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2 years ago

What quantity of heat is required to change 20g of ice at -25°c to steam at 120°c

Physics

1 answer:

Oksana_A [137]2 years ago

6 0

Answer: 5289 joules

Explanation:

The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since, Mass of ice = 20 g

C is not provided (Recall that the specific heat capacity of ice is 2.010J/g°C)

Φ = (Final temperature - Initial temperature)

(120°C - (-25°C) =

120°C + 25°C = 145°C

Thus, Q = 20g x 2.010J/g°C x 145°C

Q= 5829J

Thus, 5829Joules of heat energy is required to heat ice

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Work is the product of force and an object's

Neporo4naja [7]

Answer:

C. displacement

Explanation:

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The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t

KatRina [158]

Answer:

T₂ = 123.9 N, θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

T₁ₓ - $T_{2x}$ = 0

T_{2x}= T₁ₓ

Y axis y

T_{1y} + T_{2y} - 200N = 0

T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

sin 60 = T_{1y} / T₁

cos 60 = t₁ₓ / T₁

T_{1y} = T₁ sin 60

T1x = T₁ cos 60

T_{1y}y = 100 sin 60 = 86.6 N

T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

T_{2y} = T₂ sin θ

T₂ₓ = T₂ cos θ

we substitute

T₂ sin θ= 200 - 86.6 = 113.4

T₂ cos θ = 50 (1)

to solve the system we divide the two equations

tan θ = 113.4 / 50

θ = tan⁻¹ 2,268

θ = 66.2º

we caption in equation 1

T₂ cos 66.2 = 50

T₂ = 50 / cos 66.2

T₂ = 123.9 N

8 0

2 years ago

This problem has been solved!

lana66690 [7]

Answer:

Charge on each metal sphere will be $8\times 10^{8}C$

Explanation:

We have given number of electron added to metal sphere A $n=10^{12}electron$

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres $=\frac{10^{12}}{2}=5\times 10^{11}electron$

We know that charge on each electron $e=1.6\times 10^{-19}C$

So charge on both the spheres will be equal to $q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C$

So charge on each metal sphere will be equal to $8\times 10^{8}C$

6 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

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2 years ago

How many electrons can carbon accept from other atoms while bonding? (1 point)

solong [7]

Answer: Four electrons

Explanation:

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