A roller coaster at the top of a 45 meter hill has a mass of 5000 kilograms. What is the potential energy?

What material structure explanation lies behind the fact that the propagation velocity of longitudinal waves is the lowest in ga

Lina20 [59]

Answer:

What material structure explanation lies behind the fact that the propagation velocity of longitudinal waves is the lowest in gases and the highest in solids?

8 0

3 years ago

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at t

il63 [147K]

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

6 0

3 years ago

The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will

ExtremeBDS [4]

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

The thermal expansion of materials is described by the expression

ΔL = α Lo ΔT

α = $\frac{\Delta L}{L_o \ \Delta T}$

in the case of the bar the expansion is

ΔL = L_f - L₀

ΔL= 1.002 -1

ΔL = 0.002 m

the temperature variation is

ΔT = 100 - 30

ΔT = 70º C

we calculate

α = 0.002 / 1 70

α = 2,857 10⁻⁵ ºC⁻¹

5 0

3 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

3 years ago

La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit

katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

v = dr / dt

the function of position

r = 2 cos πt i^ + 3 sin πt j^

let us note that it is a movement in two dimensions

let's perform the derivative

v = -2π sin πt i^ + 3π cos πt j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

v = (-4.44 i^ + 6.66 j^ ) m/s

To calculate the mean acceleration we use the expression

a = ( $v_{f} - v_{o}$ ) / Δt

indicates that the time is the first 3 s

we look for the initial velocity t = 0 s

v₀ = 0 i ^ + 3π j ^

we look for the fine velocity, t = 3 s

v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

Δt = (3 -0) = 3 s

a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

a_average = (0 i ^ -2π j ^ ) m/s²

6 0

3 years ago