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AfilCa [17]
2 years ago
14

A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?

Physics
1 answer:
12345 [234]2 years ago
7 0

The block is made of A) Tin, as its specific heat capacity is 0.225 J/(g^{\circ}C)

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by \Delta T, according to the following equation :

Q=mC_s \Delta T

where  C_s is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

Q = 450 J is the amount of energy supplied to the block

\Delta T = 1^{\circ}C is the change in temperature of the material

Solving the equation for C_s, we can find the specific heat capacity of the unknown sample:

C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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Given the indices of refraction n1 and n2 of material 1 and material 2, respectively, rank these scenarios on the basis of the p
lisov135 [29]

Answer:

c>d>f=a>b>e

Explanation:

When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.

According to the Snell's law:

\rm refractive\ index\ (n)=\frac{speed\ of\ light\ in\ the\ incident\ medium}{speed\ of\ light\ in\ the\ refracted\ medium}

a)

n_1-n_2=1.33-1.00\\=0.33

b)

n_2-n_1=1.46-1.33

=0.23

c)

n_2-n_1=2.42-1.33\\=1.09

d)

n_2-n_1=1.46-1.00\\=0.46

e)

n_1-n_2=1.50-1.33\\=0.17

f)

n_2-n_1=1.33-1.00\\=0.33

c>d>f=a>b>e

5 0
2 years ago
1. A diffraction grating with 5.000 x 103 lines/cm is used to examine the sodium
Nuetrik [128]

Answer:

0.0002°, 0.1691°, 0.338°

Explanation:

Difference between the two line = 5.97 * 10-⁸m

d = 1 / N

N = 5.0 * 10³

d = 2.0 * 10⁴m

nL = Nsin¤

For first order

588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤

Sin¤ = 2.944*10-³

¤ = sin-¹ 0.002944

¤ = 0.1687°

First order ¤ =

Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)

Sin-¹ (0.002947) = 0.1689°

Angular separation = 0.1689 - 0.1687 = 0.0002°

Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)

Second order ¤ = 0.3378°

Angular difference = 0.3378° - 0.1687° = 0.1691°

Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°

Angular difference = 0.5067° - 0.1687° = 0.338°

7 1
2 years ago
Read 2 more answers
Ram has power of 550 watt. What does it mean?
WARRIOR [948]
It means you can do 550 Newton Meters of work every second. Power is the rate of doing work, I hope this helps
4 0
2 years ago
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
kati45 [8]

Answer:

The  coefficient of kinetic friction  \mu_k =  0.724

Explanation:

From the question we are told that

   The  length of the lane is  l =  36.0 \  m

    The speed of the truck is  v  =  22.6\  m/s

     

Generally from the work-energy theorem we have that  

    \Delta KE  =   N  *  \mu_k *  l

Here N  is the normal force acting on the truck which is mathematically represented as

     \Delta KE is the change in kinetic energy which is mathematically represented as

        \Delta KE =  \frac{1}{2} *  m *  v^2

=>     \Delta KE =  0.5  *  m *  22.6^2

=>      \Delta KE =  255.38m

        255.38m =    m *  9.8  *  \mu_k *   36.0

=>     255.38  =    352.8  *  \mu_k

=>   \mu_k =  0.724

 

6 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
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