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3 years ago

Help me please

Physics

2 answers:

vodka [1.7K]3 years ago

7 0

#4 and #9 are true.
#10 is true IF it's an atom of Hydrogen.
The other 7 statements are false.

yarga [219]3 years ago

3 0

1. False
2.False
3.False
4.True
5. True
6. False
7.True
8.False
9. True
10. False

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Read 2 more answers

Paco pulls a 67 kg crate with 738 N and of force across a frictionless floor 9.0 M how much work does he do in moving the crate

olga_2 [115]

Answer:

W = 6642 J

Explanation:

Given that,

Mass of a crate, m = 67 kg

Force with which the crate is pulled, F = 738 N

It is moved 9 m across a frictionless floor

We need to find the work done in moving the crate. Let the work done is W. It is given by :

W = F d

W = 738 N × 9 m

= 6642 J

So, the work done is 6642 J.

5 0

3 years ago

A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed

GrogVix [38]

Answer:

a. $\tau = 2.1161 s$

b. $V(18.8 \ s) = 0.0256 \ V$

Explanation:

The equation for the voltage V of discharging capacitor in an RC circuit at time t is:

$V(t) = V_0 e^{(- \frac{t}{\tau}) }$

where $V_0$ is the initial voltage, and $\tau$ is the time constant.

For our problem, we know

$V_0 = 185 \ V$

and

$V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }$

$ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })$

$- \frac{10 \ s}{\tau} = ln (\frac{1.64 \ V}{ 185 \ V })$

$\tau = \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }$

This gives us

$\tau = 2.1161 s$

and this is the time constant.

At t = 18.8 s we got:

$V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 0.0256 \ V$

4 0

3 years ago