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2 years ago

Freeeeeeeeeeeeeeeeeeeeeeeee points

Physics

2 answers:

Klio2033 [76]2 years ago

5 0

Thanks!!!!!!!!!!!!!!!

andreyandreev [35.5K]2 years ago

3 0

Answer:

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A factory has 1200 workers of which 720 are male and the rast are female what percent of workers are female

umka21 [38]

Answer:

Percent of Female Workers = 40%

Explanation:

The percentage of the female workers in the given group of workers can be easily found by the following formula:

$Percent\ of\ Female\ Workers = \frac{No.\ of\ Female\ Workers}{Total\ No.\ of\ Workers}\ x\ 100\%$

where,

Total No. of Workers = 1200

No. of Female Workers = Total Workers - No. of Male Workers

No. of Female Workers = 1200 - 720 = 480

Therefore,

$Percent\ of\ Female\ Workers = \frac{480}{1200}\ x\ 100\%$

<u>Percent of Female Workers = 40%</u>

7 0

2 years ago

What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?

Artist 52 [7]

Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

$U=k\frac{q_1q_2}{r}$ (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

$U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J$

Hence, the energy between charges is 0.413 J

3 0

2 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .