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andrezito
5 days ago
12

An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste

ad it were already moving at speed v and the same amount of work W was done on it. What would be its final speed
Physics
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A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Sav [3153]
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
1 month ago
A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the base at a velocity of 15.5 m/s
Yuliya22 [3333]
This can be determined using the principle of energy conservation. The ski lift begins with a velocity of v= 15.5 m/s, and all of its kinetic energy Ek converts into potential energy Ep, thus we set Ep equal to Ek.
Because Ek is given by (1/2)*m*v², where m denotes mass and v represents speed, while Ep equals m*g*h, where m is mass, g is 9.81 m/s², and h is height. Now:

Ek=Ep 

(1/2)*m*v²=m*g*h, canceling out the mass,

(1/2)*v²=g*h, rearranging for height by dividing by g,

(1/2*g)*v²=h and substituting the values:

h=12.245 m. The hill's height rounded to the nearest tenth is h=12.25 m.

5 0
2 months ago
Read 2 more answers
Determine the mass of a ball with a velocity of 40.0 m/s and a wavelength of 8.92 Ã 10-34 m.
Sav [3153]
The wavelength can be calculated as Planck's constant divided by the momentum of the ball.
This translates to:
lambda = h / p.............> equation I
Momentum is equal to mass times velocity............> equation II

By substituting equation II into equation I, we obtain:
lambda = h / mv
Here are the values provided:
lambda = 8.92 * 10^-34 m
Planck's constant = 6.625 * 10^-34
velocity = 40 m/sec

Substituting these values into the previous equation, we calculate the mass as follows:
8.92*10^-34 = (6.625*10^-34) / (40*m)
mass = 0.0185678 kg

4 0
1 month ago
The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2
serg [3582]

Respuesta:

Explicación:

Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!

0 0
3 months ago
How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov
Keith_Richards [3271]

Answer:

x = v₀ cos θ   t,   y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.

Now, we will apply trigonometry to determine the initial velocity components.

              sin θ = v_{oy} / v₀

              cos θ = v₀ₓ / v₀

              v_{y} = v_{oy} sin θ

              v₀ₓ = v₀ cos θ

Next, let's formulate the equations of motion.

X axis

         x = v₀ₓ t

         x = v₀ cos θ   t

        vₓ = v₀ cos θ

Y axis

        y = y₀ + v_{oy} t - ½ g t2

        y = y₀ + v₀ sin θ t - ½ g t2

        v_{y} = v₀ - g t

       v_{y} = v₀  sin θ - gt

        v_{y}^{2} = v_{oy}^2 sin² θ - 2 g y

It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components

7 0
1 month ago
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