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3 years ago

A cubic tank holds 1,000.0 kg of water. what are the the dimensions of the tank in meters? explain your reasoning

Physics

1 answer:

Scilla [17]3 years ago

8 0

Given: A cubic tank holds 1,000.0 kg of water.

Mass of water in tank (m) = 1000.0 kg

Density of water (d) = 1000.0 kg /m³

Concept: Volume(V) = Mass / Density

Since the tank holds these water in it so the volume of water will be equal to the volume of the tank.

Hence, the volume of the tank = Mass of water / Density of water

or, = 1000.0 kg / 1000.0 kg m⁻³

or, = 1.0 m³

Since tank is cubical in shape. Let its side be 'x'

The volume of tank (x³) = 1.0 m³

or. side of tank (x) = 1.0 m

Hence, the dimensions of the tank will be 1.0 m.

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A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago

Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would

Andre45 [30]

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

$W = mg$

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

$W_{x} = mg_{x}$

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

Therefore,

$34.65 N = (1.5 kg)g_{x}$

$g_{x} = \frac{34.65 N}{1.5 kg}$

7 0

3 years ago

An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad

Anna11 [10]

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a) where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

5 0

3 years ago

PLEASE ANSWER ASAP, WILL MARK BRAINLIEST***

Andreyy89

The answer is to your question is c

7 0

2 years ago

a ball of mass 0.5 kg moving at 10 m/s collides with another ball of equal mass at rest. if the two balls move off together afte

nexus9112 [7]

Answer:

5 m/s

Explanation:

Here we can see there is no external force acted on a two masses when we consider the motion. If there is no external forces then momentum is conserved.

Initial momentum = Final momentum

0.5 × 10 = 1 × V

V = 5 m/s

7 0

3 years ago