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Zanzabum
2 years ago
6

A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by

a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)
Physics
1 answer:
Marizza181 [45]2 years ago
6 0

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

f_o = Focal length of objective = 0.9 cm

f_e = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52

The angular magnification of the compound microscope is -252.52

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En la etiqueta de un bote de fabada de 350 g, leemos que su aporte energético es de 1630 kj por cada 100 g de producto a) La can
fredd [130]

Answer:

(a) 153.37 g

(b) 5705 kJ

Explanation:

(a) To find the amount of bean needed by a man you first calculate the equivalence in beans to 2500kJ

2500kJ*\frac{100g}{1630kJ}=153.37\ g

Thus, 153.37 g has the energy needed by a man that needs 200kJ per day.

(b) The amount of energy per pot of bean is given by:

E=350g*\frac{1630kJ}{100g}\\\\E=5705\ kJ

Thus, the energy is 5705kJ

6 0
2 years ago
A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N What is the value of t
nadya68 [22]

1) weight of the box: 980 N

The weight of the box is given by:

W=mg

where m=100.0 kg is the mass of the box, and g=9.8 m/s^2 is the acceleration due to gravity. Substituting in the formula, we find

W=(100.0 kg)(9.8 m/s^2)=980 N


2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

N=W cos \theta

where W is the weight of the box, calculated in the previous step, and \theta=50^{\circ} is the angle of the ramp. Substituting, we find

N=(980 N)(cos 50^{\circ})=630 N


3) Acceleration: 7.5 m/s^2

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

a_p = g sin \theta

Substituting, we find

W_p = (9.8 m/s^2)(sin 50^{\circ})=7.5 m/s^2

5 0
2 years ago
Consider an alcohol and a mercury thermometer that read exactly 0 oC at the ice point and 100 oC at the steam point. The distanc
Alexeev081 [22]

Answer:

No, both the thermometers will give the different reading.

Explanation:

Given,

  • Both thermometer has same ice point = T_i\ =\ 0^o C
  • Both thermometer has same steam point = T_s\ =\ 100^o C
  • Distance between the ice point and steam point in both the thermometer is same of 100 division,

All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion.  Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.

7 0
2 years ago
A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope
pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

L=\dfrac{\lambda}{4}

\lambda=4\times L

Put the value into the formula

\lambda=4\times31.2\times10^{-2}

\lambda=1.248\ m

We need to calculate the frequency of sound in pipe

Using formula of frequency

f=\dfrac{v}{\lambda}

Put the value into the formula

f=\dfrac{340}{1.248}

f=272.4\ Hz

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

L=\dfrac{l}{3}

L = \dfrac{129}{3}

L=43\ cm

We need to calculate the wavelength of the wire

Using formula of length

L=\dfrac{\lambda}{2}

\lambda=L\times 2

Put the value into the formula

\lambda=43\times2

\lambda=86\ cm

We need to calculate the speed of the wave

Using formula of frequency

f=\dfrac{v}{\lambda}

v=f\times\lambda

Put the value into the formula

v=272.4\times86\times10^{-2}

v=234.26\ m/s

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0
2 years ago
Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
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