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Citrus2011 [14]
2 years ago
9

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a = a

₁ + Fm where a₁ = 3.00 meter/second², F = 12.0 kilogram⋅meter/second² and m =7.00 kilogram. First, which of the following is the correct step for obtaining a common denominator for the two fractions in the expression in solving for a?
a. (m/m times a₁/1) + (1/1 times F/m)
b. (1/m times a₁/1) + (1/m times F/m)
c. (m/m times a₁/1) + (F/F times F/m)
d. (m/m times a₁/1) +(m/m times F/m )
Physics
1 answer:
Nikitich [7]2 years ago
8 0

Answer:

The correct option is option (a).

The acceleration of an object is \frac{33}{7} m/s².

Explanation:

Given expression is

a=a_1+\frac Fm

[ divide a whole number by 1 to turn into a fraction. Since a_1 is a whole number, so  a_1  is divided by 1 to turn into a fraction]

\Rightarrow a=\frac{a_1}{1}+\frac Fm

[The l.c.m of the denominators 1 and m is m. Now multiply both numerator and denominator by m of \frac{a_1}1 and multiply both numerator and denominator by 1 of \frac Fm]

\Rightarrow a=(\frac mm\times\frac{a_1}{1})+(\frac 11\times\frac Fm)

The correct option is option (a)

Given that,

F= 12.0 kg.m/s² , m=7.00 kg and a_1 = 3.00 m/s²

\therefore a=(\frac mm\times\frac{a_1}{1})+(\frac 11\times\frac Fm)

     =(\frac{7.00}{7.00}\times \frac{3.00}{1})+(\frac11\times \frac{12.0}{7.00})

    =\frac{21}{7}+\frac{12}{7}

    =\frac{21+12}{7}

    =\frac{33}{7} m/s²

The acceleration of an object is \frac{33}{7} m/s².

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serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr  and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

  1. db = 2dr
  2. L + (L - x) = 2x
  3. 2L - x = 2x
  4. 2L = 3x
  5. x = \frac{2}{3}L

Insert it in x = \frac{2}{3}L

\frac{2}{3}(2.4km) = 1.6km

Now we use formula db = 2dr

  1. db = 2L - x
  2. db = 2(2.4km) - 1.6km
  3. <u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

  1. Vr = L/ Δt ⇒ Δt = \frac{L}{Vr}
  2. \frac{2.4km}{6.8km/hr}
  3. \frac{6}{17}hr

(Km cancel each other)

  1. Vb = db/Δt ⇒ db = VbΔt
  2. 13.6km/hr(\frac{6}{17}hr )
  3. <u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

6 0
2 years ago
A brick is resting on a rough incline as shown in the figure. The friction force acting on the brick, along the incline, is
Tasya [4]
B) equal to the gravitational force of the brick
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2 years ago
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Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and
yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

v=\dfrac{dx}{dt}....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

a=\dfrac{dv}{dt}....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

F=\dfrac{d(p)}{dt}

Put the value of p

F=\dfrac{d(mv)}{dt}

F=m\dfrac{dv}{dt}

Put the value from equation (II)

F=ma

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

F=\dfrac{Gm_{1}m_{2}}{r^2}

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

3 0
2 years ago
A parachute works because the canvas of the parachute is acted upon by __________.
Ne4ueva [31]
The question is asking to choose among the following choices that could complete the question about the inertia, base on my research and further investigation, the possible answer would be letter B. Gravity. I hope you are satisfied with my answer and feel free to ask for more 
8 0
2 years ago
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A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al
anzhelika [568]

Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

          vₓ = v₀ₓ = 1.10 m / s

          aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

         v_{y}= v_{oy} -ay t

          ay = (v_{oy} -v_{y}) / t

          ay = (0 -10.9) / 1.6

          ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

8 0
2 years ago
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