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1 year ago

An empty glass beaker has a mass of 103 g. When filled with water, it has a total mass of 361g.

Physics

1 answer:

aivan3 [116]1 year ago

7 0

Answer:

0.96 gcm¯³

Explanation:

From the question given above, the following data were obtained:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of beaker + oil = 351 g

Density of water = 1 gcm¯³

Density of cooking oil =?

Next, we shall determine the mass of water. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of water =?

Mass of water = (Mass of beaker + water) – (Mass of empty beaker)

Mass of water = 361 – 103

Mass of water = 258 g

Next, we shall determine the volume of the beaker. This can be obtained by calculating the volume of water in the beaker.

Density of water = 1 gcm¯³

Mass of water = 258 g

Volume of water =?

Density = mass /volume

1 = 258 / volume

Cross multiply

1 × volume = 258

Volume of water = 258 cm³

Thus the volume of the beaker is 258 cm³.

Next, we shall determine the mass of the cooking oil. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + oil = 351 g

Mass of cooking oil =?

Mass of cooking oil = (Mass of beaker + oil) – (Mass of empty beaker)

Mass of cooking oil = 351 – 103

Mass of cooking oil = 248 g

Finally, we shall determine the density of the cooking oil. This can be obtained as follow:

Mass of cooking oil = 248 g

Volume of the beaker = 258 cm³

Density of cooking oil =?

Density = mass / volume

Density = 248 / 258

Density of cooking oil = 0.96 gcm¯³

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Answer:

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Explanation:

given data

before it hits the ground = 46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt² ..............1

here d is distance and g is acceleration and t is time

so when object falling it will be

h = 4.9 t² ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)² ...................3

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²

t = 0.576 s and 3.771 s

we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h = 69.68 m

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2 years ago

A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha

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Answer:

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ = 25*0.75*sin30°

τ = 25*0.75*0.5

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Natali5045456 [20]

Answer:

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Explanation:

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Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is $s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8$

As both of them cover the same distance

$45000a = 3at_J^2/8$

$t_J^2 = 45000*8/3 = 120000$

$t_J = \sqrt{120000} = 346.4 s$

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2 years ago

Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p

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Answer:

Diode Lasers

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is

4. The optical gain bandwidth (as measured between half intensity points) will

normally depend on the pumping current (diode current) but for this problem

assume that it is 2 nm.

(a) What is the mode integer m of the peak radiation?

(b) What is the separation between the modes of the cavity? Please express your

answer as Δλ.

(d) What is the reflection coefficient and reflectance at the ends of the optical

cavity (faces of the InGaAsP crystal)?

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the

beam is a Gaussian beam with the waist located at the output. And the beam

waist size is approximately the x-y dimensions of the cavity.)

Solution:

(a) The wavelength λ of a cavity mode and length L are related by

λ = , where m is the mode number, and n is the refractive index.

So the mode integer of the peak radiation is

1290

1055.1

10250422

= ×

××× == −

−

m .

(b) The mode spacing is given by nL

c f 2

=Δ . As

c f = , λ

Δ−=Δ 2

c f .

Therefore, we have nm

nL f

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .

bandwidth could fit in two possible modes.

For mode integer of 1290, nm

nL 39.1550

1290

10250422 6

= ××× ==

−

Take m = 1291, nm

nL 18.1549

1291

10250422 6

= ××× ==

−

Or take m = 1289, nm

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

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2 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

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0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

m1 = 1.29 kg

m2 = 0.3 kg

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Plugging known information to get :

F_r =F_g

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l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

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2 years ago