A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after
1 answer:
<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
Explanation:
<h2>Given:</h2>m = 1,250 kg
= 30 km/hr
F = 1,700 N
t = 10 s
<h2>Required:</h2>Final velocity
<h2>Equation:</h2><h3>Force</h3>F = ma
where: F - force
m - mass
a - acceleration
<h3>Acceleration</h3>a =
where: a - acceleration
- initial velocity
- final velocity
t - time elapsed
<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>F = ma
Substitute the value of F and m
(1700 N) = (1250 kg)(a)
a =
a = 1.36 m/s²
<h3>Solve for final velocity using the formula for acceleration</h3>- Convert 30 km/hr to m/s
=
=
- Substitute the value of a,
and t
a =
- Convert to km/hr
=
=
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Answer:
total width bandwidth = 8kHz
Explanation:
given data
transmitter operating = 3.9 MHz
frequencies up to = 4 kHz
solution
we get here upper side frequencies that is
upper side frequencies = 3.9 × + 4 × 10³
upper side frequencies = 3.904 MHz
and
now we get lower side frequencies that is
lower side frequencies = 3.9 × - 4 × 10³
lower side frequencies = 3.896 MHz
and now we get total width bandwidth
total width bandwidth = upper side frequencies - lower side frequencies
total width bandwidth = 8kHz