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3 years ago

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What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copp

er is 0.39 J/gºC. A. 115 J B. 150 J C. 250 J D. 300 J

1 answer:

Lina20 [59]3 years ago

3 0

The amount of heat required is B) 150 J

Explanation:

The amount of heat energy required to increase the temperature of a substance is given by the equation:

$Q=mC\Delta T$

where:

m is the mass of the substance

C is the specific heat capacity of the substance

$\Delta T$ is the change in temperature of the substance

For the sample of copper in this problem, we have:

m = 25 g (mass)

C = 0.39 J/gºC (specific heat capacity of copper)

$\Delta T = 15^{\circ}C$ (change in temperature)

Substituting, we find:

$Q=(25)(0.39)(15)=146 J$

So, the closest answer is B) 150 J.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a

gizmo_the_mogwai [7]

Answer:

Acceleration, $767.08\ m/s^2$

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

$v^2=u^2+2ah$

u = initial velocity=0 m/s

a = acceleration=g

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s$

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2$

So, the average acceleration of the ball during the time it is in contact is $767.08\ m/s^2$ .

4 0

3 years ago

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

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3 years ago

Một mẫu vật liệu có khối lượng riêng là 2500kg/m3

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Explanation:

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2 years ago

Plz help difference between closed system and open system?

frozen [14]

A closed system is a system that is completely isolated from its environment. The physical universe, as we currently understand it, appears to be a closed system. An open system is a system that has flows of information, energy, and/or matter between the system and its environment, and which adapts to the exchange.

D makes the most sense, but you just have to put two and two together and go with your gut feeling, first cross out the answers that don't make sense (A didnt make sense) and go from there! I hope the little bit above me helped you answer or decide :) Good luck!

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3 years ago

Determine the components of reaction at the fixed support

denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are

$A_{(x)} = 400 \ N$ , $A_{(y)} = -500 \ N$ , $A_{(z)} = 600 \ N$ , $M_x = 1225 \ N\cdot m$ , $M_y = 750 \ N\cdot m$ , $M_z = 0 \ N\cdot m$

Explanation:

Looking at the diagram uploaded we see that there are two forces acting along the x-axis on the fixed support

These force are 400 N and $A_{(x)}$ [ i.e the reactive force of 400 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(x)} - 400 = 0$

=> $A_{(x)} = 400 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the y-axis on the fixed support

These force are 500 N and $A_{(y)}$ [ i.e the force acting along the same direction with 500 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(y)} + 500 = 0$

=> $A_{(y)} = -500 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the z-axis on the fixed support

These force are 600 N and $A_{(z)}$ [ i.e the reactive force of 600 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(z)} - 600 = 0$

=> $A_{(z)} = 600 \ N$

Generally taking moment about A along the x-axis we have that

$\sum M_x = M_x - 500 (0.75 + 0.5) + 600 ( 1 ) = 0$

=> $M_x = 1225 \ N\cdot m$

Generally taking moment about A along the y-axis we have that

$\sum M_y = M_y - 400 (0.75 ) + 600 ( 0.75 ) = 0$

=> $M_y = 750 \ N\cdot m$

Generally taking moment about A along the z-axis we have that

$\sum M_z = M_z = 0$

=> $M_z = 0 \ N\cdot m$

8 0

3 years ago