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2 years ago

9

Calculate the force between two objects that have masses of 75. Kilograms and 2,100. Kilograms. Their centers of gravity are sep

arated by a distance of 1.00 meter.

2 answers:

iris [78.8K]2 years ago

4 0

Force = gravity x mass1 x mass2 / distance^2

Force = (6.67x10 ^-11) x (75) x (2100) / 1^2

Force = 1.050525 x 10 ^-5 Newtons

scoundrel [369]2 years ago

3 0

Answer:

use the formula above it's easy

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Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.

Andrei [34K]

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

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1 year ago

The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Colum

TEA [102]

"The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Column 2 would be on the y-axis."

typical plot is speed or velocity on the y-axis n time on the x-axis so the ans is Column 1 should be titled “Time,” and Column 2 should be titled “Velocity.”

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3 years ago

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The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit

Helga [31]

Answer:

$3.4\cdot 10^{-19} J$

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

$1 eV = 1.6 \cdot 10^{-19} J$

In our problem, the work function of cesium is

$E=2.1 eV$

so, we can convert it into Joules by using the following proportion:

$1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J$

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3 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

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3 years ago

_h2o2(aq)-_h2o(l) +o2(g)

zvonat [6]

Answer:

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Explanation:

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2 years ago