A measure of the quantity of matter is called mass. D.
This is how much of matter is contained in an object. It is different from weight, which is the pull of gravity on an object.
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
Answer:
The value is 
Explanation:
From the question we are told that
The average acceleration of the bicycle is 
The initial forward velocity is 
Generally from kinematic equations we have that
Here v is the final velocity of the bicycle and the value is v = 0 m/s
So
=> t = 18.6 s
The average acceleration is -5.21 m/s² and the stopping time is 5.2 sec.
Given,
Final velocity, v = 0
Initial velocity, u = 61 mph
mph to km/h
u = 97.6 km/h
Or, u = 27.11 m/s
Displacement, s = 231 feet
Or, s = 70.4 meter
By using the third equation of motion, we get
v² - u² = 2as
0 - (27.1)² = 2 × a × 70.4
Or, acceleration(a) = -5.21 m/s²
Now, by the first equation of motion,
v = u + at
Or, - u = at
-27.11 = -5.21 × t
Or, t = 27.11/5.21
t = 5.2 sec
The stopping time (t) is 5.2 sec
Hence, the average acceleration is -5.21 m/s², and the time is 5.2 sec.
To learn more about acceleration and time, visit: brainly.com/question/25007278
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Object is stopped when Velocity is equal to zero
