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2 years ago

A small aircraft accelerated down a runway at 4.0 m/s²

Physics

1 answer:

Radda [10]2 years ago

4 0

Given data in the problem :-

Acceleration (a) = 4.0 m/s^2
Initial velocity (u) = 0 m/s
Final velocity (v) = 34 m/s
Distance travelled by aircraft (S) = ?

From Newton's Laws of Motion we know that ,

v = u + at [t = Time taken by aircraft to cover the distance]

⇒ 34 = 0 + 4t

⇒ t = 34/4 s

∴ t = 8.5 s

From Newton's Laws of Motion we also know that ,

S = u.t + 1/2a.t^2

⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m

∴ S = 144.50 m

Thus the distance travelled by the aircraft while accelerating is 144.50 meter .

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What will be the average velocity of a body falling in free fall on Earth for 3 s?

SpyIntel [72]

Answer:

29.4m/s

Explanation:

Given parameters:

Time = 3s

Unknown:

Average velocity = ?

Solution:

To solve this problem, we use the expression below:

v = u + gt

v is the average velocity

u is the initial velocity = 0m/s

g is the acceleration due to gravity = 9.8m/s²

t is the time

So;

v = 0 + (9.8 x 3) = 29.4m/s

6 0

3 years ago

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

monitta

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

7 0

3 years ago

A car has a mass of 850 kg. By pushing on the car, Evan increases its speed

Dmitry_Shevchenko [17]

Answer:

I=1,2•10³ kg•m/s

Explanation:

v¹=3.5m/s

vf=5m/s

v=5-3.5=1.5m/s

I=p

I=mv=850•1.5=1275 kg•m/s=1,2•10³ kg•m/s

8 0

4 years ago

Read 2 more answers

awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall

pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

$KE = 6.66 J$

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

$m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}$

now plug in all values in it

$1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22$

so we have

$3.01 = 0.375v_1 + 0.927v_2$

similarly in Y direction we have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}$

now plug in all values in it

$0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22$

so we have

$0 = 0.927v_1 - 0.375v_2$

$v_2 = 2.47 v_1$

now from 1st equation we have

$3.01 = 0.375 v_1 + 0.927(2.47 v_1)$

$v_1 = 1.13 m/s$

$v_2 = 2.78 m/s$

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

$KE = \frac{1}{2}(1.47)(3.01^2)$

$KE = 6.66 J$

5 0

4 years ago

When opening a door, you push on it perpendicularly with a force of 57.0 n at a distance of 0.470 m from the hinges. what torque

TiliK225 [7]

<span>Torque = force * distance from axis of rotation. Multiply. If it's not perpendicular, you multiply that with the sine of the angle between the force and the perpendicular direction. T= 0.470*57*Sin 90 = 26.79 N</span>

3 0

3 years ago