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3 years ago

A 3.0 kg rock is dropped from rest. What will its velocity be in 6.0 s if a 7.0 N force of air resistance acts on it?

Physics

2 answers:

OlgaM077 [116]3 years ago

6 0

Answer:

14 m/s^2

Explanation:

$v_{f}=V_{i} +at$

$V_{f}=$ final velocity

$V_{i}=$ initial velocity

a= acceleration

t=time

Initial velocity is zero, since the rock was originally at rest and time is 6 seconds.

You are looking for the final velocity, but you need acceleration to find it. Remember that:

$F=m*a$

Where F is force, m is mass, and a is acceleration. You know force is 7.0 N and mass is 3.0 kg. Use these to find a.

$7.0 N= 3*a$

$\frac{7}{3} = a$

So now...

$v_{f}=V_{i} +at$

$v_{f}=0 +(\frac{7}{3}) (6)$

$v_{f}= 14 m/s^{2}$

kicyunya [14]3 years ago

5 0

Explanation:

F=ma

m=3kg

F=7N

7=3a

a=7/3

acceleration=velocity/time taken

7/3=velocity/6

velocity=7/3×6/1

velocity=14ms-¹

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4 years ago

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches

yaroslaw [1]

Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

$H = ut - \frac{1}{2}gt^{2}$

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴ $-63 = 0(t) - \frac{1}{2}(0.98) t^{2} \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s$

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

$v = u - gt$

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴ $v = - 9.8(3.59)\\v = - 35.18m/s$

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

$v = u + at$

Here, the initial velocity will be the final velocity just before the parachute opens, that is

$u = - 35.18 m/s$

From the question,

$v = - 3.3 m/s$

$a = 1.5 m/s^{2}$

We then get

$-3.3 = - 35.18 + (1.5)t$

$-3.3 + 35.18 = 1.5t\\31.88 = 1.5t$

$t = \frac{31.88}{1.5}$

$t = 21.25 secs$

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

$x = ut + \frac{1}{2}at^{2} \\$

$x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2} \\x = -747.58 + 338.67\\x = -408.91m\\$

$x$ ≅ $- 409m$

Hence, the height the fall begins is 63m + 409m

= 472 m

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Explanation:

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