Answer:
oxygen silicon aluminun iron
Answer:
Object could only be moving with increasing speed.
Explanation:
Let us consider the general formula of acceleration:
a = (Vf - Vi)/t
Vf = Vi + at -------- equation 1
where,
Vf = Final Velocity
Vi = Initial Velocity
a = acceleration
t = time
<u>FOR POSITIVE ACCELERATION:</u>
Vf = Vi + at
since, both acceleration and time are positive quantities. Hence, it means that the final velocity of the object shall be greater than the initial velocity of the object.
Vf > Vi
It clearly shows that if an object moves with positive acceleration. <u>It could only be moving with increasing speed.</u>
Solving the same equation for negative acceleration shows that the final velocity will be less than initial velocity and object will be moving with decreasing speed.
And for the constant velocity final and initial velocities are equal and thus, acceleration will be zero.
Answer:
15.7m/s
Explanation:
To solve this problem, we use the right motion equation.
Here, we have been given the height through which the ball drops;
Height of drop = 14.5m - 1.9m = 12.6m
The right motion equation is;
V² = U² + 2gh
V is the final velocity
U is the initial velocity = 0
g is the acceleration due to gravity = 9.8m/s²
h is the height
Now insert the parameters and solve;
V² = 0² + 2 x 9.8 x 12.6
V² = 246.96
V = √246.96 = 15.7m/s
Answer:
No, not at all. This kinetic energy of car is converted/lost in the form of sound and heat under the tyre when driver apply brakes.
Explanation:
This law states that energy can neither be created nor destroyed but it changes from one form to the other.
In this case reduction of kinetic energy of car does not contradict the law but it changes into heat and sound energy. When driver apply brakes we hear the sound of Tyre due to friction with the road. The tyre become hot too.
So it is not in contradiction to the law of conservation of energy (First Law of Thermodynamics).
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Answer:
Wl = 1740 N
Explanation:
maximum lift weight unaided = force exerted (F) = 650 N
length of the wheelbarrow (L) = 1.4 m
weight of the wheelbarrow (w) = 80 N
distance of center of gravity of the wheel barrow from the wheel = 0.5 m
distance of center of gravity of the load from the wheel = 0.5 m
find the weight of the load (Wl)
from the diagram attached we can see that there is going to be a rotation about the axis A. let clockwise rotation be positive
ΣT = (F x 1.4) - ((Wl x 0.5) + (w x 0.5) = 0
(F x 1.4) = ((Wl x 0.5) + (w x 0.5)
Wl =
Wl =
Wl = 1740 N