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skad [1K]
3 years ago
10

Question 6 (10 points)

Physics
1 answer:
horsena [70]3 years ago
7 0

Answer:

a= g = - 9.81 m/s2.

The following equations will be helpful:

a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad

When you substitute the specific acceleration due to gravity (g), the equations are as follows:

g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd

If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:

g = vf /t d = 1/2 gt2 vf2 = 2gd

The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.

hope this info helps you out!

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Answer:

The situation given in option A and B are examples for moving an object toward left side while the option C and option D are examples for moving an object toward right side. Option B will also be an example for not moving the object.

Explanation:

As per the option A statement, the force acting towards left is greater than the force acting toward right side. So the net force will be towards the direction having maximum magnitude. Thus, the box will move toward left side in option A. The same situation arises for the object in option B. But here the difference in the forces is only 1 N, so the change in the position of the object will be very less. Thus it may look like there is no acceleration in the box of option B.

Similarly, the force acting on the objects given in option C and D have magnitude greater towards the right side than towards the left side.  So these two will be accelerated toward the right side.

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Based on the information in the table, which mineral is LEAST likely to be found in Earths rocks
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C. Magnesium

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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that
Harrizon [31]

Answer:

The time taken is   t = 40007 sec  

Explanation:

From the question we are told that

   The diameter of the egg is d_e = 5.5cm  = \frac{5.5}{100} = 5.5*10^{-2}m

    The initial temperature of egg the T_e = 4.3^{o}C

     The temperature of the boiling water T_b = 100^oC

    The heat transfer coefficient is  H  = 800 W/m^2 \cdot K

    The  final temperature is T_e_f = 74^oC

     The  thermal  conductivity of water is k = 0.607 W/m^oC

     The diffusivity of the egg \alpha = 0.146 * 10^{-6} m^2 /s

Using one term approximation

We have the

            \frac{T_e_f - T_b}{T_e - T_b}  = Ae^{-\lambda ^2 \tau}

The radius is  r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m     Note that this radius is approximation to that of  a real egg

    Now we need to obtain the Biot number which help indicate the value of A  \ and \ \lambda to use in the above equation

     The Biot number is mathematically represented as

               Bi = \frac{H r}{k}

Substituting values  

               Bi = \frac{800 * 2.75 *10^{-2}}{0.607}

                    = 36.24

So for this value  which greater than 0.1 the  coefficient \lambda_1 \ and  \ A_1 is  

        \lambda = 3.06632

        A = 1.9942

Substituting this into equation 1 we have

          \frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-9.383 \tau}

           0.13624 =  e^{-9.383 \tau}

Taking natural log of both sides

           -1.993 =  -9.383\  \tau

          \tau =  0.2124

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          t = \frac{\tau r^2}{\alpha }

substituting value  is  

         = \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}

         t = 40007 sec  

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Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

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