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olga2289 [7]
3 years ago
10

The sides of a rectangle are 6.01 meters and 12 meters. Taking the significant figures into account, what is the area of the rec

tangle?
OA. 70 square meters

OB. 72 square meters

OC. 72.00 square meters

OD. 72.1 square meters

OE. 72.12square meters
Physics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer: The area of the rectangle is 72.12 square meters

Explanation:

The area of the rectangle is  length x width  

Area = 6.01 m x 12 m = 72.12 m²

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Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time
Ber [7]

Answer:

The value is  \Delta  t = 4.0 *10^{-7} \  s

Explanation:

From the question we are told that

  The current is \Delta  I  = 0.100 \  A

  The  inductor  is L  =  2.0mH  =  2.0*10^{-3} \  H

  The voltage induced is  \epsilon   =  500 V

Generally the emf induced is mathematically represented as

      \epsilon =  L  *  \frac{\Delta I }{\Delta  t }

Here  \Delta  t is the time taken  

=>  \Delta  t =  \frac{L  * \Delta  I }{\epsilon }

=>  \Delta  t =  \frac{2*10^{-3}  * 0.100 }{500  }

=>  \Delta  t = 4.0 *10^{-7} \  s

8 0
3 years ago
A particle moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.
mrs_skeptik [129]

The particle's acceleration is 5.1 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²

Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s

The given parameters are;

  • V1 = 2.35 m/s
  • V2 = - 8.72 m/s
  • T1 = 3.42s
  • T2 = 5.59s

Acceleration a = ΔV ÷ ΔT

a = (2.35 + 8.72) / (5.59 - 3.42)

a = 11.07 / 2.17

a = 5.1 m/s²

Therefore, the particle's acceleration is 5.1 m/s²

Learn more about Acceleration here: brainly.com/question/9069726

#SPJ1

4 0
2 years ago
Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
An unknown fluid has a specific gravity of 0.750. What is the volume of 22.5 kg of this fluid?
Andru [333]
<h2>Option C is the correct answer.</h2>

Explanation:

Specific gravity of fluid = 0.750

Density of fluid = Specific gravity of fluid x Density of water

Density of fluid = 0.750 x 1000

Density of fluid = 750 kg/m³

Mass of fluid = 22.5 kg

We have

         Mass = Volume x Density

         22.5 = Volume x 750

         Volume = 0.03 m³ = 30 L

Option C is the correct answer.

5 0
3 years ago
If the only forces acting on a 2.0kg mass are F1 = (3i-8j)N and F2 = (5i+3j)N, what is the magnitude of the acceleration of the
belka [17]

Answer: 4.7m/s²

Explanation:

According to newton's first law,

Force = mass × acceleration

Since we are given more the one force, we will take the resultant of the two vectors.

Mass = 2.0kg

F1+F2 = (3i-8j)+(5i+3j)

Adding component wise, we have;

F1+F2 = 3i+5i-8j+3j

F1+F2 = 8i-5j

Resultant of the sum of the forces will be;

R² = (8i)²+(-5j)²

Since i.i = j.j = 1

R² = 8²+5²

R² = 64+25

R² = 89

R = √89

R = 9.4N

Our resultant force = 9.4N

Substituting in the formula

F = ma

9.4 = 2a

a = 9.4/2

a = 4.7m/s²

Therefore, magnitude of the acceleration of the particle is 4.7m/s²

3 0
3 years ago
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