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Alex_Xolod [135]

4 years ago

12

How much time would it take for an

1 answer:

Elina [12.6K]4 years ago

4 0

36 and a half hours

Explanation:

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Calculate the speed of an 8.0*10^4kg airliner with a kinetic energy of 1.1*10^9J.

Burka [1]

Answer:

The velocity will be v = 165.83[m/s]

Explanation:

This is a problem where the definition of kinetic energy can be applied, which can be determined with the following equation.

$E_{k}=\frac{1}{2}*m*v^{2}\\ where:\\m = mass = 80000[kg]\\v = velocity [m/s]\\E_{k}= kinetic energy [J]=1100000000[J]\\Replacing:\\v=\sqrt{\frac{2*E_{k} }{m} } \\v=\sqrt{\frac{2*1100000000 }{80000} }\\v=165.83[m/s]$

8 0

3 years ago

A car travelled a distance of 200 m with initial velocity of 216 km/hr. Calculate the acceleration

pogonyaev

Answer:

$a = 16\ m/s^2$

Explanation:

When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

$v_f^2=v_o^2+2ax$

Where:

vf = final velocity

vo = initial velocity

a = acceleration

x = distance

Solving for a:

$\displaystyle a=\frac{v_f^2-v_o^2}{2x}$

The car travels a distance of x=200 m and the velocities are:

vo = 216 Km/h

vf = 360 Km/h

Both velocities must be converted to meters by seconds.

vo = 216 Km/h *1000/3600 = 60 m/s

vf = 360 Km/h *1000/3600 = 100 m/s

The acceleration is:

$\displaystyle a=\frac{100^2-60^2}{2*200}$

$\displaystyle a=\frac{10000-3600}{400}$

$\displaystyle a=\frac{6400}{400}$

$\mathbf{a = 16\ m/s^2}$

7 0

3 years ago

A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s

Jlenok [28]

Answer:

The applied torque is 3.84 N-m.

Explanation:

Given that,

Moment of inertia of the wheel is $2\ kg-m^2$

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

$\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m$

So, the applied torque is 3.84 N-m.

4 0

3 years ago

Which terms refer to the smallest pure substances? A. mixtures and alloys B. molecules and atoms C. elements and isotopes D. met

MariettaO [177]

B. molecules and atoms

7 0

3 years ago

How long would it take you too run the 100m sprint if from rest you accelerate at 2m/s/s for the whole race?

fredd [130]

Answer:

10 s

Explanation:

Given:

Δx = 100 m

v₀ = 0 m/s

a = 2 m/s²

Find: t

Δx = v₀ t + ½ at²

100 m = (0 m/s) t + ½ (2 m/s²) t²

t = 10 s

4 0

3 years ago