Answer:
A. 200 J
Explanation:
The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.
Answer:42.4m/s^2
Explanation:
Velocity(v)=6m/s
Radius(r)=0.85 meter
Centripetal acceleration=(v x v) ➗ r
Centripetal acceleration=(6 x 6) ➗ 0.85
Centripetal acceleration=36 ➗ 0.85
Centripetal acceleration=42.4
Answer:
f.The period is independent of the suspended mass.
Explanation:
The period of a pendulum is given by

where
L is the length of the pendulum
g is the acceleration due to gravity
From the formula, we see that:
1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length
2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation
So, the only correct statements are
f.The period is independent of the suspended mass.
Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.
Answer:
The image distance is 30 cm
image height = - 5 cm
Explanation:
The formula for calculating the image distance is expressed as
1/f = 1/u + 1/v
where
f is the focal length
u is the object distance
v is the image distance
From the information given,
u = 30
f = 15
By substituting these values into the formula,
1/15 = 1/30 + 1/v
1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30
Taking the reciprocal of both sides,
v = 30
The image distance is 30 cm
magnification = image height/object height = - v/u
Given that object height = 5 cm, then
image height/5 = - 30/30 = - 1
image height = - 5 * 1
image height = - 5 cm
Answer:
true
Explanation:
a wheelbarrow has its load situated between the fulcrum and the force the wheel Barrow is 2nd class because of its resistance between the force and the axis