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3 years ago

If constants aren't given in an experiment, what can be constant in the experiment?

Physics

1 answer:

mario62 [17]3 years ago

6 0

Answer:

Anything in an experiment that remains unchanged.

Explanation:

An example could be the temperature of the laboratory room. If there is something that has an effect on an experiment that is not variable, it is a constant. Another constant could be, say, if you were doing calculations with the same amount and kind of fluid throughout the experiment, then that fluid would also be a constant.

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You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is

Alisiya [41]

Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

Weight in air

W (air) = mg, where g is the gravitational acceleration

Weight with submerged with only one mass

m(s)g + Fb = mg + m(b)g, consider this to be equation 1

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, consider this to be equation 2

equation 1 – equation 2 would give us

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

Substituting V into the above equation we get

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

8 0

2 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago

A ball of mass 2 kg is kept on the hill of height 3 km. Calculate the potential energy possessed by it ?

zysi [14]

We know that -

$P.E=m*g*h$

Where,

$m = mass$

$g = acceleration$ $due$ $to$ $gravity$

$h=height$

First we convert height into meters.

1 km = 1000 meters

3 km = 1000 * 3 meters = 3000 meters

So, putting the values in the above formula, and by taking 'g' = 9.8 m/s², we get-

$P.E.= 2*3000*9.8$

$P.E.= 58800$ $Joules$

$P.E.= 58.8$ $kJ$

7 0

3 years ago

Read 2 more answers

Which of the following is not a reason fluorescent lamps are advantages over incandescent lamps?

iren2701 [21]

It’s because flourecent lights operate at higher temperatures than incadecent lights.

3 0

3 years ago

A test charge is Introduced into the electric field of a charge. It feels a force of 2 F where the electric field is 4 E. What w

grandymaker [24]

I think Its 1/4 F I guess

4 0

3 years ago