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3 years ago

14

Pop up spring Lab

1 answer:

Nikitich [7]3 years ago

5 0

Answer:

I honestly don't know but that thing looks cool lol

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When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical

laila [671]

Answer:

I = 18 x 10⁻⁹ A = 18 nA

Explanation:

The current is defined as the flow of charge per unit time. Therefore,

I = q/t

where,

I = Average Current passing through nerve cell

q = Total flow of charges through nerve cell

t = time period of flow of charges

Here, in our case:

I = ?

q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C

t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s

Therefore,

I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)

<u>I = 18 x 10⁻⁹ A = 18 nA</u>

6 0

3 years ago

balu736 [363]

Answer:

B

Explanation:

7 0

2 years ago

Read 2 more answers

Apply a force of 50N to the left describe the motion of the box

GenaCL600 [577]

When a force is applied to the box , this will cause an acceleration to the box.
(force =mass×acceleration)

So the box has a constant acceleration and a changing velocity.

4 0

3 years ago

Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi

Snowcat [4.5K]

Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ $V=20\ \Omega,\ V_1=30\ \Omega$

⇒ $G=1680\ \Omega,\ G_1=2930\ \Omega$

Concept to be used:

Conversion of galvanometer into voltmeter.

Let $G$ be the resistance of the galvanometer and $I_g$ the maximum deflection in the galvanometer.

To measure maximum voltage resistance $R$ is connected in series .

So,

⇒ $V=I_g(R+G)$

We have to find the value of $R$ we know that in series circuit current are same.

For $G=1680$ For $G_1=2930$

⇒ $I_g=\frac{V}{R+G}$ equation (i) ⇒ $I_g=\frac{V_1}{R+G_1}$ equation (ii)

Equating both the above equations:

⇒ $\frac{V}{R+G} = \frac{V_1}{R+G_1}$

⇒ $V(R+ G_1) = V_1 (R+G)$

⇒ $VR+VG_1 = V_1R+V_1G$

⇒ $VR-V_1R = V_1G-VG_1$

⇒ $R(V-V_1) = V_1G-VG_1$

⇒ $R =\frac{V_1G-VG_1}{(V-V_1)}$

⇒ Plugging the values.

⇒ $R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}$

⇒ $R =\frac{(50400 - 58600)}{(-10)}$

⇒ $R=\frac{-8200}{-10}$

⇒ $R=820\ \Omega$

The coil resistance of the circuit is 820 Ω .

4 0

3 years ago

A child of mass 25 kg is skating fast, +10 m/s, and tries to get revenge by colliding with the 60 kg adult who is sitting still.

tankabanditka [31]

Answer: -4.4 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1}V_{o}+m_{2}U_{o}$ (2)

$p_{f}=m_{1}V_{f}+m_{2}U_{f}$ (3)

$m_{1}=25 kg$ is the mass of the child

$V_{o}=10 m/s$ is the initial velocity of the child

$m_{2}=60 kg$ is the mass of the adult

$U_{o}=0 m/s$ is the initial velocity of the adult (it is sitting still)

$V_{f}$ is the final velocity of the child

$U_{f}=6 m/s$ is the final velocity of the adult

Substituting (2) and (3) in (1):

$m_{1}V_{o}+m_{2}U_{o}=m_{1}V_{f}+m_{2}U_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{o}-m_{2}U_{f}}{m_{1}}$ (5)

$V_{f}=\frac{(25 kg)(10 m/s)-(60 kg)(6 m/s)}{25 kg}$ (6)

Finally:

$V_{f}=-4.4 m/s$ This means the velocity of the child is in the opposite direction

4 0

3 years ago