70 POINTS and BRAINLIEST! Which segment represents freezing? E-D D-C C-B B-A

Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t

lorasvet [3.4K]

Answer:

$V_3\approx 4.28\,\,V$

$I_1=0.0572\,\,amps$

$I_3\approx 0.171\,\,amps$

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

$\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega$

By knowing this, we can estimate the total current through the circuit,:

$Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps$

So approximately 0.17 amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

$V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V$

So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

$I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps$

4 0

4 years ago

Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails

Inessa [10]

Answer: 6.48m/s

Explanation:

First, we know that Impulse = change in momentum

Initial velocity, u = 19.8m/s

Let,

Velocity after first collision = x m/s

Velocity after second collision = y m/s

Also, we know that

Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).

5700 = 1500(19.8 - x)

5700 = 29700 - 1500x

1500x = 29700 - 5700

1500x = 24000

x = 24000/1500

x = 16m/s

Also, at the second guard rail. impulse = ft, so that

Impulse = 79000 * 0.12

Impulse = 9480

This makes us have

Impulse = m(x - y)

9480 = 1500(16 -y)

9480 = 24000 - 1500y

1500y = 24000 - 9480

1500y = 14520

y = 14520 / 1500

y = 9.68

Then, the velocity decreases by 3.2, so that the final velocity of the car is

9.68 - 3.2 = 6.48m/s

5 0

3 years ago

CAN SOMEONE PLZ HELP 5,6,7,8,9 plzzz show work too I’m not understanding this much appreciated

soldi70 [24.7K]

#5 is supposed to have a little graph showing for each choice. They're not there. Nothing to choose from.

6-C

7-A

8-B

9-D

7 0

4 years ago

Alexandra drops an egg from 30 m above the ground to hit Vanessa on the head. If Vanessa stands 3.5 meters tall. How fast is the

maksim [4K]

This is the problem. It's worth saying that I'm working with the magnitudes and their respective scalar equations. That is, I'm not using vectors.
v_e_g_g_,_i = 0m/s
g = 9.81m/s^2
h_i = 30m
h_f = 3.5m
v_e_g_g_,_f = ?
t = ?
|h_f-h_i| = v_e_g_g_,_i*t + (g*t^2)/2
26.5m = 0m/s*t + ((9.81m/s^2)*(t^2))/2
t = \sqrt{2*26.5m/(9.81m/s^2)} = 2.32 s
v_e_g_g_,_f = v_e_g_g_,_i + g*t
v_e_g_g_,_f = 0m/s+(9.81m/s^2)*(2.32 s)
v_e_g_g_,_f = 22.76 m/s

6 0

3 years ago

A +0.0129 C charge feels a 4110 N

Otrada [13]

Answer:

r = 14.13 m

Explanation:

Given that,

Charge 1 = +0.0129 C

Charge 2 = -0.00707 C

The force between charges, F = 4110 N

We need to find the distance between charges. Let the distance be r. The force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\4110=\dfrac{9\times 10^9\times 0.0129\times 0.00707}{r^2}\\\\r=\sqrt{\dfrac{9\times 10^9\times 0.0129\times 0.00707}{4110}} \\\\r=14.13\ m$

So, the distance between charges is equal to 14.13 m.

6 0

3 years ago

70 POINTS and BRAINLIEST! Which segment represents freezing? E-D D-CC-B B-A​

70 POINTS and BRAINLIEST! Which segment represents freezing? E-D D-CC-B B-A