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2 years ago

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A constant eastward horizontal force of 70 newtons is applied to a 20 -kilogram crate moving toward the east on a level floor. I

f the frictional force on the crate has a magnitude of 10 newtons, what is the magnitude of the crate's acceleration?

1 answer:

matrenka [14]2 years ago

7 0

Answer:

a = 3 m/s²

Explanation:

given,

mass of crate = 20 Kg

horizontal force on crate = 70 N

frictional force on the crate = 10 N

acceleration of crate = ?

now, calculating net force acting on the crate.

F = horizontal force - frictional force

F = 70 - 10

F = 60 N

net force on the crate is equal to 60 N.

We also know that

F = m a

60 = 20 x a

a = 3 m/s²

Hence, the acceleration of the crate is equal to 3 m/s²

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Answer:195 J

Explanation:

Given

mass of ball $m=0.0570\ kg$

ball leaves the hand with $u=15\ m/s$

maximum height reached by ball $h=8\ m$

Initial Mechanical energy when ball just leaves the hand

$M.E._1=(P.E.+K.E.)_1$

$M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1$

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

$M.E._1=\frac{1}{2}\times m\times (15)^2$

$M.E._1=6.41\ J$

Mechanical Energy at highest point

$(M.E.)_2=(P.E.+K.E.)_2$

at highest Point velocity is zero

$(M.E.)_2=mgh_2+0$

$(M.E.)_2=0.0570\times 9.8\times 8$

$(M.E.)_2=4.46\ J$

Decrease in Mechanical energy

$(M.E.)_1-(M.E.)_2=6.41-4.46$

$(M.E.)_1-(M.E.)_2=1.95\ J$

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2 years ago

A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$

Where $g = 9.8 m/s^2$

Similarly, the distance y the object has traveled is calculated as follows:

$\displaystyle y=\frac{g\cdot t^2}{2}$

If we know the height h from which the object was dropped, we can solve the above equation for t:

$\displaystyle t=\sqrt{\frac{2\cdot y}{g}}$

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

$\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec$

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

$t_2=2.56 - 2 = 0.56\ sec$

Therefore, it has traveled down a distance:

$\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m$

Thus, the height of the pen is:

$h_p = 32 - 1.54\Rightarrow h_p=30.46\ m$

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You are playing a game called "Will It Float?" In this game, you are given a large, square can of tuna. If you know the density

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Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o

ipn [44]

Answer:

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence, E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

U = E_p + E_k + E_g

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GenaCL600 [577]

Answer:

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Explanation:

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