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3 years ago

How much potential energy does a 40n block medicine ball gain when it is lifted 5.m

2 answers:

5 0

By definition, the gain in PE (potential energy) is
ΔPE = m*g*h

Given:
mg = 40 N (Note that m*g = weight)
h = 5 m

ΔPE = (40 N)*(5 m) =200 J

Answer: 200 J

Ksivusya [100]3 years ago

5 0

The answer is 200J I hope it helps ^_^

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Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4

Zepler [3.9K]

Answer: $5.214(10)^{11} N$

Explanation:

According to Coulomb's Law:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them".

Mathematically this law is written as:

$F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}$

Where:

$F_{E}$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}=5.8 C$ and $q_{2}=6.4 C$ are the electric charges

$d=80 cm \frac{1 m}{100 cm}=0.8 m$ is the separation distance between the charges

Solving:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}$

$F_{E}=5.214(10)^{11} N$

7 0

4 years ago

In longitudinal waves the places where the coils are bunched together are called *

Firdavs [7]

In longitudinal waves the places where the coils are bunched together are called *
Compressions

3 0

3 years ago

A green croquet ball of mass 0.50 kg is rolling at +12 m/s. It collides with a blue croquet ball that also

Svetach [21]

Answer:

a) 9.6 m/s

b) 11.7 m/s

c) 12 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{g} V_{o} + m_{b} U_{o}$ (2)

$p_{f}=m_{g} V_{f} + m_{b} U_{f}$ (3)

$m_{g}=0.5 kg$ is the mass of green ball

$m_{b}=0.5 kg$ is the mass of the blue ball

$V_{o}=12 m/s$ is the initial velocity of the green ball

$U_{o}=0 m/s$ is the initial velocity of the blue ball

$V_{f}$ is the final velocity of the green ball

$U_{f}$ is the final velocity of the blue ball

Substituting (2) and (3) in (1):

$m_{g} V_{o} + m_{b} U_{o}=m_{g} V_{f} + m_{b} U_{f}$ (4)

Isolating $U_{f}$ :

$U_{f}=\frac{m_{g} V_{o} - m_{g} V_{f}}{m_{b}}$ (5)

$U_{f}=\frac{m_{g} (V_{o} - V_{f})}{m_{b}}$ (6) This is the equation we will use for the next cases

Knowing this, let's begin with the answers:

a) In this case $V_{f}=2.4 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 2.4 m/s)}{0.5 kg}$ (7)

$U_{f}=9.6 m/s$ (8)

b) In this case $V_{f}=0.3 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 0.3 m/s)}{0.5 kg}$ (9)

$U_{f}=11.7 m/s$ (10)

c) In this case $V_{f}=0 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 0 m/s)}{0.5 kg}$ (11)

$U_{f}=12 m/s$ (12)

4 0

3 years ago

A gymnast whose weight is 5 10 N hangs from the middle of a bar supported by two vertical strands of rope. What is the tension i

german

A 765 N because when the gravitational pull of the weight balances hypothetically in the range of the atmosphere, your weight is not doubles but the force extends it by an uprange of up to 150%

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3 years ago

When two oceanic plate boundaries meet, this crustal feature forms?

fgiga [73]

The correct answer is B

4 0

3 years ago