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3 years ago

Question

Physics

1 answer:

levacccp [35]3 years ago

5 0

I uploaded the answer t $^{}$ o a file hosting. Here's link:

bit. $^{}$ ly/3tZxaCQ

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Se coloca agua en un recipiente de aluminio y se pone a calentar en una estufa que le suministra 230 kj, lo cual hace que la tem

tensa zangetsu [6.8K]

Answer:

32 °C.

Explanation:

Hola.

En este caso, debemos entender que la relación entre el calor y la temperatura viene dada por:

$Q=mCp\Delta T$

De este modo, dado que estamos estudiando la misma sustancia (agua) con masa constante, la relación calor-temperatura es lineal y directamente proporcional, por tal razón, si se duplica el calor suministrado, la temperatura también será duplicada, de modo que:

$\Delta T_{nuevo}=2*16\°C\\\\\Delta T_{nuevo}=32\°C$

¡Saludos!

3 0

3 years ago

Whats a good string length for a parachute

Yuri [45]

Answer: Hope This Helps!

Explanation:

The length of the string should be equal to the radius of the desired circle. Attaching the suspension lines: Creator of parachutes Use 4 suspension lines for each parachute. And Attatch the suspension lines onto the canopy.

3 0

2 years ago

From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South A

Nikolay [14]

Answer:

0.146 m/s

Explanation:

We can see it in the pic.

3 0

3 years ago

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0

3 years ago

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

3 years ago