Answer:
Explanation:
There are two hypotheses she could test:
A cat's heart rate changes while it is napping.
A cat's heart rate does not change while it is napping.
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
Answer:
Correct answer: C. 50 cm
Explanation:
Given data:
The distance of the object from the top of the concave mirror o = 50.0 cm
The magnitude of the concave mirror focal length 25.0 cm.
Required : Image distance d = ?
If we know the focal length we can calculate the center of the curve of the mirror
r = 2 · f = 2 · 25 = 50 cm
If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.
We conclude that the image distance is 50 cm.
We will now prove this using the formula:
1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50
1/d = 1/50 => d = 50 cm
God is with you!!!
Answer : The momentum of ball is, 15 kg.m/s
Explanation :
Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.
Formula of momentum is:
where,
p = momentum = ?
m = mass = 1.5 kg
v = velocity = 10 m/s
Now put all the given values in the above formula, we get:
Therefore, the momentum of ball is 15 kg.m/s
Explanation:
Let us assume that the mass of a pitched ball is 0.145 kg.
Initial velocity of the pitched ball, u = 47.5 m/s
Final speed of the ball, v = -51.5 m/s (in opposite direction)
We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :
So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.
Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :
Hence, this is the required solution.
