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2 years ago

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The different in size of each of the rope's pullers, correspond to a difference in the magnitude of the applied force, such that

: le 150N 100N SON SON 4. Apply a 200N force to the left rope, and 150 N to the right rope. What is the magnitude and direction of the Resultant Force?

1 answer:

olga55 [171]2 years ago

5 0

Answer:

F = - 50 N

Hence, the magnitude of resultant force is 50 N and its direction is leftwards.

Explanation:

The magnitude of the resultant force is always equal to the sum of all forces. While, the direction of resultant force will be equal to the direction of the force with greater magnitude:

$Resultant\ Force = F = F_{1} - F_{2}$

considering right direction to be positive:

F₁ = Force applied on right rope = 150 N

F₂ = Force applied on left rope = 200 N

Therefore, the resultant force can be found by using these values in equation:

$F = 150\ N - 200\ N$

<u>F = - 50 N</u>

<u>Hence, the magnitude of resultant force is 50 N and its direction is leftwards.</u>

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Lauren wants to know which location in her apartment is best for growing African violets. She has three African violets. She put

Harrizon [31]

First, before determining which variable is which, we go over the definition of each.
The independent variable is the one which is intentionally changed in order to investigate its effect on the dependent variable.
The dependent variable is monitored and changes occur in it due to the changing conditions of the independent variable.

In this case, the location of the African violets is the independent variable as it is intentionally changed, while the rate of growth of the African violets is the dependent variable as it is being measured.

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2 years ago

The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 × 10–15 s–1 at 298 K and a rate constant of 8

netineya [11]

Answer:

The activation energy for this reaction, Ea = 159.98 kJ/mol

Explanation:

Using the Arrhenius equation as:

$ln\frac {K_2}{K_1}=-\frac {E_a}{R}\times (\frac {1}{T_2}-\frac {1}{T_1})$

Where, Ea is the activation energy.

R is the gas constant having value 8.314 J/K.mol

K₂ and K₁ are the rate constants

T₂ and T₁ are the temperature values in kelvin.

Given:

K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K

K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K

Applying in the equation as:

$ln\frac {8.66\times 10^{-7}}{3.61\times 10^{-15}}=-\frac {E_a}{8.314}\times (\frac {1}{425}-\frac {1}{298})$

Solving for Ea as:

Ea = 159982.23 J /mol

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7 0

2 years ago

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

$Ek = 70\% \ Ek_o$

$\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )$

$v^2 = 70 \% \ (v_o)^2$

$v = - v_o \sqrt{70 \%}$ → <em>negative sign due to ball rebounds</em>

$v = - v_o \sqrt{0.7} \texttt{ m/s}$

$\texttt{ }$

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

$\Delta p_{stone} = - \Delta p_{ball}$

$\Delta p_{stone} = - [ mv - mv_o ]$

$\Delta p_{stone} = m[ v_o - v ]$

$\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]$

$\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Average Speed of Plane : brainly.com/question/12826372
Impulse : brainly.com/question/12855855
Gravity : brainly.com/question/1724648

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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2 years ago

The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The co

Marat540 [252]

Answer:

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Explanation:

Given:

mass of the system, $m_s=3.3\times 10^{3}\ kg$

velocity of librarian relative to the ground, $v_l=2.5\ m.s^{-1}$

velocity of the cart relative to the ground, $v_c=0.5\ m.s^{-1}$

N<u>ow using the principle of elastic collision:</u>

Net momentum of the system is zero.

$m_l\times v_l=(3300-m_l)\times v_c$

$m_l\times 2.5=(3300-m_l)\times 0.5$

$m_l=550\ kg$ is the mass of librarian.

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2 years ago

Read 2 more answers

A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s

lbvjy [14]

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

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I= moment of inertia (= 0.5 mr^2 )

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Tr= 0.5mr^2(a/r)

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a(1/2+1)=g

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4 0

2 years ago