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2 years ago

A cyclist accelerates from 0m/s to 8m/s in 3 seconds.

Physics

1 answer:

garik1379 [7]2 years ago

5 0

Answer:

the rate of acceleration is 2.6666(and so on) if thats your question

Explanation:

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A toy rocket is fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the time it wil

miskamm [114]

Answer: 16.3 seconds

Explanation: Given that the

Initial velocity U = 80 ft/s

Let's first calculate the maximum height reached by using third equation of motion.

V^2 = U^2 - 2gH

Where V = final velocity and H = maximum height.

Since the toy is moving against the gravity, g will be negative.

At maximum height, V = 0

0 = 80^2 - 2 × 9.81 × H

6400 = 19.62H

H = 6400/19.62

H = 326.2

Let's us second equation of motion to find time.

H = Ut - 1/2gt^2

Let assume that the ball is dropped from the maximum height. Then,

U = 0. The equation will be reduced to

H = 1/2gt^2

326.2 = 1/2 × 9.81 × t^2

326.2 = 4.905t^2

t^2 = 326.2/4.905

t = sqrt( 66.5 )

t = 8.15 seconds

The time it will take for the rocket to return to ground level will be 2t.

That is, 2 × 8.15 = 16.3 seconds

8 0

3 years ago

How does the direction of friction relate to the direction of motion

lesya692 [45]

It reacts to the motion. If the mass hanging from the pulley was overwhelmingly heavier than the mass on the ramp, it'll obviously pull the ramp mass up and thus friction would be trying to oppose this and vice versa.

5 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .