Acceleration=force/mass=28/(10+4)=2m/s^2
force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block
vf=vi+at
-10=20*28/14 * t
t=30/2=15sec
i hope this can help you.
Answer:
Object 3 has greatest acceleration.
Explanation:
Objects Mass Force
1 10 kg 4 N
2 100 grams 20 N
3 10 grams 4 N
4 1 kg 20 N
Acceleration of object 1,

Acceleration of object 2,

Acceleration of object 3,

Acceleration of object 4,

It is clear that the acceleration of object 3 is
and it is greatest of all. So, the correct option is (3).
Kate's recollection of these different events along her life best exemplifies the use of her episodic memory.
<h3>What is episodic memory?</h3>
The term episodic memory makes reference to conscious personal background experiences that were collected along life.
Episodic memory is also defined as the collection of all life-day experiences collected by a person.
The episodic memory may be, for example, the first day when a person drove his/her car or a bike.
Learn more about episodic memory here:
brainly.com/question/25040884
Answer:
a) 2.33 m/s
b) 5.21 m/s
c) 882 m³
Explanation:
Using the concept of continuity equation
for flow through pipes

Where,
A = Area of cross-section
V = Velocity of fluid at the particular cross-section
given:


a) 
substituting the values in the continuity equation, we get

or

or

b) 
substituting the values in the continuity equation, we get

or

or

c) we have,
Discharge
thus from the given value, we get


Also,
Discharge
given time = 1 hour = 1 ×3600 seconds
substituting the value of discharge and time in the above equation, we get

or

volume of flow = 
Answer: b the energy of light...
Explanation: