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2 years ago

15

How many joules of kinetic energy does a pendulum have when it has 100 joules of potential energy

2 answers:

Juliette [100K]2 years ago

6 0

Answer :100J of KE.

Explanation:

zloy xaker [14]2 years ago

4 0

Answer:

The maximum kinetic energy is 100 j.

Explanation:

<h3>The kinetic energy = (potential energy) + (kinetic energy) and the potential energy of 0 J implying its kinetic energy is 100 J, which is its maximum. </h3>

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Lithium has 3 protons. how many neutrons are in the isotope lithium -8

Oksi-84 [34.3K]

Answer:

5 neutrons

Explanation:

Both protons and neutrons have a mass of about 1 atomic mass unit (amu). The number in the notation for an isotope is its mass in amu. That means the number is essentially the sum of protons and neutrons in the isotope (electrons don't contribute significantly to the mass).

If lithium has 3 protons, lithium-8 has 8 - 3 = 5 neutrons.

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2 years ago

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Two objects, with different sizes, masses, and temperatures, are placed in thermal contact. in which direction does the energy t

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<span>(c) energy travels from the object at higher temperature
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2 years ago

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

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3 years ago

If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m

algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

$H=ut+\dfrac{1}{2}at^2$

Here, a = g

$H=\dfrac{1}{2}gt^2$

$H=\dfrac{1}{2}\times 9.8\times (2.261)^2$

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So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

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