# Physics Question Essay

1-What is the total distance travelled by the particle?

Express your answer in meters.

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Total Distance Travelled by the particle is 30 m and the total time taken to travel that distance is 50 seconds.

2- What is the average velocity of the particle over the time interval ?

Express your answer in meters per second.

Average Velocity = ( Total Distance Travelled by the particle)/(Total time taken)

= 30/50

=0.6 meters per second.

3- What is the instantaneous velocity of the particle at ?

Express your answer in meters per second.

Gradient of the line = Velocity at any point of time.

V= (40-10)/(50-0)

V=30/50

V=3/5 m/s at t = 10.0 seconds.

Another common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. In this graph, time is plotted on the horizontal axis and velocity on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight-line motion, however, these vectors have only one nonzero component in the direction of motion.

Thus, in this problem, we will call the velocity and the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion

4- Which of the graphs shown is the correct v vs. t plot for the motion described in the previous parts?

‘D’ is the correct answer because in this graph the velocity is constant at 0.6 m/s which was exactly the gradient of the distance-time graph. The Distance- time graph also showed 0 acceleration which is also depicted in the Choice ‘D’.

1- What is the initial velocity of the particle, ?

Express your answer in meters per second.

Answer:

Initial velocity of the particle is 0.5 meters per second.

2- What is the total distance travelled by the particle?

Express your answer in meters.

Answer:

Area of Trapezium = ½(sum of parallel sides)x Height

= ½(50+20)x2

= 70 meters

3- What is the average acceleration of the particle over the first 20.0 seconds?

Express your answer in meters per second per second.

Gradient of the line = Acceleration at any point of time .

Answer:

For t= 20 sec,

A= (2-0.5)/20-0)

A= 0.075 meter per second per second.

4- What is the instantaneous acceleration of the particle at ?

Answer:

For t= 45 sec,

A= (2-0)/40-50)

A= -0.2 meter per second per second.

5- Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts?

‘C’ is the correct answer. It shows that the particle was moving at a constant acceleration of 0.2 for 20 seconds , then the acceleration dropped to 0 from t=20 till t=40 and then the particle decelerated for 10 seconds with an acceleration of -0.2.

………………………………………………..

6- Electrons that “paint” the picture in a TV tube undergo constant acceleration over a distance of 3.8 . If they reach a final speed of , what are (a) the electrons’ acceleration and (b) the time spent accelerating?Electrons that “paint” the picture in a TV tube undergo constant acceleration over a distance of 3.8cm .if they reach a final speed of 1.2 x 107 m/s, what are:

(a) The electrons acceleration? Express your answer using two significant figures

Answer:

Data :

U=0

S=3.8 cm or 0.038 meters

V=1.2x 107 m/s

A = ??

V2 = U2 + 2as

(1.2x 107)2=02 + 2(a)( 0.038)

1.44 x 1014 = 0.076a

1.9 x 1015 m/s2

(b) The time spent accelerating? Express your answer using two significant figures

Answer:

V = u + at

1.2 x 107 = 0 + 1.9 x 1015(t)

T = 6.3 seconds.

7- Starting from rest, a car accelerates at a constant rate, reaching 90km/h in 13s.

What is its acceleration?

Express your answer using two significant figures.

Answer:

Data :

U=0

V=90 km/h = 25 m/s

T=13

A = ?

V= u +at

25= 0 + 13a

A= 1.9 m/s2

How far does it go in this time?

Express your answer using two significant figures.

Answer:

s=ut + ½at2

s=0 + ½ (1.9)(13)2

s=160 meters

9-Uncompressed, the spring for an automobile suspension is 50cm long. It needs to be fitted into a space 34cm.

If the spring constant is 3.5 , how much work does a mechanic have to do to fit the spring?

Express your answer using two significant figures.

Answer:

Data :

Spring Constant 3.5 kN/m

X = 16 cm = 0.16 meters

Work Done = Area under graph = (0.5)(F)(x)

Where F = kx

Therefore Us = (0.5)(kx)(x)

And Finally, Us = 0.5kx2

= 0.5(3.5)(.016)2

=4.5 x 102 Joules

10- A 120g arrow is shot vertically from a bow whose effective spring constant 400N/m

If the bow is drawn 55 before shooting the arrow, to what height does the arrow rise?

Express your answer using two significant figures.

Us = 0.5kx2

Us = 0.5(400)(.55)2

Us = 60.5 Joules

Potential Energy = Mechanical Energy

Mgh = 60.5

0.12(9.81)(h)= 60.5

=51 meters

The bow rose to 51 meters height.

11- Estimate your power output as you do deep knee bends at the rate of one per second??

Answer:

Data:

Mass = Since the whole body does not move around 60 % of it moves therefore mass will be(.60×70) = 42 kg

Gravity =9.81 meter per square per square second

Height = .45 m/s

Power for one repetition per second = 2mgh

Power for one repetition per second = 2(9.81)(42)(.45)= 370 Watts.

12- A 44kg child stands at rest on ice skates. She catches a 1.5kg ball moving horizontally at 8.7m/s.

What is her speed after she’s caught the ball?

Express your answer using two significant figures.

Answer:

Kinetic Energy of the ball = ½ mv2

=½(1.5)(8.7) 2

= 57 Joules

Since kinetic enery of the girl was 0 so the total Energy of the girl and the ball will be 57 Joules.

Kinetic energy of the girl and ball = ½(45.5)(v) 2

57= ½(45.5)(v) 2

v=1.6 m/s

13- An astronaut in an orbiting spacecraft is “weighed” by being strapped to a spring of constant

K= 400N/m and set into simple harmonic motion.

If the oscillation period is 2.5 , what is the astronaut’s mass?

Express your answer using two significant figures.

Answer:

T=2.58 seconds

K= 400N/m

M = ?

Using the formula and substituting the values will give

In this standard form (k/m) is equal to omega squared. Where omega is the angular frequency of oscillation. That is:

omega = 2*pi / period of oscillation

Thus:

k/m = (2*pi / T)^2

Rearranging:

m = 400 / (2*pi / 2.5)^2

m =63 kg

References:

Krowne, A.(2005). Hooke’s Law, Retrieved on August 4 , 2009 from http://planetphysics.org/encyclopedia/HookesLaw.html.