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torisob [31]
2 years ago
9

1) Calculate the potential energy of a 5.00 kg object sitting on a 3.00 meter high ledge.

Physics
1 answer:
maria [59]2 years ago
5 0

Answer:

15kg

Explanation:

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When is the kinetic energy of the ball zero and when is it at its highest? When is its potential energy at its lowest and at its
liubo4ka [24]

Answer:

if there is only one planet in the universe and the ball is there it will have 0 kinetic energy if the ball is in the very center of that planet only if the planet itself is absolutely motionless. its at its highest if the planet is moving away from the ball at a slightly faster speed forever. Between point A and B both potential energy and kinetic energy are at perfect 0.

Explanation:

never will have a measurable kinetic or potential energy status unless every single object is included in the calculation.

3 0
2 years ago
A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar
BlackZzzverrR [31]

Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)

q2 = 4.507 × 10^-8 C

8 0
2 years ago
Which of the following is a consequence of the special theory of relativity?
aleksandr82 [10.1K]

Answer:

D.

Explanation:

Specifically, Special Relativity showed us that space and time are not independent of one another but can be mixed into each other and therefore must be considered as the same object, which we shall denote as space-time. The consequences of space/time mixing are: time dilation. and length contraction.

3 0
2 years ago
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

5 0
1 year ago
If the same pressure is exerted over a greater area will more of less force result? Group of answer choices :
Zigmanuir [339]

Answer:

More force

Explanation:

Pressure and force are related by the equation:

p=\frac{F}{A}

where

p is the pressure

F is the force

A is the area

We can re-arrange the equation as

F=pA

In this problem, the pressure is kept the same (p' = p) while the area is increased. As we can see from the previous equation, the force applied is directly proportional to the area: therefore, a greater area means also a greater force.

8 0
3 years ago
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