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Viefleur [7K]
2 years ago
10

What is the surface area of an object weighing 100M which exerts a pressure of 20nm-2​

Physics
1 answer:
vovangra [49]2 years ago
6 0

lovely question hope 7 solve it

Explanation:

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When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity \rightarrow 20m/s

\lambda = Wavelength \rightarrow 35*10^{-2}m

Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

The frequency is directly proportional to the angular velocity therefore

\omega = 2\pi f

\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

Where

A = Amplitude

Then replacing,

V_{max} = (1*10^{-2})(359.03)

V_{max} = 3.59m/s

Therefore the maximum speed of a point on the string is 3.59m/s

8 0
3 years ago
Acceleration and Force
olga55 [171]

Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

6 0
2 years ago
Un camion de envios se encuentra detenido en una señal de pare, permitiendo que pase una ambulancia. Inicia su recorrido y al ca
Nesterboy [21]

Answer:

0.741\ \text{m/s}^2

Explanation:

v = Velocidad final = 40\ \text{km/h}=\dfrac{40}{3.6}\ \text{m/s}

u = Velocidad inicial = 0

t = Tiempo empleado = 15 s

a = Aceleración

De las ecuaciones cinemáticas tenemos

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{40}{3.6}-0}{15}\\\Rightarrow a=0.741\ \text{m/s}^2

La aceleración del camión en el primer intervalo de tiempo es 0.741\ \text{m/s}^2.

4 0
3 years ago
The 88-lb force P is applied to the 210-lb crate, which is stationary before the force is applied. Determine the magnitude and d
Marina86 [1]

Answer:

F=-88Ib

Explanation:

From the question we are told that:

Force P=88Ib

Mass of crate M_c=210Ib

Generally the equation for Frictional force F is mathematically given by

Friction\ force (f) = friction\ coefficient\ (u) * Normal\ reaction (N)

F=u*N

with \mu =0.47

F=98.7Ib

Therefore since Static Friction supersedes applied force body remains at rest.

Frictional force =88Ib (negative)

F=-88Ib

5 0
2 years ago
I need this asap pls ​
user100 [1]

Answer:

2 i think

Explanation:

3 0
3 years ago
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