Answer:
The average speed would be 279.167 meters per second.
Explanation:
Speed is the magnitude that expresses the variation in position of an object and as a function of time. In other words, speed is defined as the relationship established between the distance traveled and the time elapsed in the movement of a mobile.
Speed can be calculated using the following expression:
In this case:
- distance= 7,035 km
- time= 7 hours
Since the required unit of speed is meter per second, then you convert from km to meters for the case of distance, and from hours to seconds for the case of time. So:
- distance= 7,035 km= 7,035,000 m (being 1 km= 1,000 m)
- time= 7 hours= 420 minutes= 25,200 seconds (being 1 hour=60 minutes and 1 minute=60 seconds)
Replacing in the definition of speed:
Solving:
Then, <u><em>the average speed would be 279.167 meters per second.</em></u>
20,000,000 kg*m/s
p=mv
10,000,000=m(v/2)
20,000,000=mv
Answer:
the kinetic energy of A will be twice that of B
Explanation:
The formula for calculating kinetic energy is expressed using the formula
KE = 1/2mv²
m is the mass of the object
v is the velocity
For object A:
KEA = 1/2mAvA²
For object B:
KEB = 1/2mBvB² ... 1
If object A is twice the mass of object B, then mA = 2mB
From 1:
KEA = 1/2mAvA²
Substitute mA = 2mB
KEA = 1/2(2mB)vA² .... 2
Divide 1 by 2
KEB/KEA = 1/2mBvB²/mBvA²
KEB/KEA = 1/2vB²/vA²
Assuming they have the same velocities then vA ,= VB
The equation becomes:
KEB/KEA = 1/2vB²/vA²
KEB/KEA = 1/2vB²/vB²
KEB/KEA = 1/2
KEA = 2EB
Hence the kinetic energy of A will be twice that of B
We can solve both parts of the problem by using the lens equation:
where
f is the focal length
is the distance of the object from the lens
is the distance of the image from the lens
Let's also keep in mind that for a converging lens, the focal length is taken as positive, so f=+19.0 cm.
(a) In this case,
. So, the lens equation becomes
And so the distance of the image from the lens is
where the positive sign means the image is real, i.e. is located behind the lens (opposite side of the real object)
(b) In this case,
. So, the lens equation becomes
And so the distance of the image from the lens is
where the negative sign means the image is virtual, i.e. is located in front of the lens (i.e. on the same side of the real object)