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son4ous [18]
2 years ago
13

Using the formula F = M* A. What is the acceleration of a .5 kg

Physics
1 answer:
Katyanochek1 [597]2 years ago
7 0

Answer:32 m/s/s

Explanation: since F=M*A, F=16N, M=0.5kg, A= F/M

A=16/0.5

A=32 m/s/s

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Explain why ventilation is very important if there is risk of exposure to random gas in your home school
bonufazy [111]
Ventilation is very important because it helps remove the gas form people’s homes and schools and it redirects the random gas outside so it is less likely to hurt people
3 0
3 years ago
When measuring an object with a ruler, the end of the object should be lined up with the _______of the ruler.
KatRina [158]
0 mark is your answer because you want to start at 0
8 0
2 years ago
Read 2 more answers
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
2 years ago
A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?
Studentka2010 [4]

A charge of 12 c passes through an electroplating apparatus in 2.0 min, then the average current is 0.1 ampere.

<h3>What is an electric charge?</h3>

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.

Electric current is defined as the charge per unit of time.

The mathematical relation between current and the electric charge

I =Q/T

where I is the current flowing

Q is the total electric charge

T is the time period for which the current is flowing

As given in the problem A charge of 12 c passes through an electroplating apparatus in 2.0 min

Let us first convert the time period of minutes into seconds

1 min = 60 seconds

2 min = 2*60 seconds

         =120 seconds

By using the above relation between electric current and electric charge

and by substituting the respective values of the charge and the time period

I =Q/T

I = 12c/120 seconds

I = 0.1 Ampere

Thus, the average current flowing through the apparatus would be 0.1 Ampere.

Learn more about an electric charge from here

brainly.com/question/8163163

#SPJ4

5 0
1 year ago
Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat
STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}

Where,

a = Radius

z = Distance to the magnetic field

I = Current

\mu_0 = Permeability constant in free space

Our values are given as

D=2a = 16cm \rightarrow diameter of the sphere then,

a = 0.08m

Thus z = a

B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}

B = \frac{\mu_0I}{2(2^{3/2})a}

B = \frac{\mu_0 I}{2^{5/2}a}

Re-arrange to find I,

I = \frac{2^{5/2}Ba}{\mu_0}

I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}

I = 1.08*10^{-6}A

Therefore the current at the pole of this sphere is 1.08*10^{-6}A

5 0
2 years ago
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