Answer:
1.76m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 65m/s
Distance traveled = 1200m
Unknown:
Acceleration = ?
Solution:
This is linear velocity and we apply the appropriate motion equation to solve this problem.
V² = U² + 2as
S is the distance
u is the initial velocity
V is the final velocity
a is the acceleration
Now, insert the parameters and solve;
65² = 0² + 2 x a x 1200
4225 = 2400a
a = 1.76m/s²
Answer
given,
mass = 100 kg
acceleration = 10 m/s²
A mass 20 kg slides over 100 kg block
acceleration = 3 m/s²
horizontal friction exerted by the 100 kg block on 20 kg
using newton's second law
F - f = 0
F = f
f = ma
f = 20 × 3
f = 60 N
now net force acting on the 100 kg block
F_net = m a
F_net = 100 x 10
F_net = 1000 N
after 20 kg block falls the acceleration of the bock
F = 1000 +60
F = 1060 N
acceleartion on the block


a = 10.60 m/s²
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Answer:
The beam of light is moving at the peed of:
km/min
Given:
Distance from the isalnd, d = 3 km
No. of revolutions per minute, n = 4
Solution:
Angular velocity,
(1)
Now, in the right angle in the given fig.:

Now, differentiating both the sides w.r.t t:

Applying chain rule:


Now, using
and y = 1 in the above eqn, we get:

Also, using eqn (1),


Answer:

Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by
on q. Then we can know the magnitude of the force exerted by
about q, finally this will allow us to know the magnitude of 
exerts a force on q in +y direction, and
exerts a force on q in -y direction.

The net force on q is:

Rewriting for
:
