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Liono4ka [1.6K]
3 years ago
8

Once a baseball has been hit into the air, what forces are acting upon it? How can you tell that any forces are acting upon the

ball?
Physics
1 answer:
Tom [10]3 years ago
8 0
Well, its in the air, so the air is "upon" the ball. and when it comes down...you catch it, and throw it, and get someone out, and win the game, and just keep doing that, and boooommm you're and pro baseball player. Life is good
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Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the
dangina [55]

Answer:

a. The spheres will attract each other.

Explanation:

When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.

  • When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.
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4 0
3 years ago
Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so
lisabon 2012 [21]

As per kinematics equation we are given that

v^2 = v_o^2 + 2ax

now we are given that

a = 2.55 m/s^2

v_0 = 21.8 m/s

v = 0

now we need to find x

from above equation we have

0^2 = 21.8^2 + 2(2.55)x

0 = 475.24 + 5.1 x

x = 93.2 m

so it will cover a distance of 93.2 m

7 0
3 years ago
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GrogVix [38]

Answer:

r =  \frac{v}{i}  = v = ri \\  i =  \frac{v}{r}

4 0
3 years ago
A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0
2 years ago
What happens to the velocity of a sound wave in air if the temperature of the air increases?
Yuki888 [10]
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