Ask question

Ask question

All categories

2 years ago

10

An airplane refuels in Bakersfield and continues on to Fresno. It travels an average speed of 610 km/h. IF the trip takes 2.75 h

ours, what is the flight distance between Bakersfield and Fresno?

1 answer:

rusak2 [61]2 years ago

8 0

2.75x610km

1220+7.5x61

1220+427 ... just over

1647

You might be interested in

If there was no frictional force acting on the ball in question 2 what would happen to the player? Use inertia and Newton’s 1st

laila [671]

Newwton's law of inertia states that an object will not be able to move unless force is applied to it

6 0

3 years ago

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is

mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate = 150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

$h= \frac{1}{2}at^2$

$h= \frac{1}{2}4.71\times4.96^2$

h= 58.012 m

h≅ 58 m

4 0

3 years ago

A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The

cluponka [151]

Answer:

v = 4.1 m/s

Explanation:

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now for pure rolling condition we know that

$v = R\omega$

so we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})$

$mgh = \frac{7}{10}mv^2$

now we will have

$v^2 = \sqrt{\frac{10}{7}gh}$

$v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}$

$v = 4.1 m/s$

7 0

3 years ago

If the net force acting on a stationary object is zero then the object will

Drupady [299]

If the net force is zero, then there is no acceleration, which means no change in velocity. The velocity starts at zero, so it must remain at zero. Therefore, the object will not move.

7 0

3 years ago

Is it possible to create a visual model (stop motion) of a wave motion using the differential equation for simple harmonic oscil

Alexeev081 [22]

Answer:

Yes.

Explanation:

A particular solution for the 1D wave equation has the form

$\Psi(x,t) \ = \ A \ sin ( \ k x \ + \omega t \ + \phi \ )$

where A its the amplitude, k the wavenumber, ω the angular frequency and φ the phase angle.

Now, for any given position $x_0$ , we can use:

$\phi_0 \ = \ \phi \ + \ k x_0$

so, the equation its:

$\Psi(x_0,t) \ = \ A \ sin ( \omega t \ + \ \phi_0 )$ .

This is the equation for a simple harmonic oscillation!

So, for any given point, we can use a simple harmonic oscillation as visual model. Now, when we move a $\delta$ distance from the original position, we got:

$x_1 = x_0 + \delta$

and

$\phi_1 = \phi \ + k x_1$

now, this its

$\phi_1 = \phi \ + k ( x_0 + \delta)$

$\phi_1 = \phi \ + k x_0 + k \delta$

$\phi_1 = \phi_0 + k \delta$

So, there its a phase angle difference of $k \delta$ . We can model this simply by starting the simple harmonic oscillation with a different phase angle.

8 0

3 years ago