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mixer [17]
3 years ago
12

A 2-kg bowling ball is 1 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 4.4 m/

s. Assuming that there is no air resistance, which statement is true?
The mechanical energy is not conserved.

The mechanical energy is conserved.

The intial potential energy is greater than the final kinetic energy.

The initial potential energy is less than the final kinetic energy.
Physics
2 answers:
elena55 [62]3 years ago
5 0

Answer:The correct answer is 'The mechanical energy is conserved'.

Explanation:

Mechanical energy of the body is defined a sum of potential energy and kinetic energy possessed by the body in motion. It is the energy associated with position and motion of the body.

Mechanical energy = Potential energy + Kinetic Energy

The principle of Conservation of Mechanical Energy states that the mechanical energy of the body remains constant as long as the forces acting on the body are conservative forces.

From, this we can say that mechanical energy of the bowling ball with mass of 2 kg falling at speed of 4.4 m/s will remain conserved.

Ira Lisetskai [31]3 years ago
3 0

The mechanical energy is conserved.

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3 years ago
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
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