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3 years ago

Turning up the Heat

Physics

1 answer:

ella [17]3 years ago

3 0

Answer:

harem jath

Explanation: t=i dont know anything about how to speak english thats the answer for your question

You might be interested in

The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a)

Lunna [17]

Answer:

The number of radians turned by the wheel in 2s is $\theta= 8\ radians$

The angular acceleration is $\alpha =14 rad/s^2$

Explanation:

The angular velocity is given as

$w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3$

Now generally the integral of angular velocity gives angular displacement

So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

This is mathematically evaluated as

$\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt$

$= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.$

$= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0$

$= 4 +4$

$\theta= 8\ radians$

Now generally the derivative of angular velocity gives angular acceleration

So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

This is mathematically evaluated as

$\frac{dw}{dt} = \alpha (t) = 2 + 3t^2$

so at t=2

$\alpha (2) = 2 +3(2)^2$

$\alpha =14 rad/s^2$

7 0

3 years ago

Un resorte se alarga 5 cm bajo la acción de una fuerza de 39,2 N. ¿Cuál es la constante del resorte? Si ahora la fuerza es 68,6

Lorico [155]

Answer:

$k=784 N/m$

$\Delta x=8,8 cm$

Explanation:

Usando la ley de Hook tenemos:

$F=k\Delta x$

Solving it for k we have:

$k=\frac{F}{\Delta x}$

$k=\frac{39,2}{0,05}$

$k=784 N/m$

Usando la misma ecuación y sabiendo k tenemos:

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{68,6}{784}$

$\Delta x=8,8 cm$

Espero esto te ayude!

6 0

3 years ago

At a wedding reception, you notice a child who looks like their mass is about 25 kg running across the dance floor then sliding

Vitek1552 [10]

Mass of the object m = 25 kg

Coefficient of friction Uk = 0.15

Frictional force Ff = Uk x F => Ff = Uk x m x g

Ff = 0.15 x 25 x 9.8

Frictional Force Ff = 36.75 N

4 0

3 years ago

Read 2 more answers

A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, land

erastovalidia [21]

Answer:

1) 0.51 seconds.

2) 1.45 m/s.

Explanation:

given, height from which cat falls = 1.3 m

we know that, s = ut + $\frac{1}{2}$ at².

here if we consider cat moment only in downward direction,

intial velocity of cat in downward direction , u = 0.

so, time, t = $\sqrt{\frac{2h}{g} }$ .

⇒ t = $\sqrt{\frac{2(1.3)}{9.81} }$ = 0.51 seconds.

t = 0.51 seconds.

now, consider cat moment only in forward direction

s = ut , since acceleration is zero in forward direction

⇒ u = $\frac{s}{t}$ .

so, u = $\frac{0.75}{0.51}$ = 1.45 m/s .

6 0

3 years ago

The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

5 0

3 years ago