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trasher [3.6K]
3 years ago
11

The graph shows the heating curve of water.

Physics
2 answers:
Lesechka [4]3 years ago
8 0
The answers I got was D and B
Bogdan [553]3 years ago
8 0
1. B
WORKING ON 2....
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The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel
azamat

The final speed of the nickel at the given quantity of heat is determined as 202.1  m/s.

<h3>Final speed of the nickel</h3>

Apply the principle of conservation of energy.

Q = mcΔθ

Q = (18)(0.444)(66 - 20)

Q = 367.63 J

Q = K.E = ¹/₂mv²

2K.E = mv²

v = √(2K.E/m)

where;

  • v is the final speed

v = √(2 x 367.63)/(0.018))

v = 202.1 m/s

Learn more about speed here: brainly.com/question/4931057

#SPJ1

5 0
1 year ago
What happens to electric charges within a conductor when a charged object is brought close to it
Agata [3.3K]
Well, if a charger conductor is touched to another object or close enough to touching the object then the conductor can transfer its charge to that object. Conductors allow for electrons to be transported from particle to particle, so a charged object will always distribute its charge until the repulsive forces are minimized.
8 0
3 years ago
A horizontal compass is placed 21 cm due south from a straight vertical wire carrying a 36 a current downward. in what direction
Anit [1.1K]

 <span>
The needle of a compass will always lies along the magnetic field lines of the earth. 
A magnetic declination at a point on the earth’s surface equal to zero implies that 
the horizontal component of the earth’s magnetic field line at that specific point lies along 
the line of the north-south magnetic poles. </span>

The presence of a current-carrying wire creates an additional <span>
magnetic field that combines with the earth’s magnetic field. Since magnetic 
<span>fields are vector quantities, therefore the magnetic field of the earth and the magnetic field of the vertical wire must be combined vectorially. </span></span>

<span>
Where:</span>

B1 = magnetic field of the earth along the x-axis = 0.45 × 10 ⁻ ⁴ T

B2 = magnetic field due to the straight vertical wire along the y-axis

We can calculate for B2 using Amperes Law:

B2 = μ₀ i / [ 2 π R ]

B2 = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 36 A ) / [ 2 π (0.21 m ) ] <span>
B2 = 5.97 × 10 ⁻ ⁵ T = 0.60 × 10 ⁻ ⁴ T </span>

The angle can be calculated using tan function:<span>
tan θ = y / x = B₂ / B₁ = 0.60 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T <span>
tan θ = 1.326</span></span>

θ = 53°

<span>
<span>The compass needle points along the direction of 53° west of north.</span></span>

8 0
3 years ago
2. Can you place three forces of 5g, 6g, and 12g so they are in equilibrium. Justify your answer.
Bond [772]

Answer:

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

Explanation:

Equilibrium means their sum must be zero.

Here the forces are 5g, 6g, and 12g.

For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.

Here

        Magnitude of largest force = 12 g

        Sum of magnitudes of other forces = 5g + 6g = 11g

       Magnitude of largest force >   Sum of magnitudes of other forces

So this forces cannot form equilibrium.

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

4 0
3 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
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