A man pushes a lawn mower with a force of 200 N across 50 meters of his lawn. How much work did the man perform?

The question will be on one of the comments on the answers you give just now take the points

Gennadij [26K]

Answer:

Explanation:

hi

6 0

2 years ago

What is the magnitude of the acceleration vector which causes a particle to move from velocity −5i−2j m/s to −6i+ 7j m/s in 8 se

shusha [124]

Answer:

Acceleration, $a=\dfrac{1}{8}(-i+9j)\ m/s^2$

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

Final velocity of particle in vector form, v = (-6i + 7j) m/s

Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}$

$a=\dfrac{(-i+9j)}{8\ s}\ m/s^2$

or

$a=\dfrac{1}{8}(-i+9j)\ m/s^2$

Hence, the value of acceleration vector is solved.

4 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 $x \ge 0$ :

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

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

2 years ago

Formula for deriving initial temperature in linear expansivity

jekas [21]

Answer:

∆L=aL∆T

Explanation:

that's the answer for your Question

5 0

2 years ago

One of the key disadvantages of buying fitness equipment is __________. A. privacy B. flexibility C. maintenance D. choice Pleas

Alex Ar [27]

Answer:

<h2>C. maintenance </h2>

Explanation:

I personally believe one key disadvantages is the cost of maintaining the equipment unlike the gym where you have to subscribe for the month or the year and forget about anything, owning the gym equipment comes with the extra cost and responsibilities of maintenance to the owners.

5 0

3 years ago