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GuDViN [60]
3 years ago
7

En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8

y 1.4 segundos de haberse iniciado el movimiento
Physics
1 answer:
Ivanshal [37]3 years ago
6 0

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

A=10cm\\T=2s

the equation of the motion is

x=Acos(\omega t)\\

we can calculate w by using

\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi

Hence, we have that

x=10cm*cos(\pi t)\\

the speed will be

v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}

hope this helps!

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A Ray of light falling on rough surface follows the laws of reflection but no image of the object placed before it is C explain
stepladder [879]

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The light rays falling on a rough surface does follow the laws of reflection. The light rays are incident parallel on the rough surface but due to uneven surface the light rays are not reflected parallel rather they are reflected in different direction. Hence, no image is formed.

4 0
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How much time does it take a rhino running at 40 m/s to run 250 m?​
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Answer:

Rhinoceroses are odd-toed ungulates native to sub-Saharan Africa and southern Asia, though all five living species have hugely contracted in range and number due to the influence of humans. Despite their titanic, tank-like bulk, rhinos can be amazingly swift: The fastest may reach at least 50 kilometers per hour (31 mph).

Explanation:

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8 0
3 years ago
Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10
nydimaria [60]

Answer:

Explanation:

The formula for this is

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2} and moving things around to solve for r:

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8 0
3 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
4 years ago
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