

12 m/s ∧2

Explanation:

The picture attached explains it all and i hope it helps. Thank you

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A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

On a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the grou

Dafna1 [17]

The mechanical energy is lost due to friction between the two peak is 78,458.688 J

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the ground and peak L is at 24.4 m. At location K the speed of the cart is 16.4 m/s, and at location L it is 12.4 m/s.

Total energy at peak K,

TE₁ = 1/2 mv₁² +mgh₁

Substitute the values, we get

TE₁ = 1/2 x319 x 16.4² +319 x 9.81 x 43.6

TE₁ = 179,340. 524 J

Total energy at peak L,

TE₂ = 1/2 mv₂² +mgh₂

Substitute the values, we get

TE₂ = 1/2 x319 x 12.4² +319 x 9.81 x 24.4

TE₂ = 100,881. 836 J

The mechanical energy lost is

M.E = TE₁ -TE₂

M.E = 179,340. 524 J - 100,881. 836 J

M.E = 78,458.688 J

Thus, the mechanical energy is lost due to friction between the two peak i

Learn more about mechanical energy.

brainly.com/question/13552918

#SPJ1

7 0

2 years ago

A truck travels at a constant speed of 45 kilometers per hour. How far does the truck travel in 20 minutes?

Mekhanik [1.2K]

The answer is 15 kilometers in 20 minutes.

6 0

3 years ago

What is the disorder caused when cells lose the ability to control growth and continue to divide?

8090 [49]

The answer is cancer that is the reason there is no complete cue in people

7 0

3 years ago

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

Gennadij [26K]

Answer:

18.1 × 10⁻⁶ A = 18.1 μA

Explanation:

The current I in the wire is I = ∫∫J(r)rdrdθ

Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.

So, dI = J(r)rdrdθ

dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²

Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m

I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA

3 0

4 years ago