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3 years ago

Can a body have two acceleration at a time?

Physics

1 answer:

bixtya [17]3 years ago

8 0

Answer:

No.

Explanation:

Acceleration of an object is defined as the rate of change of velocity per unit time. It can be given by :

a = dv/dt

dv is the change in velocity i.e. final velocity-initial velocity

dt is change in time

At a particular time, its change in velocity can be one only. It would imply that a body cannot have two acceleration at a particular time.

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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

3 0

3 years ago

When the speed of an object changes, what happens to its velocity?

vova2212 [387]

Answer:

Its velocity also changes

Explanation:

5 0

3 years ago

7. If F1 is the magnitude of the force exerted by the Earth on a satellite in orbit about the Earth and F2 is the magnitude of t

natita [175]

Answer:

F1 is equal to F2

Explanation:

Here

F1 is the gravitational force exerted by the earth on the satellite.

F2 is the gravitational force exerted by the satellite on the earth.

Now these two forces are equal but opposite in nature. This is given by the Third law of motion by Newton. According to this law, when there is force exerted between two objects, one force is balanced the other force which is equal in magnitude and opposite in nature.

Thus the gravitational force of the earth exerted on the satellite is equal to the force exerted by the satellite on the earth.

Hence F1 = F2.

8 0

4 years ago

Read 2 more answers

A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he e

stiv31 [10]

Answer:

1120 N

Explanation:

The velocity with which he hits the water can be found with kinematics:

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)

v = -13.3 m/s

Or it can be found with conservation of energy.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × -9.8 m/s² × -9.00 m)

v = -13.3 m/s

Sum of forces on the diver after he hits the water:

∑F = ma

F − mg = m Δv/Δt

F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)

F = 1120 N

6 0

3 years ago

A spring (k = 20 N/m) sits on a frictionless surface. The spring is compressed 20 cm and a 2 kg mass is placed before the spring

Olenka [21]

Answer:

E.)none of the above

Explanation:

Elastic energy = kinetic energy

½ kx² = ½ mv²

v = x √(k / m)

v = 0.2 m √(20 N/m / 2 kg)

v = 0.632 m/s

Time to travel 2.0 m:

t = (2.0 m) / (0.632 m/s)

t = 3.16 s

7 0

4 years ago

Can a body have two acceleration at a time?​

Can a body have two acceleration at a time?