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Basile [38]
1 year ago
12

What would happen to the number of home runs a player could hit if the air was removed from above the field? Use your understand

ing of Newton's laws to explain.
Physics
1 answer:
Rama09 [41]1 year ago
7 0

Answer:

number of home runs a player would hit would drastically increase

Explanation:

In a hypothetical scenario, if this were to happen the number of home runs a player would hit would drastically increase. When a player hits the ball force is applied to the ball which sends it flying, then the ball meets an opposing force of air resistance and a downward force of gravity which ultimately causes the ball to fall to the ground. This is explained in Newton's first law which states that "objects in motion remain in uniform motion and in a straight line unless they are met by an opposing unbalanced force." Without the force of air resistance, the ball will only be met with a downward force of gravity which would allow it to travel at a much greater distance before hitting the ground, thus resulting in a drastic increase of home runs.

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A handbag weighing 162N is carried by two students each holding the handle of the bag next to him. If each handle is inclined at
Mashcka [7]

Answer:

162N

Explanation: resolving into components and equating both vertical components equal to 162 as explained in attachment

Download pdf
3 0
2 years ago
Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in
notsponge [240]

Answer:

speed of boat as

v_b = 10 mph

river speed is given as

v_r = 2 mph

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

since the boat and river is in same direction so we will have

v_1 = v_r + v_b

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

v_2 = v_b - v_r

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

v_1 = \frac{24}{2} = 12 mph

also when it moves in upstream then it covers same distance in 3 hours of time

v_2 = \frac{24}{3} = 8 mph

v_b + v_r = 12 mph

v_b - v_r = 8 mph

so we have speed of boat as

v_b = 10 mph

river speed is given as

v_r = 2 mph

8 0
2 years ago
Read 2 more answers
A force f = bx 3 acts in the x direction, where the value of b is 3.7 n/m3. how much work is done by this force in moving an obj
Elodia [21]
The answer would be 21.6 but rounded up it would be 22J.
8 0
2 years ago
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

6 0
2 years ago
A ball is dropped from the rest from a height of 6. 0 meters above the ground. The ball falls freely and reaches the ground at t
igor_vitrenko [27]
What is the velocity of the ball<span> when it </span>reaches<span> its highest point? ... Then double it for the "hang time"-the time one's feet are off the </span>ground<span>. ... Calculate the </span>speed<span> of a bowling </span>ball<span> that travels 5.0 </span>meters<span> in 2.4</span>seconds<span>. ... An object </span>falls freely<span> from </span>rest<span> on a planet where the acceleration due to gravity is twice as much as it is ...</span>6';"'A ball is dropped<span> from </span>rest from a height<span> 6.0 </span>meters above<span> t e </span>ground<span>. The hall </span><span>falls freely and reaches the ground 1.1 seconds later.

</span>
8 0
2 years ago
Read 2 more answers
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