Answer:
a = 2.72 [m/s2]
Explanation:
To solve this problem we must use the following kinematics equation:

where:
Vf = final velocity = 1200 [km/h]
Vo = initial velocity = 25 [km/h]
t = time = 2 [min] = 2/60 = 0.0333 [h]
1200 = 25 + (a*0.0333)
a = 35250.35 [km/h2]
if we convert these units to units of meters per second squared
![35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]](https://tex.z-dn.net/?f=35250.35%5B%5Cfrac%7Bkm%7D%7Bh%5E%7B2%7D%20%7D%5D%2A%28%5Cfrac%7B1%7D%7B3600%5E%7B2%7D%20%7D%29%2A%5B%5Cfrac%7Bh%5E%7B2%7D%20%7D%7Bs%5E%7B2%7D%20%7D%20%5D%2A%28%5Cfrac%7B1000%7D%7B1%7D%20%29%2A%5B%5Cfrac%7Bm%7D%7Bkm%7D%20%5D%20%3D%202.72%20%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D)
Answer:
<u>Magnitude</u>
Explanation:
Each value in nature has a number part, called its magnitude and a dimension called its unit.
For example,
The length of an object is 10 cm. It means that 10 shows the magnitude of length and cm shows its unit.
Answer:
Zero work done,since the body isn't acting against or by gravity.
Explanation:
Gravitational force is usually considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.
Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.
The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.
Answer:
P = 0.25 W
Explanation:
Given that,
The emf of the battry, E = 2 V
The resistance of a bulb, R = 16 ohms
We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

So, 0.25 W power is delivered to the bulb.