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3 years ago

What is the electric force acting between two charges of −0.0085 C and −0.0025 C that are 0.0020 m apart?

Physics

2 answers:

dsp733 years ago

3 0

$F=k \frac{ Q_{1} Q_{2} }{ r^{2} }$
where k=8.99e9
So (8.99e9)(-0.0085)(-0.0025)/(0.002)^2=4.7759e10N

katen-ka-za [31]3 years ago

3 0

Answer:

4.8•10^10 N

Explanation:

(9.00•10^9)(-0.0085)(-0.0025)

————————————————= 4.8•10^10

0.0020^2

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An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of

IrinaVladis [17]

Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

$r_1$ = Distance between center of mass of person and pole = 0.45 m

$r_2$ = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

$\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s$

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

$L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s$

The new angular speed is 1.70344 rad/s

7 0

2 years ago

A sculpture is suspended in equilibrium by two cables, one from a wall and the other

aleksandrvk [35]

Answer:

$T_1=6655.295917 \approx 6655.3N$

Explanation:

From the question we are told that

Angle of cable 2 $\theta=37.0\textdegree$

Weight of sculpture $W=5000 N$

Generally the Tension from cable 2 $T_2$ is mathematically given by

$T_2sin37\textdegree=5000N$

$T_2=5000N/sin37\textdegree$

$T_2=8308.2N$

Generally the Tension from Cable 1 $T_1$ is mathematically given by

$T_1=T_2 cos37\textdegree$

$T_1=8308.2* cos 37\textdegree$

$T_1=6655.295917 \approx 6655.3N$

7 0

3 years ago

Which way does a river flow??

spin [16.1K]

Downwards - from uphill towards the lowlands and eventually into the sea.

8 0

2 years ago

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of grav

erastovalidia [21]

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = \frac{7.4}{cos(70)}$

$Fm = 21.636N$

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that $\beta = \theta$

3 0

2 years ago

Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so it's not doing anything but changing the units.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (we can elevate any conversion factor to any number and they still will be 1) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

2 years ago