All categories

3 years ago

How should you Chang the voltage if the resistance stay the same?

Chemistry

1 answer:

Tcecarenko [31]3 years ago

7 0

Voltage = Current x Resistance

If you want to change the voltage without changing the resistance, we must change the current instead.

You might be interested in

I’m stuck on this can I get a little help please

arsen [322]

Answer:

table A

Explanation:

its table A because it requires less force push. the less force used the less friction there is.

the less friction there is the more smooth a surface is.

3 0

3 years ago

9. At STP, how many liters of any gas are occupied by one mole?

BlackZzzverrR [31]

Answer:

The answer is: 22, 4 liters

Explanation:

We use the formula PV=nRT

At STP: 1 atm of pressure and 273 K of temperature.

PV=nRT

V=nRT/P =1 mol x 0,082 l atm/L mol x 273 K /1 atm

V=22, 386 l

8 0

4 years ago

Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov

love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

$V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3$

Mass of the Astroturf = m

Density of the Astroturf = d = 187 $oz/ft^3$

$d=\frac{m}{V}$

$m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz$

1 oz = 0.0283495 kg

$m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg$

Weight of the Astroturf = W

W = mg

$=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N$

The weight of the Astroturf is 179,684.31 Newtons.

8 0

3 years ago

CuCl2 + Al ---> AlCl3 + Cu

Ket [755]

For a balance chemical equation:

The number of atoms of a molecule should remain conserved that is total number of atom of a molecule in reactant side should equal to the number of atoms of a molecule in the product side.

Here also the number of atoms of an element should remain conserved on both sides, So the answer is option B.

That is

B) 3CuCl₂ + 2Al ---> 2AlCl₃ + 3Cu

Here the number of atoms of Cu, Cl and Al are equal in both side , reactant as well as product.

5 0

4 years ago

Read 2 more answers

105 mL of H2O is initially at room temperature (22.0∘C). A chilled steel rod at 2.0∘C is placed in the water. If the final tempe

Amiraneli [1.4K]

Answer : The mass of the steel bar is, 35.2 grams.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of steel = $0.452J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of steel rod = ?

$m_2$ = mass of water = $Density\times Volume=1.00g/mL\times 105mL=150g$

$T_f$ = final temperature of mixture = $21.3^oC$

$T_1$ = initial temperature of steel = $2.0^oC$

$T_2$ = initial temperature of water = $22.0^oC$

Now put all the given values in the above formula, we get

$m_1\times (0.452J/g^oC)\times (21.3-2.0)^oC=-(105g)\times 4.18J/g^oC\times (21.3-22.0)^oC$

$m_1=35.2g$

Therefore, the mass of the steel bar is, 35.2 grams.

6 0

3 years ago