Can anyone answer this question its a science question NO LINKS !!!!

You throw a ball vertically upward, and as it leaves your hand, its speed is 10.0 m/s.

8090 [49]

The answer is A bc I did the quiz and I got it right

5 0

3 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

a

$F =326.7 \ N$

b

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

3 years ago

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$