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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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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0
3 years ago
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Where does the name basketball come from?
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Definitely ball and basket
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3 years ago
The distance-time graph for a faster moving object has a smaller slope than the graph for a slower moving object. True or false
allsm [11]
False because faster moving slope would be going up and slower would be going down because its decreasing 
6 0
3 years ago
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A horizontally oriented pipe has a diameter of 5.8 cm and is filled with water. The pipe draws water from a reservoir that is in
Gnoma [55]

Answer:

v₂ = 97.4 m / s

Explanation:

Let's write the Bernoulli equation

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Index 1 is for tank and index 2 for exit

We can calculate the pressure in the tank with the equation

        P = F / A

Where the area of ​​a circle is

       A = π r²

E radius is half the diameter

      r = d / 2

      A = π d² / 4

We replace

    P = F 4 / π d²2

    P₁ = 397 4 /π  0.058²

    P₁ = 1.50 10⁵ Pa

The water velocity in the tank is zero because it is at rest (v1 = 0)

The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa

Since the pipe is horizontal y₁ = y₂

We replace on the first occasion

   P₁ = P₂ + ½ ρ v₂²

  v₂ = √ (P1-P2) 2 / ρ

  v₂ = √ [(1.50-1.013) 10⁵ 2/1000]

  v₂ = 97.4 m / s

6 0
4 years ago
A traveler pulls on a suitcase strap at an angle 36° above the horizontal. If of work are done by the strap while moving the sui
ollegr [7]

Answer:

The tension in the strap is 74.82 N.

Explanation:

Given that,

Angle between the horizontal and the suitcase is 36 degrees.

The distance traveled by the suitcase is 15 meters.

Let the work done by the suitcase is 908 J. We know that the work done in the vector form is given by :

W=Fd\ \cos\theta\\\\908=F\times 15\times cos(36)\\\\F=74.82\ N

So, the tension in the strap is 74.82 N. Hence, this is the required solution.

6 0
3 years ago
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