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Paladinen [302]
1 year ago
9

if a Firebird travels at a velocity of 0 to 60 mph in four seconds traveling east what was the acceleration of the Firebird

Physics
1 answer:
Tresset [83]1 year ago
3 0

Answer:

6.7 m/s^2

Explanation:

The formula of acceleration is:

\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{v_2 - v_1}{t_2-t_1}}

where \displaystyle{\vec{a}} is acceleration, \displaystyle{\vec{v}} is velocity and \displaystyle{t} is time. \displaystyle{v_2} means final velocity. \displaystyle{v_1} means initial velocity, \displaystyle{t_2} means final time and \displaystyle{t_1} means initial time.

We are given that the Firebird travels at velocity of 0 to 60 mph in four seconds. Therefore:

  • Our initial velocity starts at 0 mph.
  • Our final velocity is at 60 mph.
  • Our initial time is 0 second.
  • Our final time is 4 seconds.

Since it travels to the east then our vector will be positive. However, acceleration has to be in m/s^2 unit (Sl unit) so we'll have to convert from mph (miles per hours) to m/s (meters per second) first.

We know that:

  • A mile equals to 1609.344 meters.
  • An hour equals to 60 minutes which a minute equals to 60 seconds. So 60 minutes will equal to 3600 seconds.

Now we divide 1609.344 by 3600 to find a unit rate of m/s:

\displaystyle{\dfrac{1609.344}{3600} \ \, \sf{m/s}}\\\\\displaystyle{= 0.44704 \ \, \sf{m/s}}

Now multiply 0.44704 m/s by 0 and 60 to get velocity in m/s unit:

  • Initial velocity = 0 m/s
  • Final velocity = 60 * 0.44704 = 26.82 m/s

Time is already in second so no need for conversion. Substitute known information in the formula:

\displaystyle{\vec{a} = \dfrac{26.82-0}{4-0}}\\\\\displaystyle{\vec{a} = \dfrac{26.82}{4}}\\\\\displaystyle{\vec{a} = 6.7 \ \, \sf{m/s^2}}

Therefore, the Firebird will accelerate at the rate of 6.7 m/s^2.

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Determine the value of the resultant and its location from O.<br>see attach image.​
xxTIMURxx [149]

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L  

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L  

the magnitude of the resultant will be

F = w₀L  √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

     (√⅛)[d] = ½         [⅔L] + ¼      [⅔L/2] - ¼      [L - ⅓L/2]

     (√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

     (√⅛)[d] =      ⅓L  +    ⅟₁₂L     -  ¼L + ⅟₂₄L  

     (√⅛)[d] = 5L/24

               d = 5L/24 / (√⅛)

               d = 5√⅛L/3

8 0
3 years ago
If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?
aivan3 [116]

Initial velocity = \(v_0\)

acceleration in the downward direction = -9.8 \(\frac {m}{s^2}\)

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

\(v^2 = {v_0}^{2} + 2aH

\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410

\({v_0}^{2} = 2 \times 9.8 \times 0.410\)

\(v_0 = 2.834 \frac {m}{s}\)

Speed with which the flea jumps = \(2.834 \frac {m}{s}\)

4 0
3 years ago
A pendulum consists of a 0.5 kg mass attached to the end of 1-meter-long rod of negligible mass. When the rod makes an angle of
Alenkinab [10]

Answer:

T=4.24 N.m

Explanation:

Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot

F=0.5 kg* 9.8 m/s^{2}= 4.9 N

and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus

so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N

and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m

anothe way to do it is,

T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result

7 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
2 years ago
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