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Masja [62]
3 years ago
5

Suppose you increase your walking speed from 5m/s to 15m/s in a period of 1 s. What is your acceleration?

Physics
1 answer:
olchik [2.2K]3 years ago
7 0
Acceleration=change in velocity divided by change in time or a=\frac{\Delta v}{\Delta t}

we are given that it goes from 5m/s to 15m/s
\Delta v=v_f-v_0
in this case, \Delta v=15\frac{m}{s}-5\frac{m}{s}=10\frac{m}{s}

and the change in time or \Delta t is 1 second because this change occurs over a period of 1 second

so the acceleration is a=\frac{10\frac{m}{s}}{1s}=10\frac{m}{s^2}
the acceleration is 10\frac{m}{s^2}
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Answer:

emotions

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2 years ago
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Cuando una persona sube y baja una escalera, Cuanto vale su desplazamiento y cual es la medida de su trayectoria.
adoni [48]

Answer:

Primero, definimos el desplazamiento como la distancia entre la posición final y la posición inicial.

Así, si comenzamos abajo, luego subimos la escalera, y luego bajamos, la posición final y la posición inicial serán la misma

por lo que el desplazamiento es igual a cero.

La medida recorrida es el espacio total recorrido.

Es decir, si entre el principio y el final de la escalera hay una distancia D.

La persona que sube y baja, recorre esta distancia dos veces.

Entonces cuando una persona sube y baja la escalera, la medida de su trayectoria será 2*D.

8 0
3 years ago
A cylindrical vessel with water is rotated about its vertical axis with a constant angular velocity co. Find:
Nataly [62]

Answer:

Explanation:

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4 0
3 years ago
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The work energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. If
sattari [20]

Answer:

v=\sqrt{2gh}\ m/s

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From work energy theorem

Work done by all forces = Change in kinetic energy

Lets take

m= mass of object

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The final velocity of object is v

Work done by gravitational force = m g . h

The final kinetic energy = 1/2 m v²

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6 0
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The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th
bagirrra123 [75]

The velocity of the wave on the string is given by

v=\sqrt{\frac{T}{\frac{m}{L}}}  \\  v=\sqrt{\frac{TL}{m}}

Solving the above equation,

v^2=\frac{TL}{m} \\  L=\frac{v^2m}{T}

The frequency of the wave f=300 and wave length is 0.76

The velocity is v=(300)(0.76)=228

Substituting numerical values,

L=\frac{228^2(0.0059)}{200}\\ T=1.534

The length of the string is 1.534 m

4 0
3 years ago
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