If a charge of 12 C flows past any point along a circuit in _____ seconds, the current at that point would be 3 A.

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0

3 years ago

Like his parents, Thomas is short, stocky, and heavyset. Although he exercises regularly and eats a well-balanced meal, his body

Viktor [21]

Answer:

Genetics?

Explanation:

8 0

3 years ago

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The wheels of an automobile are locked as it slides to a stop from an initial speed of 30.0 m/s. If the coefficient of kinetic f

Amiraneli [1.4K]

Answer:

x = 76.5 m

Explanation:

Let's use Newton's second law at the point of contact between the wheel and the floor.

fr = m a

fr = miy N

N-W = 0

N = W

μ mg = m a

a = miu g

a = 0.600 9.8

a = 5.88 m / s²

Having the acceleration we can use the kinematic relationships to find the distance

$v_{f}$ ² = v₀² + 2 a x

$v_{f}$ = 0

x = -v₀² / 2 a

Acceleration opposes the movement by which negative

x = - 30²/2 (-5.88)

x = 76.5 m

8 0

3 years ago

What are the two ost important processes in the oxygen in and out of the atmospher

BaLLatris [955]

Decomposing - When plants and animals die, they decompose. This process uses up oxygen and releases carbon dioxide. Rusting - This is also called oxidation. When things rust they use up oxygen.
$hope \: i \: helped \\ \\ \\ \\ \\ \\ darklynx$

3 0

3 years ago

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Block A weighs 14 lb and block B weighs 5 lb If B is moving downward with a velocity (vB)1 = 3 ft/s at t = 0, determine the velo

kolezko [41]

Velocity, va2 = 10.5 ft/s

<u>Explanation:</u>

From the figure:

Length of the cable = Sa + 2Sb = l

∴ vₐ = -2vb

Applying the principle of Impulse and momentum in x-direction

$mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2$

Limit is t1 to t2

$-(\frac{14}{32.2}) (2) (3) - T = \frac{14}{32.2} v_y_2$ -(1)

Applying the principle of Impulse and momentum in y-direction

$mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2$

Limit is t1 to t2

$(\frac{5}{32.2}) (3) - 2T + 3 = -\frac{5}{32.2} \frac{v_y_2}{2}$ -(2)

Solving equation (1) and (2), we obtain

T = 1.6lb

va2 = 10.5 ft/s

8 0

3 years ago