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2 years ago

Tissues types Please help

Physics

1 answer:

Harlamova29_29 [7]2 years ago

6 0

Answer:

Explanation: what grade are u in?

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what happens to the current in a circuit if a 1.5 volt battery is removed and is placed by a 9 volt battery?

juin [17]

Answer:

The current would increase.

Explanation:

The current would increase. 10. The relationship between resistance and current in a circuit is that the greater the resistance the less the current and the greater the current the less the resistance is.

8 0

2 years ago

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

The terminal speed of a sky diver is 130 km/h in the spread-eagle position and 326 km/h in the nosedive position. Assuming that

SIZIF [17.4K]

Answer:

$\frac{A_{slow} }{A_{fast} }=6.2885$

Explanation:

Given data

Terminal velocity for spread eagle position vt=130 km/h

Terminal velocity for nosedive position vt=326 km/h

The terminal speed of the diver is given by

$v_{t}=\sqrt{\frac{2mg}{CpA} }$

Therefore the area is given by

$A=\frac{2mg}{Cp(v_{t})^{2} }\\$

Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is

$\frac{A_{slow} }{A_{fast} }=\frac{326^{2} }{130^{2} }\\\frac{A_{slow} }{A_{fast} }=6.2885$