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julia-pushkina [17]
2 years ago
14

. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the the perso

n do on the book? How much work does gravity do on the boo
Physics
2 answers:
viktelen [127]2 years ago
5 0

Answer:

a) 29.4 J

b) - 29.4 J

Explanation:

Given:

Mass of the book, m = 2 kg

Height above the floor, h = 1.5 m

Now,

the work done by the person will be = Force applied on the book × displacement of the book

thus,

Work done by the person = mg × h

where, g is the acceleration due to gravity

thus, on substituting the values, we get

Work done by the person = 2 × 9.8 × 1.5 = 29.4 J

now,

for the force applied by the gravitational pull (downwards) the displacement is in opposite direction (upwards) to the force of the gravity.

Thus,

work done by the gravity will be negative

therefore, the work done by the gravity = - mg × h

or

work done by the gravity = - 29.4 J

photoshop1234 [79]2 years ago
4 0

Answer:

29.4 J  -29.4 J

Explanation:

The work done by the person on the book is given by =mgh

where m =mass

          g= acceleration due to gravity

          h=height

Here m=2 kg given g=9.8 kg m/sec^2 h=1.5 meter

So work done by the person =2×9.8×1.5=29.4 J

As the gravitational force work downward and person lift the book upward so work done by gravity = -mgh = -2×9.8×1.5=  -29.4 J  

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is  \theta =   24.2^o

Explanation:

From the question we are told that

     The length of eire let out is  L = 250 \ m

      The emf generated is \epsilon = 40 V

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The emf generated is mathematically represented as

                             \epsilon = L\ v\ B\ sin \ \theta

making \theta  the subject of the formula

                        \theta =   sin ^{-1}[ \frac{\epsilon}{L  * B  *v} ]

substituting values

                        \theta =   sin ^{-1}[ \frac{40}{250  * (5*10^{-5})  *(7.80 *10^{3})} ]

                        \theta =   24.2^o

6 0
2 years ago
A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude
enot [183]
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity v_x=v_0 cos 30^{\circ}, where v_0=35 m/s
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2 directed downwards, and with initial velocity v_y=v_= sin 30^{\circ}. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height. 
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An elevator has a weight of 14,700 N and has an acceleration of –0.30 m/s2. The free-body diagram shows the forces acting on the
qwelly [4]

14250. I just took it

4 0
2 years ago
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The amount of pressure required to move a 6800 lb force with a 6" d piston is ___ psi.
Katena32 [7]
The pressure needed in PSI = Pounds of force needed divided by the cylinder Area
The Cylinder rod Area is 21.19  sq inches
Thus, the pressure= 6800/21.19
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7 0
2 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

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Distance = 4.10 m

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E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
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