Answer:
Velocity of the ping pong ball must be = V2= 6,035.34m/s
Explanation:
M1= momentum of the bowling ball
m1 = mass of the bowling ball= 5.8kg
v1= velocity of the bowling ball= 1.59m/s
M2= momentum of the ping pong ball
m2= mass of the ping pong ball= 1.528 g/1000= 0.001528kg
v2= velocity of the ping pong ball
Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s
Momentum of the ping pong ball = M2= M1= m2v2
= 0.001528 *v2= 9.222
v2= 9.222/0.001528= 6,035.34 m/s
Answer:
The distance traveled by the woman is 34.1m
Explanation:
Given
The initial height of the cliff
yo = 45m final, positition y = 0m bottom of the cliff
y = yo + ut -1/2gt²
u = 20.0m/s initial speed
g = 9.80m/s²
0 = 45.0 + 20×t –1/2×9.8×t²
0 = 45 +20t –4.9t²
Solving quadratically or by using a calculator,
t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s
So this is the total time it takes for the ball to reach the ground from the height it was thrown.
The distance traveled by the woman is
s = vt
Given the speed of the woman v = 6.00m/s
Therefore
s = 6.00×5.69 = 34.14m
Approximately 34.1m to 3 significant figures.
Answer:
V=Bh
Explanation:
B h is used for rectangular solids and cylinders
Answer:
F = 2.6692 x 10⁻⁹ N
Explanation:
Given,
The mass of the rock, m = 10 kg
The mass of the boulder, M = 100 kg
The distance between them, d = 5 m
The gravitational force between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by the formula
<em> F = GMm/d² newton</em>
Where,
G - Universal gravitational constant
Substituting the given values,
F = 6.673 x 10⁻¹¹ x 100 x 10 / 5²
F = 2.6692 X 10⁻⁹ N
Hence, the force between the two bodies is, F = 2.6692 X 10⁻⁹ N
When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
To find the answer, we have to know about the rules followed by drawing ray-diagram.
<h3>What are the rules obeyed by light rays?</h3>
- If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
- The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
- The refracted ray follows the same path if the incident light passes through the center of the curve.
Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
Learn more about refraction by a lens here:
brainly.com/question/13095658
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