A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c
1 answer:
Answer:
- After throwing the snow, velocity of the thrower is 2.33 m/s
- the velocity of the receiver is 0.026 m/s
Explanation:
Given the data in the question;
Using conservation of momentum,
Initial thrower has a momentum of mv; v
(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s
Now, When he throws it at 31.5 m/s, these constitutes a momentum of;
(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s
hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s
To get his velocity, we say;
161.94035 = mv
{ he lost weight of the snow ball so, m = 69.5 kg )
161.94035 = 69.5 × v
v = 161.94035 / 69.5
v = 2.33 m/s
Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s
Next is the Receiver;
the receiver will gain momentum of 1.49625 kg.m/s
he has no momentum initially and after he catches the snow ball;
1.49625 kg.m/s = mv
1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v
1.49625 kg.m/s = 57.5475 kg × v
v = ( 1.49625 kg.m/s ) / 57.5475 kg
v = 0.026 m/s
Therefore, the velocity of the receiver is 0.026 m/s
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Explanation:
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This relationship can be expressed as
where K is a constant of value 1.38*10^-23
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converting to degree we have
A) -48.0 cm
In order to find the focal length of the lens, we can use the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we have:
p = 16.0 cm
q = -12.0 cm (the negative sign is due to the fact that the image is on the same side as the object, so it is a virtual image, so the sign of q is negative)
Substituting, we find f:
B) Diverging
We have two types of lenses:
- A converging (convex) lens is curved outwards in its center, so that the incoming rays of light parallel to the principal axis are focused into the focus of the lens, on the opposite side
- A diverging (concave) lens is curved inwards in its center, so that the incoming rays of light parallel to the principar axis are deviated away from the principal axis, and they appear to come all from the focal point of the length on the same side of the object
A converging lens is identified by a positive focal length, while the focal length in a diverging lens is negative. Here, f = -48.0 cm, so this is a diverging lens.
C) 6.38 mm
We can answer this part of the problem by using the magnification equation:
where
y' is the size of the image
y is the size of the object
Here we have:
y = 8.50 mm
p = 16.0 cm
q = -12.0 cm
So we find:
D) Erect
We can determine the orientation of the image by looking at the sign of the size of the image found in part C). In fact:
- if the image is erect, the sign of y' is positive
- if the image is inverted, the sign of y' is negative
In this situation, we see that
y' = 6.38 mm
Which is positive, so the image is erect.