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3 years ago

Choose the following types of hazards below

Physics

2 answers:

Aleksandr [31]3 years ago

6 0

Um where are the options for the answers?

sweet [91]3 years ago

6 0

The hazards are lower

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What is the part of the cell that stores food?

Alenkasestr [34]

Part of the cell that stores food is called vacuole

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3 years ago

A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

The distance the car traveled is 21.45 m

Explanation:

Motion With Constant Acceleration

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

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3 years ago

Show that the units N/kg can be written using only units of meters (m) and second (s). Is this a unit of mass ,acceleration or f

nikdorinn [45]

When the difference between two results is larger than the estimates error, the result is

3 0

3 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$