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3 years ago

What is the fleas acceleration as it extends its legs

Physics

1 answer:

valina [46]3 years ago

3 0

When a flea jumps it reaches a takeoff speed of 1.0 m/s over a distance of 0.50mm

hope its right :)

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A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s

bulgar [2K]

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

$E = \frac{F}{m}$

now in order to find the acceleration of the satellite we know by Newton's II law

$F = ma$

so we will have

$a = \frac{F}{m}$

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

4 0

3 years ago

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While running around the track at school, Milt notices that he runs due East on the 100m homestretch and due West on the 100m ba

Yuri [45]

Answer a would be correct since velocity is a vector and has a magnitude and a direction. In this case v₁ = - v₂.

8 0

3 years ago

Read 2 more answers

Which travels fastest through empty space? A. electromechanical waves B. transverse waves C. longitudinal waves D. electromagnet

Nady [450]

The answer is B hope it helps:)

7 0

3 years ago

A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck trav

aniked [119]

You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

for part b)
You can use the formula
Delta x/Delta t = average velocity

Need to find xf, knowing xi = 0

Thus, use the formula
x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
x_f = 100m

so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

20m/s = xm/20s 20*20 = x
x = 400

thus we have 100+400 = 500m then it slows down from 500m to x_f

thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5) x_f = 500 + 50
x_f = 550

therefore the total distance traveled is 550m

to calculate average velocity
550/35 = 16m/s

thus V_xavg = 16m/s

8 0

3 years ago

A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L

GaryK [48]

Answer:

a) L₂ = 0.676 m

b) σ₁ = 2.28*10⁸ N/m²

σ₂ = 9.62*10⁸ N/m²

c) ε₁ = 0.00253678

ε₂ = 0.00457875

Explanation:

Given info

L₁ = 1.22 m

A₁ = 2.19 cm² = 2.19*10⁻⁴ m²

L₂ = ?

A₂ = 0.52 cm² = 0.52*10⁻⁴ m²

P = 5.00*10⁴ N

E₁ = 9*10¹⁰ N/m²

E₂ = 2.1*10¹¹ N/m²

In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that

ΔL₁ = ΔL₂ ⇒ P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)

⇒ L₂ = A₂*E₂*L₁ / (A₁*E₁)

⇒ L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)

⇒ L₂ = 0.676 m

The stress in the brass rod is obtained as follows

σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²

The stress in the niquel rod is obtained as follows

σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²

The strain in the brass rod is obtained as follows

σ₁ = E₁*ε₁ ⇒ ε₁ = σ₁ / E₁

⇒ ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678

The strain in the niquel rod is obtained as follows

σ₂ = E₂*ε₂ ⇒ ε₂ = σ₂ / E₂

⇒ ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875

3 0

3 years ago