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yawa3891 [41]
3 years ago
15

Sarah wanted to know the velocity of an airplane traveling from Houston to Dallas. After recording the distance (in miles) and t

ime (in hours) it took to travel to Dallas, what other measurement does Sarah need to determine the airplanes velocity?
A) acceleration
B) direction
C) force
D) inertia
Physics
1 answer:
Mazyrski [523]3 years ago
5 0

Answer:

Option B is the correct answer.

Explanation:

 Velocity of a body = Displacement of the body/Time taken

Displacement and velocity are a vector. Both have direction and magnitude.

Displacement is generally distance with direction.

If we divide distance with time taken we will get speed of the airplane. Speed with direction is called velocity. So we need distance between Houston and Dallas, time taken by plane to travel from Houston to Dallas and direction of displacement from Houston to Dallas.

So we need direction.

Option B is the correct answer.

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3 years ago
You can enter compound units that are combinations of other units that are multiplied together.Torque can be calculated by multi
Paraphin [41]

Answer:

3.0 x 10¹ Nm

Explanation:

Torque = F x r

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is also called lever arm

Here F = 15 N and r = 2.0 m

Torque

= 15 N X 2.0 m

= 3.0 10¹ Nm.

4 0
3 years ago
Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

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p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

\frac{m_A g}{A_A}=\frac{m_B g}{A_B}

Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0
3 years ago
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

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   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
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Karolina [17]

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8 0
2 years ago
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