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SashulF [63]
3 years ago
15

A 500-n parachutist opens his chute and experiences an air resistance force of 800 n. the net force on the parachutist is then

Physics
2 answers:
Tasya [4]3 years ago
6 0

Force of 500 N is acting on the parachutist.

Parachutist applies 500 N force in downward direction.

Answer:

300 N upward

Solution:

Parachutist feels air resistance of 800 N.

Thus, 800 N of force is acting in upward direction.

Total force acting on the parachutist is given by,

F_{net} = air resistance force - force of parachutist

F_{net} = 800-500

F_{net} = 300 N

Direction of force is in upward direction because the air resistance force is more than force of parachutist.

andriy [413]3 years ago
5 0
The net force of the parachuting guy is 300 N upward.
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A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca
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Answer:

The correct solution is:

(a) 1.375\times 10^{-2} \ J

(b) 4.43\times 10^{-3} \ C

(c) 1.42\times 10^{-3} \ F

(d) 178.57 \ \Omega

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

P_{avg}=55 \ mW

       =55\times 10^{-3} \ W

Average voltage,

V_{avg}=3.1 \ V

Now,

(a)

⇒ E=P_{avg}\times \zeta

On substituting the values, we get

⇒     =55\times 10^{-3}\times 0.25

⇒     =1.375\times 10^{-2} \ J

(b)

⇒ E=Q\times V_{avg}

then,

⇒ Q=\frac{E}{V_{avg}}

On substituting the values, we get

⇒     =\frac{1.375\times 10^{-2}}{3.1}

⇒     =4.43\times 10^{-3} \ C

(c)

⇒ C=\frac{Q}{V}

⇒     =\frac{4.43\times 10^{-3}}{3.1}

⇒     =1.42\times 10^{-3} \ F

(d)

As we know,

⇒ R=\frac{1}{4C}

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5 0
2 years ago
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown
nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

v_x = 15.0 m/s

and with an initial vertical velocity of

v_{y0} = -20.0 m/s

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

y(t) = h + v_{0y} t + \frac{1}{2}gt^2 (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

v_y = -20 m/s

So the vertical position of the balloon after a time t is

y(t) = h + v_y t

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m

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In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

v_x = 15.0 m/s

So the horizontal distance travelled in t = 6.00 s is

d_x = v_x t = (15.0 m/s)(6.00 s)=90 m

Considering also that the vertical height of the balloon after t=6.00 s is

d_y = 176.6 m

The distance between the balloon and the rock can be found by using Pythagorean theorem:

d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m

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For an observer at rest in the basket, the rock is moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer at rest in the basket is

v_y (t) = gt

Substituting time t=6.00 s, we find

v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s

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v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer on the ground is now given by

v_y (t) = v_{0y} + gt

Substituting time t=6.00 s, we find

v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s

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