Answer:
the net force applied to the car is zero.
Explanation:
According to Newton's second law, the acceleration of an object (a) is directly proportional to the net force applied (F):
where m is the object's mass.
In this problem, the car is moving with constant velocity: this means that the acceleration is zero, a = 0. Therefore, according to the previous equation, the net force must also be zero: F = 0. So, the correct answer is
the net force applied to the car is zero.
Answer:
0.021 V
Explanation:
The average induced emf (E) can be calculated usgin the Faraday's Law:
<u>Where:</u>
<em>N = is the number of turns = 1 </em>
<em>ΔΦ = ΔB*A </em>
<em>Δt = is the time = 0.3 s </em>
<em>A = is the loop of wire area = πr² = πd²/4 </em>
<em>ΔB: is the magnetic field = (0 - 1.04) T </em>
Hence the average induced emf is:
Therefore, the average induced emf is 0.021 V.
I hope it helps you!
Answer:
Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.
Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.
The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.
Hope that helps.
Explanation:
Answer:
a) 4.485 kg b) 3.94 kg
Explanation:
since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .
44 = m * 9.81m/s^2
m = 44/9.81 = 4.485kg
b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)
44 = m (a +g)
44 = m (1.37 + 9.81)
44/11.18 = m
m = 3.94 kg
<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B
<em>I hope this helps you </em>
