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Ede4ka [16]
3 years ago
13

Answer the attached question.No Spam!⚔️⚔️⚔️​

Physics
1 answer:
RoseWind [281]3 years ago
6 0
<h3>Question:5</h3>

The given data can be written in following way in coordinates axes.

(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).

a) Average velocity for first 4 seconds

Average velocity = Total Displacement/ Time taken

= (4-0)/(4 - 0)

=4/4

= 1/1

<h3> =1 m/s</h3>

b) Average velocity for 4 to 8 seconds

Average velocity = Total Displacement/ Time taken

= (4 - 4)/(8-4)

= 0/4

<h3> = 0</h3>

c) Average velocity for last 6 seconds

Last 6 seconds = from 8 to 14 seconds

Average velocity = Total Displacement/Time taken

= (2 - 4)/(14 - 8)

= -2/6

<h3> = -1/3 m/s</h3>

<h3>Question 6:</h3>

The given data can be written in following way in coordinates axes.

(0,0), (10,20), (20,20), (30,20), (40,0)

a) State the kind of motion from Os to 10s and from 30s to 40s

It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.

b) What is the velocity of the body after 10s and 40s ?

It is clear from the graph and table as well that,

<h3>Velocity after 10s is 20m/s</h3>

and

<h3>Velocity after 40s is 0 m/s</h3>

c) Calculate the distance covered by the body between 10s to 30s.

Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD

Area of rectangle ABCD= (30 - 10) × (20 - O)

=20×20

<h3> = 400 m</h3>

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Read 2 more answers
How much heat (in kJ) is needed to convert 866 g of ice at −10°C to steam at 126°C? (The specific heats of ice and steam are 2.0
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Answer:

2671.3 kJ

Explanation:

m = 866 g at - 10 degree C converts into steam at 126 degree C.

The following steps are there

(1) ice at - 10 degree C converts into ice at 0 degree C

H1 = m x specific heat of ice x ΔT

H1 = 866 x 2.03 x 10 = 17579.8 J

(ii) ice at 0 degree C converts into water at 0 degree C

H2 = m x Latent heat of fusion = 866 x 334 = 289244 J

(iii) water at 0 degree C converts into water at 100 degree C

H3 =  m x specific heat of water x ΔT

ΔH3 = 866 x 4.186 x 100 = 362507.6 J

(iv) water at 100 degree C converts into steam at 100 degree C

H4 = m x Lateant heat of vaporisation

H4 = 866 x 2260 = 1957160 J

(v) seam at 100 degree C converts into steam at 126 degree C

H5 = m x specific heat of steam x ΔT

H5 = 866 x 1.99 x 26 = 44806.84 J

The total heat required is

H = H1 + H2 + H3 + H4 + H5

H = 2671298.24 J = 2671.3 kJ

8 0
3 years ago
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