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2 years ago

What is the amount of thermal energy that can be stored in an object depends on?

1 answer:

3 0

Thermal energy is the total energy of all the molecules in an object. The thermal energy of an object depends on three things: 4 the number of molecules in the object 4 the temperature of the object (average molecular motion) 4 the arrangement of the object's molecules (states of matter).

Hope this helps

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The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string

Katyanochek1 [597]

Answer:

$\dfrac{\mu_A}{\mu_G}=0.197$

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

$v_A = \sqrt{\dfrac{F_A}{\mu_A}}$ .......(1)

speed for string G

$v_G = \sqrt{\dfrac{F_G}{\mu_G}}$ ........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}$

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{\mu_A}{\mu_G}=0.197$

5 0

2 years ago

g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for

lbvjy [14]

Answer:

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

$a\text{sin}\theta=\frac{2N+1}{2}\lambda$

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

$0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}$

Solving the above equation we get

we N= 2.313≅ 2

4 0

3 years ago

A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that h

RoseWind [281]

Answer:

a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

F -W = m a

Force is electrical force

F = k q₁ q₂ / r²

k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

q₁ = q₂ = q

a = (k q² / r² - m g) / m

a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

a = -7.29 m / s²

The negative sign indicates that the direction of this acceleration is downward

6 0

2 years ago

In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w

Daniel [21]

Answer:

$v_2=-133.17m/s$ , the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before ( $p_i$ ) and after ( $p_f$ ) the explosion. We will take the east direction as positive.

Before the explosion we have $p_i=m_iv_i=Mv_i$ .

After the explosion we have pieces 1 and 2, so $p_f=m_1v_1+m_2v_2$ .

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

$Mv_i=m_1v_1+m_2v_2$

Which means (since we want $v_2$ and $M=m_1+m_2$ ):

$v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}$

So for our values we have:

$v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s$

5 0

2 years ago

a 10 kilogram steel ball is dropped from the top of a tower 100 meters high the kinetic energy of the ball just before it sttike

EleoNora [17]

Explanation:

If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.

Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.

4 0

2 years ago