Knowing Relationships

Question 1

A ship is carrying out a survey of the sea bed using ultrasound waves. The speed of ultrasound waves in the water is 1500ms-1.

One pulse of ultra sound is received back at the ship 0.36 seconds after it was transmitted.

How would you calculate the depth of the sea bed?

Answer 1

d=vt
d = 1500 x 0.18
d = 270m

Help
( half 0.36 to get 0.18 )

Question 2

The frequency of the wave is 1200 and the waves are produced in 1 minute.

How would you calculate the Frequency?

Answer 2

f = N/ t

f = 1200/ 60

f = 20Hz

Help
( if you get 1 minute, convert into seconds)
( the dash ( / ) is divide )

Question 3

A wave has a frequency of 50Hz.

How would you calculate the time period of the wave?

Answer 3

T = 1/f

T = 1/50

T= 0.02s

Help
( the top is always a 1 )

Question 4

The crest to trough height is 0.5m.

How would you calculate the amplitude?

Answer 4

Amplitude = 0.5/2 = 0.25

Help
there is no formula
( you always divide the number of waves by 2 )

Question 5

A wave can travel at 800m in one minute.

How would you calculate the speed of the wave?

Answer 5

d = vt
800 = v x 60
v = 800/60
v = 13.3ms-1

Help
( the first line of numbers is only to help you, unless you are working out the distance, in d =vt you always divide )

Question 6

A wave has a frequency of 500Hz and a wavelength of 0.5m.

How would you calculate the speed of the wave?

Answer 6

v = f λ
v = 500 x 0.5
v = 250ms-1

Question 7

A car traveled 1.5km in 1 minute.

How would you calculate the average speed of the car?

Answer 7

d = ṽt
1.5 x 10 3 = ṽ x 60
1.5 ṽ x 10 3/ 60 = ṽ
ṽ = 25ms-1

Help
( 1.5km has been converted to 1.5 x 10 3 )

Question 8

Distance is the total that the cat has travelled 10m North then 3m South.

How would you calculate the distance on a vector quantity arrow?

Answer 8

Distance = 10 + 3 = 13m

Question 9

A car is travelling at 14ms-1 and accelerated to 30ms-1 in 4 seconds.

How would you calculate the acceleration of the car?

Answer 9

a = v - u/ t

a = 30 - 14/ 4

a = 16/4

a = 4ms-1

Help
( you do not need to write a = 16/4 )

Question 10

A car of mass 600g accelerated at 10ms-2.

How would you calculate the unbalanced force required for this acceleration?

Answer 10

F = ma
F = 600 x 10
F = 6000N

Help
( ms-2 means metres per second per second )

Question 11

A dog has a mass of 15kg.
How would you you calculate the mass and weight of the dog on earth’s surface and the moons surface.
( Earth g = 9.8Nkg-1 )
( Moon g = 1.6Nkg-1 )

Answer 11

W = mg
W = 15 x 9.8
W = 147N 

W = mg
W = 15 x 1.6 
W = 24N

Help
( look out for where the certain thing is. Eg. On earth or the moon )
( you will have a sheet provided with the gravitational field strengths )

Question 12

A car is towed 150m along a road. A force of 2000N is applied.

How much work is done in moving the car?

Answer 12

Ew= Fd
Ew= 2000 x 150 
Ew= 300,000J

Question 13

A crane lifts a mental girder of mass 25kg through a height of 15m.

How would you calculate the potential energy gained?

Answer 13

Ep = mgh 
Ep = 25 x 9.8 x 15
Ep = 36775J

Help
( if it does not tell you where the gravitational field strength is then automatically assume it’s earth )

Question 14

A 650kg car is travelling at 12ms-1.

Calculate the kinetic energy of the car.

Answer 14

Ek = 1/2 mv2
Ek = 0.5 x 650 x 12 squared 
Ek = 0.5 x 650 x 12 x 12 
Ek = 46800J

Help
( times the velocity by itself )
( you don’t need to include the 3rd line )

Question 15

A sample of body tissue has a mass of 0.5kg, if the energy absorbed by the tissue is 6.5J.

How would you calculate the absorbed dose?

Answer 15

D = E/m

D = 6.5 x 10-3/0.5

D = 1.3 x 10-2Gy

Help
( if your looking for prefix’s and it has kg, look for the first letter closet to the number eg. this would be ( k ) )

Question 16

What is the equivalent dose caused by alpha particles if the absorbed dose is 4.5 x 10-5Gy ?

Answer 16

H = DwR
H = 4.5 x 10-5 x 20
H = 9 x 10-4Sv

Help
( the wR is the radiation given, you will have a sheet giving you the numbers )

Question 17

What is the equivalent does rate caused by an equivalent dose of 100mSv of beta particles in 5 hours?

Answer 17

Ḣ = H/t

Ḣ = 100 x 10-3/18000

Ḣ = 5.55 x10-6Svs-1

Help
( 5 hours ( 5 x 60 x 60 seconds = 18000 )

Question 18

A Geiger counter records 1800 counts in a 90 second period.

What is the activity of the source?

Answer 18

A = N/t

A = 1800/90

A = 20Bq

Question 19

A source has an activity of 1000kBq 6 minutes later the activity is 125kBq.
What is the half- life?

1000 —> 500 —> 250 —> 125

Answer 19

(Each ARROW represents the passing of a half- life) To get from 1000kBq to 125kBq, takes 3 half-lives.

So there are 3 half- lives in 6 minutes.
The half-life is 2 minutes.

Help
( divide 6 by 3 that gives you 2 )