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3 years ago

If a flea can jump straight up to a height of 0.550 m , what is its initial speed as it leaves the ground?

1 answer:

7 0

Let u = upward initial speed of the flea.

At the maximum height of h = 0.550 m, the vertical velocity is v = 0.
Therefore, from the formula
v² = u² - 2gh,
obtain
0 = u² - 2*(9.8 m/s²)*(0.55 m)
u² = 10.78 m²/s²
u = 3.283 m/s

Answer: The initial upward speed is 3.28 m/s (nearest thousandth)

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4 0

2 years ago

Read 2 more answers

a 46 kilogram student climbs 11 meter up a rope at a constant speed if the student power output is 230 watts how long in seconds

Bezzdna [24]

230×46=10580÷11=961 second

3 0

3 years ago

A truck traveling at a constant speed of 28 m/s passes a more slowly moving car. The instant the truck passes the car, the car b

choli [55]

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

time = $\dfrac{distance}{speed}$

time = $\dfrac{545}{28}$

time =19.46 s

velocity of the car when it crosses the truck

$S = ut + \dfrac{1}{2}at^2$

$545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2$

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

5 0

3 years ago

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a

seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

6 0

3 years ago

What's the mass show the work?

Alja [10]

You do this one just like the other one that I just solved for you.

For this one ...

The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.

The question tells us: We have 2.0 cm³.

Each cm³ of space that the object occupies contains 2.5 gm of mass.

So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.

6 0

3 years ago