Plz answer #2 plz this is 8th grade science / physics

A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.

Jobisdone [24]

Answer:

(a) t = 4.52 sec

(b) X = 1,156.49 m

Explanation:

Horizontal Launching

If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity $v_o$ remains the same in time. The distance x is computed as .

$\displaystyle x=v_o.t$

(a)

The vertical component of the velocity $v_y$ starts from zero and gradually starts to increase due to the acceleration of gravity as follows

$v_y=gt$

This means the vertical height is computed by

$\displaystyle h=h_o-\frac{gt^2}{2}$

Where $h_o$ is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0

$\displaystyle t=\sqrt{\frac{2h_o}{g}}$

$\displaystyle t=\sqrt{\frac{2(100)}{9.8}}=4.52\ sec$

(b)

We now compute the horizontal distance knowing $v_o=256\ m/s$

$\displaystyle x=(256).(4.52)$

$X=1,156.49\ m$

4 0

2 years ago

In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t

grigory [225]

Answer:

3. relatively high temperature, about 10,000 K, so that significant numbers of electrons are excited from the ground state, n = 1, to the first excited state, n = 2, but not too many of them have been ejected completely from the atoms

Explanation:

If hydrogen absorption lines are very strong in the visible spectrum of a particular star that means the population of electron in n = 2 is very high so on being exited they absorb radiation in Balmer series and give rise to absorption spectrum. The average temperature required to excite electron in hydrogen atom from n=1 to n = 2 is 10000K .

4 0

3 years ago

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t

muminat

<h2>⇒Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So, m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is m/s2

Acceleration m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation ,

m/s

Therefore, the final velocity or jumping speed is m/s

Explanation:

3 0

3 years ago

Read 2 more answers

How fast would the International Space Station (ISS) have to travel to maintain a circular orbit a distance of 1400 km above the

professor190 [17]

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by $v=\sqrt{\frac{GM}{r}}$ , where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.