Question

plz aswer for 50 points​

Answer 1

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4) question

Speed, v = 28 m/s

Explanation:

It is given that,

Initial velocity of body, u = 72 km/h = 20 m/s

Acceleration of the body, a = 4 m/s²

We have to find the speed of the body after 2 seconds. It can be calculated suing first equation of motion as :

v=u+at

$v=20\ m/s+4\ m/s^2\times 2\ \:$

v = 28 m/s

So, the speed of the car after 2 seconds is 28 m/s. Hence, this is the required solution.

5) Question ji

Answer:

u = 72 km/h

= 72×5/18

= 20 m/s

a = -5 m/s²

(i) v = 0

v = u+at

0 = 20-5t

-5t=-20

t = 20/5

t = 4 s

(ii) s = ut+1/2at²

= 20×4 + 1/2×-5×4²

= 80 - 1/2×5×16

= 80 - 40

= 40 m

(iii) Speed is doubled :-

u = 20×2 = 40 m/s

v = 0 m/s

a = -5 m/s²

v²-u²=2as

0²-40²=2×-5s

-1600=-10s

s = -1600/-10

s = 160 m

6) Question

Given :  

u (initial velocity) = 0 (rest) , v (final velocity) = 36 km/ hr = 36 × (1000/3600)

v = 36× 5/18 = 10 m/s. , t = 10s

From first equation of motion , v = u + at

10 = 0 + a ×10

10 = 10a

a = 10/10

a = 1 m/s²

acceleration of the scooter = 1 m/s²

Hence, the acceleration of the scooter is 1 m/s².

Answer 2

Heya ❤

QUESTION 1

IF A PARTICLE IS MOVING WITH UNIFORM VELOCITY ITS ACCELERATION WILL BE ZERO.

QUESTION 2

Its velocity at the highest point will be 0. The object has effectively converted all of it's kinetic energy into potential energy.

QUESTION 3

ACCELERATION IS THE RATE OF CHANGE OF VELOCITY. IT IS A VECTOR QUANTITY.

QUESTION 4

Speed, v = 28 m/s

Explanation:

It is given that,

Initial velocity of body, u = 72 km/h = 20 m/s

Acceleration of the body, a = 4 m/s²

We have to find the speed of the body after 2 seconds. It can be calculated suing first equation of motion as :

v=u+at

v=20\ m/s+4\ m/s^2\times 2\ \:v=20 m/s+4 m/s

2

×2

v = 28 m/s

So, the speed of the car after 2 seconds is 28 m/s. Hence, this is the required solution.

QUESTION 5

Answer:

u = 72 km/h

= 72×5/18

= 20 m/s

a = -5 m/s²

(i) v = 0

v = u+at

0 = 20-5t

-5t=-20

t = 20/5

t = 4 s

(ii) s = ut+1/2at²

= 20×4 + 1/2×-5×4²

= 80 - 1/2×5×16

= 80 - 40

= 40 m

(iii) Speed is doubled :-

u = 20×2 = 40 m/s

v = 0 m/s

a = -5 m/s²

v²-u²=2as

0²-40²=2×-5s

-1600=-10s

s = -1600/-10

s = 160 m

QUESTION 6

Given :

u (initial velocity) = 0 (rest) , v (final velocity) = 36 km/ hr = 36 × (1000/3600)

v = 36× 5/18 = 10 m/s. , t = 10s

From first equation of motion , v = u + at

10 = 0 + a ×10

10 = 10a

a = 10/10

a = 1 m/s²

acceleration of the scooter = 1 m/s²

Hence, the acceleration of the scooter is 1 m/s².