Horse does 200 lb while pulling a sled for 3 mi .how much work does the horse do?
1 answer:
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Answer:
The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>
Explanation:
The directions in which the delivery truck travels are;
1) 2.8 km North = 2.8·, in vector form
2) 1.0 km East = 1.0·, in vector form
3) 1.6 km South = -1.6·, in vector form
Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;
Final displacement, Δd = 2.8· + 1.0·
+(-1.6·
) = 1.2·
+ 1.0·
Final displacement, = 1.0· + 1.2·
Where;
Δx = The displacement in the x-direction = 1.0·
Δy = The displacement in the y-direction = 1.2·
The magnitude of the resultant displacement vector is given as follows
= √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)
The magnitude of the resultant displacement vector ≈ 1.6 km
The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE
Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.
A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by
where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write
which can be rewritten as
The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):
So, we get:
We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B)
The frequency of revolution of a particle in uniform circular motion is
where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be: