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saveliy_v [14]
2 years ago
7

The body mass of Asaiah is 70 Kg.

Physics
1 answer:
Kazeer [188]2 years ago
8 0

Answer:

A I hope its not wrong I hope u do good

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How is an ammeter connected in a circuit to measure current flowing through it?
trasher [3.6K]

Answer:

It is connected in series with the circuit

Explanation:

This is because to measure the current in the circuit, the current in the circuit has to flow through the ammeter. As such, the ammeter must be connected in series with the circuit so as to measure the current flowing through the circuit.

So, to measure the current flowing through a circuit with an ammeter, the ammeter must be connected in series with the circuit.

4 0
2 years ago
What is the longitud of the middle of Romano swamp
OverLord2011 [107]

Answer:

37 W 47 N

Explanation:

5 0
2 years ago
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -12.0 ∘C, to steam at 109.0 ∘C?
Sunny_sXe [5.5K]
<h2>Answer:</h2>

39.699 kJ

<h2>Explanation:</h2>

In this situation, there are a few transformations as follows;

(i) Heat required to warm the ice from -12°C to its melting point.

(ii) Heat required to melt the ice.

(iii) Heat required to boil the melted ice to boiling point (i.e to steam)

(iv) Heat required to vapourize the water

(v) Heat required to heat the steam from 100°C to 109.0°C

The sum of all the heat processes gives the heat required to warm the ice to steam;

<h3><em>Calculate each of these heat processes</em></h3>

<em>From (i);</em>

Let the heat required to warm the ice from -12.0°C to its melting point (0°C) be Q₁.

Q₁ = m x c x ΔT        -----------------------(i)

Where;

m = mass of ice = 13.0g

c = specific heat capacity of ice = 2.09 J/g°C

ΔT = final temperature - initial temperature = 0°C - (-12°C) = 12°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 2.09 x 12 = 326.04 J

<em>From (ii);</em>

Let the heat required to melt the ice be Q₂. This heat is called the heat of fusion and it is given by;

Q₂ = m x L        -----------------------(ii)

Where;

m = mass of ice = 13.0g

L = latent heat of fusion of ice = 333.6 J/g

Substitute these values into equation (ii) as follows;

Q₂ = 13.0 x 333.6

Q₂ = 4336.8 J

<em>From (iii);</em>

Let the heat required to boil the melted ice from 0°C to boiling point of 100°C be Q₃.

Q₃ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of melted ice (water) which is still 13.0g

c = specific heat capacity of melted ice (water) = 4.2 J/g°C

ΔT = final temperature - initial temperature = 100°C - 0°C = 100°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 4.2 x 100 = 5460 J

<em>From (iv);</em>

Let the heat required to vaporize the water (melted ice) be Q₄. This heat is called the heat of vaporization and it is given by;

Q₄ = m x L        -----------------------(iv)

Where;

m = mass of ice = 13.0g

L = latent heat of vaporization of water = 2257 J/g

Substitute these values into equation (iv) as follows;

Q₄ = 13.0 x 2257

Q₄ = 29341 J

<em>From (v);</em>

Let the heat required to heat the steam from 100°C to 109°C be Q₅.

Q₅ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of steam which is still 13.0g

c = specific heat capacity of steam = 2.01 J/g°C

ΔT = final temperature - initial temperature = 109.0°C - 100°C = 9°C

Substitute these values into equation (i) as follows;

Q₅ = 13.0 x 2.01 x 9 = 235.17J

<em>Finally:</em>

<em>Sum all the heat values together;</em>

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 326.04 + 4336.8 + 5460 + 29341 + 235.17

Q = 39699.01 J

Q = 39.699 kJ

Therefore, the amount of heat (in kJ) required is 39.699

7 0
2 years ago
A person experiencing heat
balu736 [363]

Answer:

it will be heavy sweating

Explanation:

because if humans expierment a heavy heat they might be sweating heavly

7 0
2 years ago
A 2-kg ball is moving with a speed of 4 m/s, and a 4-kg ball is moving with a speed of 2 m/s. What can you conclude about the of
Mila [183]
<h2>Kinetic energy of mass 4 kg ball is less than kinetic energy of mass 2 kg ball</h2>

Explanation:

Kinetic energy = 0.5 x Mass x Velocity²

For ball of mass 2 kg

           Mass, m = 2 kg

           Velocity, v = 4 m/s

           Kinetic energy = 0.5 x 2 x 4² = 16 J

For ball of mass 4 kg

           Mass, m = 4 kg

           Velocity, v = 2 m/s

           Kinetic energy = 0.5 x 4 x 2² = 8 J

Kinetic energy of mass 4 kg ball is less than kinetic energy of mass 2 kg ball

4 0
3 years ago
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