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salantis [7]
2 years ago
8

For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of

the time. if an object travels 2.0 furlongs in the first 2.0 s, how far will it travel in the first 4.0 s?
Physics
1 answer:
natali 33 [55]2 years ago
8 0

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

 

<span>Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:</span>

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

 

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

<span>d = 8 furlongs</span>

<span>
</span>

<span>It traveled 8 furlongs for the first 4.0 seconds.</span>

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Answer:

x = -1.437 cm

Explanation:

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x = A sin(\omega t)          ........(1)

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x = Position of the wave

A = Amplitude of the wave

ω = Angular velocity

t = time

In this case, the amplitude is just half the range,

thus,

A =\frac{3cm}{2}=1.5cm  (Given range = 3cm)

A = 1.5 cm  

Now, The angular velocity is given as:

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

Where, T = time period of the wave =0.27s (given)

\omega=\frac{2\pi}{0.27s}

or

\omega=23.27s^{-1}

so, at time t = 55 s, the equation (1) becomes as:

x = 1.5 sin(23.27\times 55)

on solving the above equation we get,

x = -1.437 cm

here the negative sign depicts the position in the opposite direction of +x

5 0
2 years ago
How can controlling the way light bends and reflects be used to help people?
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too much sun is dangerous for humans and can cause cancer so it's important that light is reflected for example a pool reflects water back to space that is why water sometimes is cold because it reflects light

4 0
2 years ago
The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
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Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

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final velocity = 1.5 m/s

let force be F

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we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

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Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.
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The hoop is attached.

Consider that the friction force is given by:
F = μ·N
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We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ

Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
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Now, solve for θ:
θ = tan⁻¹(2·μ)
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Ingrid is moving a box from the ground into the back of a truck. She uses 20 N of force to move the box 5 meters. If she uses an
Oduvanchick [21]

Answer:

C

Explanation:

4 0
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