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2 years ago

Which of the following mathematical operations are and are not not allowed on two quantities with different unit dimensions?

1 answer:

4 0

The following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

To answer the question, we have to know what mathematical operations are.

<h3>What are Mathematical operations?</h3>

Mathematical operations are operations which are performed to change the value of a variable. The arithmetic mathematical operations we have are

addition,
subtraction,
multiplication,
subtraction and
equality which shows the relationship between two quantities.

Now, for two quantities with different unit dimensions, addition, subtraction and equality are not allowed because these mathematical operations require the quantities to be in the same unit dimensions

Also, for two different quantities with differnt unit dimensions, multiplication and division are allowed because these mathematical operations do not require the quantities to be in the same unit dimensions.

So, the mathematical operations that are allowed are multiplication and division while the mathematical operations that are not allowed are addition, subtraction and equality.

So, the following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

Learn more about mathematical operations here:

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An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

3 0

3 years ago

2. A 55 kg woman has a momentum of 200 kg m/s. What is her velocity?

NeTakaya

Answer:

$\boxed {\tt 3.63636364 \ m/s}$

Explanation:

Velocity can be found using the following formula:

$v=\frac{p}{m}$

where p is the momentum and m is the mass.

The woman has a mass of 55 kilograms and a momentum of 200 kilogram meters per second.

$p= 200 \ kgm/s\\m=55 \ kg$

Substitute the values into the formula.

$v=\frac{200 \ kg m/s}{55 \ kg}$

Divide. Note that the kilograms, or kg, will cancel each other out.

$v=\frac{200 \ m/s}{55}$

$v= 3.63636364 \ m/s$

The woman's velocity is 3.63636364 meters per second.

6 0

3 years ago

The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7

stich3 [128]

Answer:

Electric field acting on the electron is 127500 N/C.

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Initial speed of electron, u = 0

Final speed of electron, $v=3\times 10^7\ m/s$

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}$

$a=2.25\times 10^{16}\ m/s^2$

According to Newton's law, force acting on the electron is given by :

F = ma

$F=9.1\times 10^{-31}\times 2.25\times 10^{16}$

$F=2.04\times 10^{-14}\ N$

Electric force is given by :

F = q E, E = electric field

$E=\dfrac{F}{q}$

$E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}$

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

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3 years ago

A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for

Sonja [21]

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

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3 years ago

A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total

Aloiza [94]

Answer:

a = ω^2 A formula for max acceleration (ignoring sign)

V = ω A formula for max velocity

V^2 = ω^2 A^2 = a A from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

6 0

2 years ago