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Romashka [77]
3 years ago
5

I have 2 questions left that im stuck on, and I need to do them in 10 minutes.. well im doomed. Can anyone help?

Physics
1 answer:
erma4kov [3.2K]3 years ago
7 0
I can help what is the questions
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A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.
Jobisdone [24]

Answer:

<em>(a) t = 4.52 sec</em>

<em>(b) X = 1,156.49 m</em>

Explanation:

<u>Horizontal Launching </u>

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
3 years ago
An athletes heart beats 62 times per minute. What is the frequency of her heart beat?
PilotLPTM [1.2K]

Answer:

22

Explanation:

bc it just is

8 0
3 years ago
In a lab, the mass of object a is 52 kg. object a weighs
aksik [14]
If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
F=52*9.8=509.6
therefore the weight of the object is 509.6N
5 0
3 years ago
Read 2 more answers
. Friction is a rubbing force that ___________ a spinning yo-yo.
Advocard [28]
The yo-yo speeds up when you rub it
3 0
3 years ago
Read 2 more answers
A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and
victus00 [196]

Answer:

a)  t=195.948N.m

b)  \phi=13.6 \textdegree

Explanation:

From the question we are told that:

Density \rho=1.225kg/m^2

Velocity of wind v=14m/s

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient \mu=2.05

a)

Generally the equation for Force is mathematically given by

F=\frac{1}{2}\muA\rhov^2

F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2

F=163.29

Therefore Torque

t=F*r*sin\theta

t=163.29*1.2*sin90

t=195.948N.m

b)

Generally the equation for torque due to weight is mathematically given by

t=d*Mg*sin90

Where

d=sin \phi

Therefore

t=sin \phi*Mg*sin90

195.948=833sin \phi

\phi=sin^{-1}\frac{195.948}{833}

\phi=13.6 \textdegree

5 0
3 years ago
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