Answer:
Explanation:
Let the initial velocity of small block be v .
by applying conservation of momentum we can find velocity of common mass
25 v = 75 V , V is velocity of common mass after collision.
V = v / 3
For reaching the height we shall apply conservation of mechanical energy
1/2 m v² = mgh
1/2 x 75 x V² = 75 x g x 10
V² = 2g x 10
v² / 9 = 2 x 9.8 x 10
v² = 9 x 2 x 9.8 x 10
v = 42 m /s
small block must have velocity of 42 m /s .
Impulse by small block on large block
= change in momentum of large block
= 75 x V
= 75 x 42 / 3
= 1050 Ns.
<h2>The acceleration of car is 0.2 ms⁻²</h2>
Explanation:
When the car moves , the distance covered is calculated by the relation
S = u t + 
a t²
In this question u = 0 , because car was at rest initially
Thus S = a t²
here S is displacement and a is the acceleration of car
Therefore 360 = a ( 60 )²
Because time taken is one minute or 60 seconds
Therefore a = 
or a = 0.2 m s⁻²
Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
I do believe that the answer is D.
Hope this is right! Have a great day! :-)
Answer:
(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East
(b) When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West
Explanation:
Given;
a force vector points due east, 
= 140 N
let the second force = 
let the resultant of the two vectors = F
(a) When the resultant force is pointing along east line
the second force must be pointing due east
(b) When the resultant force is pointing along west line
the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.
