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larisa86 [58]
3 years ago
12

a parachutist weighs 1000N. when she opens her parachute, it pulls upwards on her with a force of 2000N. (a) draw a diagram to s

how forces acting on the parachutist. (b) calculate the resultant forces acting on the parachutist. (c) what effect will this force have on her ?​

Physics
1 answer:
wel3 years ago
4 0

(a) See attached figure

There are only two forces acting on the parachutist:

- Its weight, downward, of magnitude 1000 N, indicated with (mg) in the diagram (where m = mass and g = acceleration of gravity)

- The air resistance on the parachute, upward, of magnitude 2000 N, indicated with R in the diagram

The magnitude of the air resistance is twice that of the weight, therefore in the diagram is shown with an arrow approximately twice as long as the arrow representing the weight.

(b) +1000 N

Taking upward as positive direction, the two forces can be written as:

R = +2000 N

mg = -1000 N

Since the two forces are along the same line, we can find their resultant by simply calculating their algebraic sum:

F=R+mg=+2000+(-1000)=+1000 N

And the positive sign indicates that the direction is upward.

(c) The parachutist will slow down (accelerates upward)

Since there is a net force acting on the parachutist, there is also an acceleration, which can be calculated using Newton's second law:

F=ma \rightarrow a=\frac{F}{m}

where

F = +1000 N is the net force

m is the mass of th parachutist

The mass can be found from the weight:

m=\frac{mg}{g}=\frac{1000}{9.8}=102 kg

Therefore, the acceleration is

a=\frac{+1000}{102}=+9.8 m/s^2

And the direction of the acceleration is upward, while the direction of motion of the parachutist is downward: therefore, the parachutist will slow down.

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3 years ago
A biker first accelerates from 0.0 m/s to 6.0 m/s in 6 s, then continues at this speed for 5 s. What is the total distance trave
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Answer:

48m

Explanation:

Given the following data;

Initial velocity = 0m/s

Final velocity = 6m/s

Time, t = 6 secs

Time, T2 = 5 secs

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

Substituting into the equation;

a = \frac{6 - 0}{6}

a = \frac{6}{6}

Acceleration, a = 1m/s²

<u>To find the distance covered in the first phase;</u>

<em>Solving for distance, we would use the second equation of motion;</em>

S = ut + \frac {1}{2}at^{2}

<em>Substituting the values into the equation;</em>

S = 0(6) + \frac {1}{2}*1*(6)^{2}

S = 0 + \frac {1}{2}*1*36

S = 0.5 *36

Distance, S1 = 18m

<u>For the second phase, time T2 = 5 secs;</u>

<em>Mathematically, speed is given by the equation;</em>

Speed = \frac{distance}{time}

<em>Making distance the subject of formula, we have;</em>

Distance, S = speed * time

<em>Substituting into the above equation;</em>

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4 0
3 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

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