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xeze [42]
3 years ago
11

A student pushes a 40-N block across the floor for a distance of 10 m. How much work was done to move the block? Question 3 opti

ons: 4 J 40 J 400 J 4,000 J
Physics
2 answers:
Arada [10]3 years ago
8 0

Answer:

The work done to move the block is 400 Joules.

Explanation:

It is given that,

Force acting on the block, F = 40 N

The distance covered by the block, d = 10 m

We need to find the work done by the block. The work done by an object is given by :

W=F\times d

W=40\ N\times 10\ m

W = 400 Joules

So, the work done to move the block is 400 Joules. Hence, the correct option is (B) "400 J".

vovangra [49]3 years ago
7 0
40x10=400 J 

Work = Force x distance 
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3 years ago
How long does it take someone to run 8 miles if they are running at a speed of 7 km/hr?
ZanzabumX [31]

Answer:

1.84 hours or 110.4 minutes

Explanation:

8 miles             1.61 km                 1 hr

------------    x   -----------------  x   -------------- = 1.84 hours or 110.4 minutes

                          1 mile                7  km    

7 0
3 years ago
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AnonymousGiantsFan
3 0
3 years ago
Read 2 more answers
A car traveling at a speed of 50.0 m/s encounters an emergency and comes to a complete stop. How much time will it take for the
ludmilkaskok [199]

Answer:

t=1.25s

Explanation:

The formula is

a= (V1-V0) /t

t = (V1-V0)/a

V1=50m/s

V0= 0 m/s

a= - 4m/s2

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6 0
3 years ago
A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive
Ivahew [28]

Answer:

a

The  x- and y-components of the total force exerted is

           F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

b

 The magnitude of the force is  

            |F_{31 +32}| = 10.25 *10^{-5} N

   The direction of the force is  

         \theta =327.43 ^o   Clockwise from x-axis

Explanation:

From the question we are told that

    The magnitude of the first charge is q_1 = +5.00nC = 5.00*10^{-9}C

      The magnitude of the second charge is q_2 = -2.00nC = -2.00*10^{-9}C

        The position of the second charge  from the first one is  d_{12} = 4.00i \  cm = \frac{4.00i}{100} = 4.00i *10^{-2} m

        The  magnitude of the third charge is q_3 = +6.00nC = 6.00*10^{-9}C

       The position of the third charge from the first one is  \= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} =  (4i + 3j) *10^{-2}m

                |d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m

                |d_{31}| =5 *10^{-2} m

        The position of the third charge from the second  one is

                \= d_{32} = 3j cm = 3j *10^{-2}m

               |d_{32}| =(\sqrt{ 3^2}) *10^{-2} m

               |d_{32}| =3 *10^{-2} m

The force acting on the third charge due to the first and second charge is mathematically represented as

           F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}

 Substituting values

          F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}  \\ \ +  \ \ \ \ \ \ \ \ \   \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}

            F_{31 +32} = 2.16 *10^{-5} (4i + 3j)  - 12*10^{-5} j

            F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

The magnitude of     F_{31 +32}  is mathematically evaluated as

            |F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}

             |F_{31 +32}| = 10.25 *10^{-5} N

The direction is obtained as

            tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}

              \theta = tan ^{-1} [-0.63889]

             \theta = - 32.57 ^o

             \theta = 360 - 32.57

            \theta =327.43 ^o

               

                         

5 0
3 years ago
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