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Neko [114]
3 years ago
13

When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed?

Physics
1 answer:
Rina8888 [55]3 years ago
4 0

Answer:

It is less than its initial speed

Explanation:

The football moves both horizontally and vertically at the same time with a diagonal velocity v. This velocity vector v has two components: the horizontal velocity v_x, and vertical velocity v_y:

v=\sqrt{v_x^2+v_y^2}

When the football reaches its maximum height, the vertical velocity is zero, leaving only the x component. So, its speed it is less than its initial speed.

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Oil having a density of 930 kg/m^3 floats on
Zolol [24]

Answer:

0.0268 m

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, buoyancy of the oil B₁ pushing up, and buoyancy of the water B₂ pushing up.

Sum of forces in the y direction:

∑F = ma

B₁ + B₂ − mg = 0

ρ₁V₁g + ρ₂V₂g − mg = 0

ρ₁V₁ + ρ₂V₂ = m

ρ₁V₁ + ρ₂V₂ = ρV

ρ₁Ah₁ + ρ₂Ah₂ = ρAh

ρ₁h₁ + ρ₂h₂ = ρh

(930 kg/m³)h₁ + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

Since the block is fully submerged, h₁ + h₂ = 4.93 cm.

(930 kg/m³) (4.93 cm − h₂) + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

h₂ = 2.68 cm

h₂ = 0.0268 m

4 0
3 years ago
What is the velocity of a motercycle traveling 13 km west in 3 hours
Soloha48 [4]
4.333333 kilometers an hour
5 0
3 years ago
Write down the SI unit of length and mass​
Vanyuwa [196]

Answer:

The SI unit for length is meters(m), for mass is kilograms(kg)

Explanation:

hope it helps

6 0
3 years ago
A wave with a large amplitude has a lot of             a.vibration  b.speed   c.energy    
adoni [48]
<span>A wave with a large amplitude has a lot of             a.vibration  b.speed  <u> c.energy</u></span>
6 0
3 years ago
Read 2 more answers
EXAMPLE
yKpoI14uk [10]

Answer:

Explanation:

a ) The volume of blood flowing per second throughout the vessel is constant .

a₁ v₁ = a₂ v₂

a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .

2A x v₁ = A x .40

v₁ = .20 m /s

b )

Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂

Applying Bernoulli's theorem

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

P₁ - P₂ = 1/2  ρ(v₂² - v₁² )

= .5 x 1060 ( .4² - .2² )

= 63.6 Pa .

5 0
3 years ago
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