Answer:
Lowering of mass of the shot has high kinetic energy which leads to covering more distance of the shot.
There is a difference in these throws because the mass of the shot decreases which increases its velocity and as a result the shot covers more distance. First, the athlete throws the shot having 16-pound weight or mass which covers 40 feet distance.
Whereas in the second throw, he uses a shot having 12-pound it covers 50 feet distance because the mass
is lower in the second throw and the athlete can easily throw the shot which covers more distance so the second shot has more kinetic energy as well as more velocity which compels it to cover more distance so we can conclude that lowering of mass leads to covering more distance of the shot.
The radius of the curved road at the given condition is 54.1 m.
The given parameters:
- <em>mass of the car, m = 1000 kg</em>
- <em>speed of the car, v = 50 km/h = 13.89 m/s</em>
- <em>banking angle, θ = 20⁰</em>
The normal force on the car due to banking curve is calculated as follows;
The horizontal force on the car due to the banking curve is calculated as follows;
<em>Divide </em><em>the second equation by the first;</em>
Thus, the radius of the curved road at the given condition is 54.1 m.
Learn more about banking angle here: brainly.com/question/8169892
Yes it is possible to add sound to a system to make it quieter
Answer:
the required angle is 0.0834879⁰
Explanation:
Given the data in the question;
slit separation; d = 1.75 mm = 1.75 × 10⁻³ m
wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m
wavelength λ₂ 510 nm = 510 × 10⁻⁹ m
Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;
tanθ = / D = mλ/d
since they both coincides;
tanθ₁ = tanθ₂
m₁λ₁/d = m₂λ₂/d
multiply both sides by d
so,
m₁/m₂ = λ₂/λ₁
we substitute
m₁/m₂ = 510 nm / 425 nm
m₁/m₂ = 510 nm / 425 nm
divide through by 85
m₁/m₂ = 6 / 5
hence m₁ and m₂ are 6 and 5
so, from the previous formula
tanθ₂ = m₂λ₂/d
we substitute
tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 0.00145714
θ₂ = tan⁻¹( 0.00145714 )
θ₂ = 0.0834879⁰
Therefore, the required angle is 0.0834879⁰