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Nezavi [6.7K]
3 years ago
5

Does anyone know what these three are? I need help!

Physics
1 answer:
VashaNatasha [74]3 years ago
3 0

Answer:

1. A

2. C

3. A

Explanation:

I'm so sorry if I am wrong

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Iron + Hydrochloric acid --> ?
marusya05 [52]
<h3>Iron - Fe</h3>

<h3>Hydrochloric Acid- HCl</h3>

<h2><u>Solution</u></h2>

\bold{Fe  +2HCl \rightarrow FeCl _{}{ \tiny2}  + H{ \tiny{2}}}

\therefore \bold{\fbox{{Balanced}}}

Iron + Hydrochloric Acid \rightarrow Ferrous Chloride + Hydrogen

<h2>Hope This Helps You ❤️</h2>
8 0
3 years ago
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A boat is traveling at a speed of 36 km/h. It has to travel a distance of 210 km. How long will it take?
oksian1 [2.3K]

Answer:

If the boat is going 36 km/h, then it will take about 5.8 hours to travel a distance of 210 km/h.

Explanation:

To calculate time, you have to divide the distance by the speed.

In this case, we take 210 and divide it by 36. The answer you will get is 5.83...

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3 years ago
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A 20-N force applied to the handle of a door produces a 44-N output force. What is the AMA of the handle?
Zepler [3.9K]
Method:

44 N / 20 N
AMA = 2.2 
4 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
4 years ago
What is the current produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 hr? Give yo
AfilCa [17]

The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

Q = 4.00C

Δt = 4.00hr = 14400s

Plug in the values and solve for I:

I = 4.00/14400

I = 0.000277777778A

I = 0.278mA

3 0
3 years ago
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