Answer:
initial velocity (u) = 200m/s
final velocity (v) = 350 m/s
time (t) = 15s
acceleration (a) = ?
NOW,
a=v-u/t
a= 350-200/15
a= 50/15
a= 3.3333
Explanation:
it's too easy just u need to understand the question . and go according to it's content .
main thing to memorize is it's simple formula.
I HAVE SOLVE THIS QNA. IN VERY
SIMPLE AND UNDERSTANDABLE FORM.
I høpë u hađ uņdērstøöď ťhìs şølutîóñ
:verý ×wəłł.
This problem uses the relationships among current
I, current density
J, and drift speed
vd. We are given the total of electrons that pass through the wire in
t = 3s and the area
A, so we use the following equation to to find
vd, from
J and the known electron density
n,
so:

<span>The current
I is any motion of charge from one region to another, so this is given by:
</span>

The magnitude of the current density is:

Being:

<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
Answer:
0.008 Kg
Explanation:
Using the equation of Moment of Inertia about an axis passing through the centre of mass of the rod and perpendicular to it's length

Where, M is the mass of the rod in Kg and L is the length of the rod in meter.


Explanation:
In first case, the forces on LHS and on RHS is the same i.e. 3 N. The force acting on the car is balanced force. As a result, the car will not move at all.
In second case,
Force on RHS = 2000 N
Force on LHS = -6000 N
Net force acting on it is given by :
F = 2000+(-6000)
= -4000 N
Hence, this is the required solution.
Answer:
1.08 m/s
Explanation:
This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.
Time taken to fall 9.5 m
vertical acceleration = a = 9.8 m/s^2.
vertical velocity = 0, (since there is only horizontal component for velocity,
)
distance traveled s = 9.5 m.
Substituting these values in the equation



⇒ t= 1.392 sec
Velocity needed
We know the time taken (1.392 s) to travel 1.5 m,
So velocity = 1.5 m / 1.392 s = 1.08 m/s
hence velocity of the diver must be at least 1.08 m/s