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mart [117]
2 years ago
13

An emergency relief plane is dropping a care package from a plane to a group of medical personnel working for a relief agency in

an African village. The package is designed to land in a small lake, inflate an attached raft upon impact, and finally resurface with the raft side down. The plane will be moving horizontally with a ground speed of 59.1 m/s. The package will be dropped a horizontal distance of 521 m from the intended target location. At what altitude above the pond must the plane be flying in order to successfully accomplish this feat?
Physics
1 answer:
ipn [44]2 years ago
7 0

Answer:

The altitude of the plane is 379.5 m.

Explanation:

Initial horizontal velocity, u = 59.1 m/s

Horizontal distance, d = 521 m

let the time taken by the packet to cover the distance is t.

Horizontal distance = horizontal velocity x time

521 = 59.1 x t

t = 8.8 s

let the vertical height is h .

Use second equation of motion in vertical direction.

h = u t  + 0.5 gt^2\\\\h = 0 + 4.9 \times 8.8\times8.8\\\\h= 379.5 m

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if a runners power is 400 watts as she runs, how much chemical energy does she convert into other forms in 10 minutes
AnnZ [28]

Answer:

Energy converted = 240000\,Joules = 240\, kJoules

Explanation:

Recall that Power is the rate at which energy is transferred therefore defined by the mathematical formula: Power\,=\,\frac{Energy\,transferred}{time}

Since the information on the power of the runner is given, as well as the time the energy conversion takes place, we can then use this equation to find how much energy is been converted. Notice that we just need to change the given time *10 minutes) into the appropriate units  (seconds)to get the answer in SI units of energy (Joules). The conversion of 10 minutes into seconds is done by multiplying : 10 minutes * 60 seconds/minute = 600 seconds.

We use this then to find the energy converted by the runner:

Power\,=\,\frac{Energy\,transferred}{time}\\400 \,W = \frac{E}{600\,sec} \\400 \,W * 600\,sec=E\\E=240000\,Joules = 240\, kJoules

3 0
3 years ago
What is the path that an electric current follows called
Gwar [14]
I’m sure it’s called a circuit:)
5 0
2 years ago
Read 2 more answers
What can fall but never get hurt
Vladimir [108]

It’s either snow or rain if it’s a riddle sort of.

4 0
3 years ago
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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
2 years ago
When measuring speed, you will measure the
qaws [65]

Answer:

Speed is the rate at which an object's position changes, measured in meters per second. The equation for speed is simple: distance divided by time

Explanation:

5 0
3 years ago
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