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2 years ago

Which has more kinetic

Physics

2 answers:

ikadub [295]2 years ago

4 0

Answer: a bowling ball

Explanation: a baseball doesn’t need as much kinetic energy because it’s smaller

If a bowling ball is rolling at the same speed as a small baseball then since the bowling ball is bigger it has more kinetic energy

kenny6666 [7]2 years ago

4 0

Answer: A Bowling Ball

Explanation: A baseball doesn’t require much kinetic energy because it is smaller than a bowling ball and much lighter.

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The significant feature of a Cepheid variable is that there is a relationship between two intrinsic parameters, one of which can

Ghella [55]

Answer:

Period of brightness variation and luminosity.

Explanation:

The Cepheid variables are used as distance indicators. This requires estimation of periods and (usually) intensity-mean magnitudes in order to establish a period—apparent luminosity relation. It is particularly important for the techniques employed to be as accurate and efficient as possible.

5 0

3 years ago

A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery

GuDViN [60]

Answer:

at the beginning: $2.3\cdot 10^{-10} F$

when the plates are pulled apart: $1.1\cdot 10^{-10} F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 \frac{A}{d}$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^{-3} m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F$

4 0

3 years ago

A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic

zhannawk [14.2K]

Answer:

The force is $F = 172 \ N$

Explanation:

From the question we are told that

The mass of the block is $m_b = 27.0 \ kg$

The coefficient of static friction is $\mu_s = 0.65$

The coefficient of kinetic friction is $\mu_k = 0.50$

The normal force acting on the block is

$N = m * g$

substituting values

$N = 27 * 9.8$

$N = 294.6 \ N$

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

$F_f = \mu_s * N$

substituting values

$F_f = 0.65 * 264.6$

$F_f = 172 \ N$

Now for this block to move the force require is equal to $F_f$ i.e

$F= F_f$

=> $F = 172 \ N$

5 0

3 years ago

Calculate the amount of heat (in kJ) required to convert 97.6 g of water to steam at 100° C. (The molar heat of vaporization of

Ratling [72]

Answer:

221.17 kJ

Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.

N = 97.6 ÷ 18

Q=molar heat *moles

Q = (40.79) * (97.6 ÷ 18)

This is approximately 221.17 kJ

4 0

3 years ago

How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10

Karo-lina-s [1.5K]

Answer:

$-2.00876\times 10^{-18}\ J$

Explanation:

v = Speed of electron = $2.1\times 10^6\ m/s$ (generally the order of magnitude is 6)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

Work done would be done by

$W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J$

The work required to stop the electron is $-2.00876\times 10^{-18}\ J$

6 0

3 years ago