All categories

3 years ago

Need help ??? Please

Physics

1 answer:

Yanka [14]3 years ago

8 0

Is there information in the previous question which relates to this one?

You might be interested in

A boy throws a stone straight upward with an initial speed of 15m/s. What maximum height will the stone reach before falling bac

Vitek1552 [10]

$height=\frac{velocity^2}{2gravity}\\\\
velocity=15\frac{m}{s}\\\\
g=9,8\frac{m}{s^2}\\\\height=\frac{15^2}{2*9,8}=\frac{225}{19,6}\\\\
\boxed{height=11,48\ meters}$

6 0

3 years ago

Where do you feel that you are traveling at the fastest speed when on the swing?

il63 [147K]

Answer:

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

7 0

3 years ago

Read 2 more answers

What is the average velocity of the person taking this walk?

Butoxors [25]

5 would be the answer

6 0

3 years ago

Read 2 more answers

What is the current in a 160V circuit if the resistance is 2Ω? V= I= R=

Alex787 [66]

Explanation:

v = IR

v= 160 R = 2

160 = 2I

2 2

I = 80A

4 0

2 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

3 years ago