Answer:
After 4 s of passing through the intersection, the train travels with 57.6 m/s
Solution:
As per the question:
Suppose the distance to the south of the crossing watching the east bound train be x = 70 m
Also, the east bound travels as a function of time and can be given as:
y(t) = 60t
Now,
To calculate the speed, z(t) of the train as it passes through the intersection:
Since, the road cross at right angles, thus by Pythagoras theorem:
Now, differentiate the above eqn w.r.t 't':
For t = 4 s:
Answer:
a = 2 m/s2
Explanation:
we know from newtons 2nd law
F = ma.
we also know that from hookes law we have
F = kx
equate both value of force to get value of acceleration
kx = ma,
where,
k is spring constant = 8.0 N/m
x is maximum displacement 0.10 m
m is mass of object 0.40 kg
a = \frac{kx}{m}
= \frac{8 *0 .10}{0.40}
a = 2 m/s2
Answer:
Force
Explanation:
The mass of an object is the quantity of matter it contains. It is measured in kilograms.
Acceleration is the ratio of the change in the velocity of an object to the change in time. It is measured in m/
.
When the mass of an object is multiplied with its acceleration, this gives the average force applied on the object. As force is defined as agent that can change the state of an object.
i.e F = m × a
where F is the force, m is the mass of the object and a its acceleration.
The two major classes of force are; contact force and field force.
The answer would be B.
(V=IR, 12-10i, 12/10=1.2)
Answer:
angular acceleration is -0.2063 rad/s²
Explanation:
Given data
mass m = 95.2 kg
radius r = 0.399 m
turning ω = 93 rpm
radial force N = 19.6 N
kinetic coefficient of friction μ = 0.2
to find out
angular acceleration
solution
we know frictional force that is = radial force × kinetic coefficient of friction
frictional force = 19.6 × 0.2
frictional force = 3.92 N
and
we know moment of inertia that is
γ = I ×α = frictional force × r
so
γ = 1/2 mr²α
α = -2f /mr
α = -2(3.92) /95.2 (0.399)
α = - 7.84 / 37.9848 = -0.2063
so angular acceleration is -0.2063 rad/s²
