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Leni [432]
3 years ago
11

What is the net force needed to lift a full grocery sack weighing 210N uniformly?

Physics
1 answer:
Alex3 years ago
8 0
If the sack weighs 210 newtons, then an upward force of 210 newtons
exactly cancels the downward force of gravity, and makes the net vertical
force on the bag zero. 

ANY upward force that's greater than 210 newtons makes the net force
act upward on the bag, and causes it to accelerate upward.
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A car traveling 90 km/hr is 100 m behind a truck traveling 50 km/hr. How long will it take the car to reach the truck?
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v = speed of car = 90 km/h

u = speed of truck = 50 km/h

d = initial separation distance = 100 m = 0.1 km

They meet at time t such that

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Your bedroom has a rectangular shape and you want to measure its size. You use a tape that is precise to 0.001 m and find that t
Alex73 [517]

Your bedroom has a rectangular shape, and you want to measure its area. You use a tape that is precise to 0.001 and find that the shortest wall in the room is 3.547 long. The tape, however, is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 . You measure the second wall to be 4.79 long. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?

b)If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?

a)30 m^2

b) 31.4 m^2

c)31.37 m^2

d)31.371 m^2

Answer:

Part A

A  =  16.99 \ m^2

Part  B

 A_r =31.37 \ m^2

Explanation:

From the question we are told that

   The  shortest wall is  x =  3.547 \ m

      The  length of the second wall is y  =  4.79 \ m

Generally the area is mathematically evaluated as

           A =  x *  y

=>    A  =  3.547 *  4.79

=>   A  =  16.99 \ m^2

Given that the diameter is  d =  6.32

Then the area is  

         A_r =  \pi \frac{d^2}{4}

         A_r =3.142 *   \frac{6.32^2}{4}

        A_r =31.37 \ m^2

3 0
3 years ago
Consider a point charge qqq in three-dimensional space. Symmetry requires the electric field to point directly away from the cha
bonufazy [111]

Answer:

 E = \frac{1}{4\pi  \epsilon_o } \  r^2

Explanation:

For this exercise let's use Gauss's law. The Gaussian surface that follows the symmetry of the charges is a sphere

           Ф = ∫ E. dA = \frac{x_{int} }{\epsilon_o}

the bold are vectors, the radii of the sphere and the electric field are parallel therefore the scalar product reduces to the algebraic product

           Ф = ∫ E dA = \frac{x_{int} }{\epsilon_o}

           E ∫ dA = \frac{x_{int} }{\epsilon_o}

           E A = \frac{x_{int} }{\epsilon_o}

the area of ​​a sphere is

            A = 4π r²

the charge inside the sphere is q = + q

           

we substitute

           E 4π r² = \frac{x }{\epsilon_o}

           E = \frac{1}{4\pi  \epsilon_o } \  r^2

4 0
3 years ago
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