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3 years ago

Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?

Physics

2 answers:

o-na [289]3 years ago

7 0

The correct answer is D. 19.3 J

Rainbow [258]3 years ago

4 0

Answer: D 19.3 J

Explanation:

The heat required to increase the temperature of a substance is given by

$Q=mC_s \Delta T$

where

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

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

In this problem, we have:

$m=5.0 g$

$C_s = 0.386 J/gC$

$\Delta T=10.0 C-0.0 C=10.0 C$

Substituting into the equation above, we find

$Q=(5.0 g)(0.386 J/gC)(10.0 C)=19.3 J$

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Sever21 [200]

The horizontal component of the velocity of the ball is calculated by multiplying the speed by the cosine of the given angle.
x-component of speed = (31 m/s)(cos 35°)
= 25.39 m/s
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3 years ago

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(a) Circular metal wires in electrical circuits may have different cross-sectional areas (different diameters) and different len

WARRIOR [948]

For the different wires the values given are

For Wire A, Area= $A_{A}$ , Resistivity= $ρ_{A}$ , Length= $L_{A}$ , Heat= $H_{A}$ , Voltage= V, Time= t, Wire B, Area= $A_{B}$ , Resistivity= $ρ_{B}$ , Length= $L_{B}$ , Heat= $H_{B}$

So, $H_{A}$ / $H_{A}$ = $\frac{V^{2} *t}{\frac{ρ(1)*L_{A} }{A_{A} } }$ / $\frac{V^{2} *t}{\frac{ρ(2)*L_{B} }{A_{B} } }$

Therefore $H_{A}$ / $H_{A}$ = $\frac{A_{A}*ρ_{B} *L_{B}}{A_{B}ρ_{A} L_{A} }$

The type of energy that moves between two materials with varying temperatures is referred to by scientists as heat. Because the average translational kinetic energy per molecule in the two materials varies, an energy transfer takes place. Up until thermal equilibrium is attained, heat is transferred from the substance with the higher temperature to the material with the lower temperature. The joule, with 1 joule equalling 1 newton meter, is the SI unit of heat. Imagine the following situation to better comprehend what happens when this energy transfer takes place: Tiny rubber balls are bouncing all over two distinct containers that are full with them. The difference between the average ball speed in one container and the second container is substantial.

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2 years ago

A cannon, elevated at 40∘ is fired at a wall 300 m away on level ground, as shown in the figure below. The initial speed of the

Gre4nikov [31]

Vo = 89 m/s
angle: 40°

=> Vox = Vo * cos 40° = 89 * cos 40°

=> Voy = Vo. sin 40° = 89 * sin 40°

x-movement: uniform => x =Vox * t = 89*cos(40)*t

x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s

y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2

y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.

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3 years ago

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What is the mass of a block concrete that gains 52,800 joules of energy when its temperature is increased by 5 degrees Celsius s

valina [46]

Answer:

12 kg

Explanation:

q = 52,800 J

c = 880 J/(kg * K) Pretty sure you missed the kg here

ΔT = 5 °C = 5 K (reminder that this only applies to ΔT and not T)

q = mcΔT ⇒ m = q/(cΔT) = 52,800 J / (880 J/(kg * K) * 5 K) = 12 kg

6 0

3 years ago

A shopper in a supermarket pushes a loaded 31 kg cart with a horizontal force of 12 N. The acceleration of gravity is 9.81 m/s 2

rosijanka [135]

Answer:

3.41334 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

m = Mass

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{12}{31}\\\Rightarrow a=0.387\ m/s^2$

Acceleration of the cart is 0.387 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 4.2+\frac{1}{2}\times 0.387\times 4.2^2\\\Rightarrow s=3.41334\ m$

The cart will move 3.41334 m in 4.2 seconds

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3 years ago