Answer:
Your answer will be 6.0kg•m/s
Explanation:
In the given question all the required details d given. Using these information's a person can easily find the momentum of the object. In the question it is already given that the mass of the object is 5 kg and the velocity at which it is traveling is 1.2 m/s.We know the equation of finding momentum asMomentum = mass * velocity = 5 * 1.2 = 6So the momentum of the object is 6 Newton.
The distance travelled during the given time can be found out by using the equations of motion.
The distance traveled during the time interval is "13810.8 m".
First, we will find the deceleration of the motorcycle by using the first <em>equation of motion</em>:
where,
vi = initial velocity = (518 km/h)
= 143.89 m/s
vf = final veocity = 60 % of 143.89 m/s = (0.6)(143.89 m/s) = 86.33 m/s
a = deceleration = ?
t =time interval = 2 min = 120 s
Therefore,
a = -0.48 m/s²
Now, we will use the second <em>equation of motion </em>to find out the distance traveled (s):
<u>s = 13810.8 m = 13.81 km</u>
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Learn more about the equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion.
The kinetic energy is 945 joules.
Kinetic energy is the energy that an object has as a result of motion. It is defined as the effort required to accelerate a mass-determined body from rest to the indicated velocity.
The speed of an object or particle, which is a scalar quantity, is the size of the change in its location over time or the size of the change in its position per unit of time.
The mass of the volleyball is 2.1 kg.
The speed of the ball when the ball leaves the hand is 30 m/s.
m = 2.1 kg
v = 30 m/s
The kinetic energy of an object is given as:
KE = (1/2 ) × m × v²
KE = (1 / 2) × 2.1 kg × ( 30 m/s)²
KE = (1 / 2) × 2.1 kg × 30 m/s × 30 m/s
KE = 2.1 kg × 15 m/s × 30 m/s
KE = 945 J
Learn more about kinetic energy here:
brainly.com/question/8101588
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Explanation:
(a) E = F/q
E = 4.8×10^-17/1.6×10^-19
E = 300 N/C
(b) same magnitude of electric field is exerted on proton
