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7nadin3 [17]
3 years ago
7

The fundamental frequency of a string fixed at both ends is 316 hz. how long does it take for a wave to travel the length of thi

s string?
Physics
1 answer:
lina2011 [118]3 years ago
8 0

yfzufsfusgsuzguugzdigidggddyi

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an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni
stiv31 [10]
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N
3 0
3 years ago
A ball is thrown upward with an initial velocity of 13 m/s. Using the approximate value of
dem82 [27]

Answer:

v=v0 - gt

Explanation:

The equation for velocity is

v=v0 - gt

where v0=14m/s, g=10m/s^2.

in 1 second:

v=14-10=4m/s

it is positive so direction is upwards

in 2 seconds:

v=14-20=-6m/s

it is negative so direction is downwards

5 0
3 years ago
I really need help.
Zielflug [23.3K]
2/5 = .4
.4*100= 40%
Alex spends more time
4 0
3 years ago
A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0
3 years ago
A toy cannon launches a 46-g golf ball straight up into the air with a kinetic energy of 6.8 J. What must the
Shkiper50 [21]

Answer : The correct option is, (C) 17 m/s

Explanation :

Formula used :

K.E=\frac{1}{2}mv^2

where,

K.E = kinetic energy = 6.8 J

m = mass of object = 46 g = 0.046 kg    (1 kg = 1000 g)

v = velocity

Now put all the given values in the above formula, we get:

K.E=\frac{1}{2}mv^2

6.8J=\frac{1}{2}\times 0.046kg\times v^2

6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2

v=17.19m/s\approx 17m/s

Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s

3 0
3 years ago
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