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Allushta [10]
1 year ago
15

could you help with question 5. Your solutions to the word problems in volving Newton's Laws should have the following features:

Each solution should begin with one or more free body diagrams with a sign convention.The solution process should follow the GRASP process.Formulae should always be used in a three-step process: formula with letter variables, substitution of values, and then the answer.Answers should be expressed with the correct number of significant digits and with the direction if required.

Physics
1 answer:
Ganezh [65]1 year ago
3 0

ANSWER:

(a) 20 m/s^2

(b) 250 m

STEP-BY-STEP EXPLANATION:

Given:

Mass (m) = 250 g = 0.25 kg

Force (F) = 20 N

Frictional force (Ff) = 15 N

(a)

The net force on the block is:

\begin{gathered} F_{net}=F-F_f \\ F_{net}=20-15 \\ F_{net}=5\text{ N} \end{gathered}

We know that:

\begin{gathered} F=m\cdot a \\ \text{ We replacing} \\ 5=0.25\cdot a \\ a=\frac{5}{0.25} \\ a=20\text{ m/s}^2 \end{gathered}

(b)

We can calculate the partial distance from the following equation:

initial velocity (u) = 0 (start from rest)

\begin{gathered} s=ut+\frac{1}{2}at^2 \\ \text{ we replacing} \\ s=0\cdot5+\frac{1}{2}\cdot20\cdot5^2 \\ s=250\text{ m} \end{gathered}

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. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat
AfilCa [17]

Answer:

a) a = 4.9 m / s²,  N = 16.97 N   and b)   F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

        Wy = W cos 30

Let's write the equations on each axis

X axis

        Wx = ma

Y Axis  

       N- Wy = 0

       N = Wy = mg cos 30

       N = 2.0 9.8 cos 30

       N = 16.97 N

We calculate the acceleration

       a = Wx / m

       a = mg sin 30 / m

       a = g sin 30

       a =9.8 sin 30

       a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N

5 0
3 years ago
What happens when a candle burns?
slega [8]

Answer:

oxygen is used up is the answer

Explanation:

These vaporized molecules are drawn up into the flame, where they react with oxygen from the air to create heat, light, water vapor (H2O) and carbon dioxide (CO2).

5 0
2 years ago
Looking at the bottom standing wave in the picture, how many complete waves are present?
Shtirlitz [24]

Answer:

the last one

Explanation:

4 0
2 years ago
Light passes through a 0.15 mm-wide slit and forms a diffraction pattern on a screen 1.25 m behind the slit. The width of the ce
Elena L [17]

Answer:

The value is  \lambda =  900 \ nm

Explanation:

From the question we are told that

   The width of the slit is  a =  0.15 \ mm = 0.00015 \  m

     The distance of the screen from the slit is D = 1.25 m

      The width of the central maximum is y =  0.75 \ cm  = 0.0075 \ m

Generally the width of the central maximum is mathematically represented as

         y   =  \frac{m *  D  *  \lambda}{a}

Here  m is the order of the fringe and given that we are considering the central maximum, the order will be  m =  1  because the with of the central maximum separate's the and first maxima

So

        \lambda     =     \frac{a y}{ m *  D }

=>     \lambda     =     \frac{ 0.000015 *  0.0075}{ 1  *  1.2 }

=>     \lambda     =   900 *10^{-9} \  m

=>      \lambda =  900 \ nm

6 0
3 years ago
It is estimated that 1kg of body fat will provide 3.8 * 10^7 J of energy. A 67kg mountain climber decides to climb a mountain 35
xz_007 [3.2K]
During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him.  The climber has to DO the positive work to haul himself up.
  
                     Work = (mass) x (gravity) x (height) .

For the guy in this problem:

                     Work = (67 kg) x (9.8 m/s²) x (3,500 meters)

                             =  2,298,100 joules.

If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off

                       (2,298,100 joules) / (3.8 x 10⁷ joules/kg)

                   =          0.06 kg of fat.

That's only about 2.1 ounces.  We KNOW he'll lose more weight than that,
climbing 11,000 feet.  That's because climbing is pretty inefficient. 
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
and sweating.
4 0
3 years ago
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