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3 years ago

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Physics

1 answer:

Stolb23 [73]3 years ago

6 0

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

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_____ have a nearly circular orbit.

WITCHER [35]

The question is poor.

It expects you to choose 'B', but things aren't nearly that simple.

We picture all of the asteroids bunched up in a neat bunch between
the orbits of Mars and Jupiter, with each asteroid following its own
nearly circular orbit. But many asteroids have wildly non-circular
'eccentric' orbits, sometimes being closer to the sun than the Earth is.
You know how you hear so much discussion about when did the Earth
get hit by an asteroid ? and when will the Earth be hit by another asteroid ?
and what will happen when the Earth is hit by an asteroid again ? None
of that would be possible if asteroids all had nearly circular orbits.

We picture comets as having these loooong skinny orbits, spending
most of every orbit waaay out in the solar system, and then dipping
close to the sun for a few days, and then going back waaaay out again.
But there are also many comets in nearly circular orbits around the sun.
You never hear anything about them, because you can never see them
without a powerful telescope, and they never do anything exciting.
So some comets could be a correct answer to this question too.

And since meteoroids are the remains of old comets, and follow the
orbit of the comet that they chipped off from, there are a lot of meteoroids
in circular orbits too, and they could also be a correct answer to this question.

8 0

4 years ago

NaOH is__<br> an acid <br> a base <br> a salt

masya89 [10]

Answer:

Sodium Hydroxide

Explanation:

4 0

4 years ago

Read 2 more answers

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago

A stream of plasma? Solar wind, coronal mass ejection

OlgaM077 [116]

Answer:

coronal mass ejection

Explanation:

Solar wind is something that has to do with magnetism,

CME is " is a significant release of plasma and accompanying magnetic field from the solar corona," that often follows solar flares.

https://en.wikipedia. org/wiki/Coronal_mass_ejection

8 0

3 years ago

(a) Calculate the self-inductance of a solenoid that is

adell [148]

Answer:

a) 0.142mH

b) 14mV

Explanation:

the complete answer is:

(a) Calculate the self-inductance of a solenoid that is <ghtly wound with wire of diameter 0.10 cm, has a cross-sec<onal area of 0.90 cm2 , and is 40 cm long. (b) If the current through the solenoid decreases uniformly from 10 to 0 A in 0.10 s, what is the emf induced between the ends of the solenoid

a) the self inductance of a solenoid is given by:

$L=\frac{\mu_o N^2 A}{L}$

μo: magnetic permeability of vacuum = 4\pi*10^{-7}N/A^2

A: cross sectional area = 0.9cm^2=9*10^{-5}m

L: length of the solenoid = 40cm = 0.4m

The N turns of the wire is calculated by using the diameter of the wire:

N = (40cm)/(0.10cm)=400

By replacing in the formula you obtain:

$L=\frac{(4\pi *10^{-7}N/A^2)(400)^2(9*10^{-5}m^2))}{(0.4m)}=1.42*10^{-4}H$

the self inductance is 1.42*10^{-4}H = 0.142mH

b) to find the emf you can use:

$emf=L\frac{\Delta I}{\Delta t}=(1.42*10^{-4}H)\frac{10A-0A}{0.10s}=0.014V=14mV$

the emf induced is 14mV

4 0

3 years ago