What is energy and object has due to its motion

Jain applies a input force of 100 N to a lever that has a Mechanical advantage of 3. What is the output force

Licemer1 [7]

Answer:

Output force = 300 N

Explanation:

Given that,

The mechanical advantage of a lever, m = 3

Input force, $F_i=100\ N$

We need to find the output force. The ratio of output force to the input force is called the mechanical advantage of the lever. So,

$m=\dfrac{F_o}{F_i}\\\\F_o=m\times F_i\\\\F_o=3\times 100\\\\F_o=300\ N$

So, the output force is 300 N.

8 0

3 years ago

Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i

inysia [295]

Answer:

$\large \boxed{\text{C. 2.3 m/s}}$

Explanation:

Data:

$m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\$

Calculation:

This is a perfectly inelastic collision. The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

$\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}$

3 0

3 years ago

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$

There is NO initial vertical velocity, so:

$\large\boxed{\Delta d= \frac{1}{2}at^2}}$

Rearrange to solve for time:

$2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}$

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

$t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s$

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

$\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}$

6 0

3 years ago

Question 1

egoroff_w [7]

C. Transform boundary

3 0

4 years ago

Read 2 more answers

A student wishes to record a 7.5-kilogram watermelon colliding with the ground. Calculate how far the watermelon must fall freel

algol [13]

Answer:

42.86m

Explanation:

The first thing we should keep in mind is that the watermelon moves with uniform acceleracion equal to gravity (9.81m / s ^ 2)

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

$\frac {Vf^{2}-Vo^2}{2.g} =Y$

where

Vf=29m/s= final speed

Vo= initial speed=0m/S

g=gravity=9.81m/s^2

Y= distance traveled(m)

solving

$\frac {Vf^{2}-Vo^2}{2.g} =Y\\\frac {(29m/s)^{2}-(0m/s)^2}{2(9.8m/s^2)} =Y\\Y=42.86m\\$

the distance traveled by watermelon is 42.86m

5 0

4 years ago