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kramer
2 years ago
15

A bunny hopping along the road can travel at an average of 0.6 m/s. Calculate far the bunny can travel in 11 seconds.

Physics
1 answer:
galben [10]2 years ago
3 0

Answer:

6.6 m

Explanation:

.6 m/s   *  11 s  = 6.6 m    ( see how the 's' cancels out and you are left with 'm'  as the answer ?)

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An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a
Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T,  by

\omega = \dfrac{2\pi}{T}

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

v = \omega A

A is the amplitude or maximum displacement from the equilibrium.

v = \dfrac{2\pi A}{T}

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

v_f^2 = v_i^2+2ah

v_f is the final velocity, v_i is the initial velocity (same as v above), a is acceleration of gravity and h is the height.

h = \dfrac{v_f^2 - v_i^2}{2a}

h = \dfrac{0^2 - 2.71^2}{2\times-9.81} = 0.37 \text{ m}

3 0
3 years ago
A wave is a _____ that transmits energy
galben [10]

Answer is disturbance

3 0
3 years ago
Read 2 more answers
In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.8 m/s. However, this speed is inade
erastova [34]

Answer:

3.95979 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow as=\frac{v^2-u^2}{2}\\\Rightarrow as=\frac{0^2-2.8^2}{2}\\\Rightarrow as=-3.92

Here s=\frac{1}{2}s

\\\Rightarrow as=-7.84\ m/s^2

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -7.84}\\\Rightarrow u=3.95979\ m/s

Initial velocity of the puck should be 3.95979 m/s

8 0
3 years ago
A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density
blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

6 0
3 years ago
what are the horizontal and vertical components of a vector that is 25units long with an angle of 130 degrees​
AlekseyPX

Answer:

The horizontal component of the vector ≈ -16.06

The vertical component of the vector ≈ 19.15

Explanation:

The magnitude of the vector, \left | R \right | = 25 units

The direction of the vector, θ = 130°

Therefore, we have;

The horizontal component of the vector, Rₓ = \left | R \right | × cos(θ)

∴ Rₓ = 25 × cos(130°) ≈ -16.06

<em>The horizontal component of the vector, Rₓ ≈ -16.06</em>

The vertical component of the vector, R_y = \left | R \right | × sin(θ)

∴  R_y = 25 × sin(130°) ≈ 19.15

<em>The vertical component of the vector, R</em>_y<em> ≈ 19.15</em>

(The vector, R = Rₓ + R_y

\underset{R}{\rightarrow} = Rₓ·i + R_y·j

∴ \underset{R}{\rightarrow} ≈ -16.07·i + 19.15j)

5 0
3 years ago
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