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2 years ago

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Students repeat the experiment but replace block X and block Y with block W and block Z , as shown in Figure 3. Block W and bloc

k Z have identical mass M . When the experiment is conducted, the students observe that the blocks do not stick together, as shown in Figure 4. The students predict the speed of block Z after the collision by assuming that the collision is perfectly elastic. They then collect data about the actual speeds of both blocks immediately after the collision, as shown in the table.
Why does the predicted speed of block Z after the collision not agree with the actual speed of block Z after the collision?

1 answer:

Lena [83]2 years ago

8 0

The block Z would be seen in figure 10 when 4 strident turn around

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Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large

diamong [38]

Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.

4 0

2 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0

2 years ago

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho

Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

$80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)$

$80=5(50-x)$

$80=250-5x$

$5x=170$

$x=\frac{170}{5}$

$x=34 cm$

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2 years ago

Una columna de mármol, cuya área de sección transversal es de 2.0 m2 sostiene una masa de 25.000 kg. Encontrar: (3 pto )a) El es

bazaltina [42]

Responder:

122,500 Pa; 2.45 × 10 ^ -6; 2.94 × 10 ^ -5m

Explicación:

Dado lo siguiente:

Área de sección transversal (A) = 2m ^ 2

Masa (m) = 25000 kg

Módulo de Young = 50 x 10 ^ 9 N / m2

(1) estrés en la columna:

Estrés = Fuerza / Área

F = masa * aceleración debido a la gravedad

F = 25000kg * 9.8m / s ^ 2 = 245,000J

Estrés = 245,000J / 2m ^ 2

Estrés = 122,500 Pa

2) Deformación de la unidad (deformación):

Usando la relación:

Módulo de Young = Estrés / tensión

50 × 10 ^ 9 = 122,500 / CEPA

Cepa = 122500 / (50 × 10^9)

Cepa = 0.00000245

C) Si la altura es de 12 m, ¿cuánto se acorta la columna?

Cepa = extensión / longitud

0.00000245 = extensión / 12

0,00000245 * 12

0,0000294 m

7 0

2 years ago