Answer:
-5.8868501529 m/s² or -5.8868501529g
0.118909090909 s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Dividing by g
The acceleration is -5.8868501529 m/s² or -5.8868501529g
The time taken is 0.118909090909 s
The correct option is (b) 5n
As a result, there is a net downward force of 5N operating on the object.
The resultant force is the force that results from adding the vector sums of all the forces operating on an item. The combined action of all the acting forces on the object produces the same effect as the resulting force. When determining the resulting force, the direction of the forces must be taken into account.
Given;
The northward force is Fn = 10N
The southward force is Fs = 15N
Required;
The net force on the mobile phone is Fnet = ?N
The object's weight exerts downward pressure, and upward resistance exerts upward pressure. The vector sum of these two forces will be the net force.
Fnet = Fs - Fn (Considering the direction downward as positive)
Fnet= 15N - 10N
Fnet = 5N
As a result, there is a net downward force of 5 N operating on the object.
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The main formula to be used here is
Force = (mass) x (acceleration).
We'll get to work in just a second. But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them. So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.
On the <u>right</u> side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration. I don't know what to do with that, because . . .
At the <em>same time</em>, both of those will be pulled <u>up</u> by the 10 kg on the other side
of the upper pulley.
I think I can handle the 10 kg, and work out the acceleration that IT has.
Let's look at only the forces on the 10 kg:
-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.
-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope.
So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.
The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.
The acceleration of 10 kg, with 49 newtons of force on it, is
Acceleration = (force) / (mass) = 49/10 = <em>4.9 meters per second²</em>
a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
from above statement we got
height = 78.4 m
since the ball is thrown, so its vertical velocity would be zero
u = 0
taking g = 9.8m/s^2
now, using the equation of motion
h = ut + gt^2/2
now putting all the values in it
we got ,
78.4 = 9.8 * t^2/ 2
by solving we got,
t = 4 sec
b) now, since along the horizontal , no force acting and accelaration is zero so
R = ut , R is RANGE
R = 5 * 4
range = 20 m
c) vertical components of the stone’s velocity just before it hits the ground = v sin θ =
horizontal components of the stone’s velocity just before it hits the ground = v cos θ
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