Mechanics Physics is a area in physics dealing with motion of objects talking about when force applied to objects what happens whether displacement or changes of the position
The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6
Explanation:
The given data is as follows.
Angular velocity (
) = 2.23 rps
Distance from the center (R) = 0.379 m
First, we will convert revolutions per second into radian per second as follows.
= 2.23 revolutions per second
= 
= 14.01 rad/s
Now, tangential speed will be calculated as follows.
Tangential speed, v = 
= 0.379 x 14.01
= 5.31 m/s
Thus, we can conclude that the tack's tangential speed is 5.31 m/s.
Answer: The velocity magnitude or the velocity direction chages.
Explanation:
According to Newton's second law of motion, the acceleration of a system moved in same direction and is also directly proportional to the external force which acts on it while inversely proportional to the mass. The formula is: a = F/m
Based on the question, since the object obtains acceleration, then it can be infered that there will be changes in the velocity magnitude or the direction as a result of the motion.
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
