Please help i don't understand

A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the

LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

f = Focal length of the ornament
u = image distance
v = object distance.

make u the subject of the equation

u = fv/(f+v)................ Equation 2

From the question,

Given:

f = 8/2 = 4 cm
v = 12 cm

Substitute these values into equation 2

u = (12×4)/(12+4)
u = 48/16
u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

M = u/v.................. Equation 3

Where

M = magnification of the ornament.

Substitute these values above into equation 3

M = 3/12
M = 0.25.

Hence, The magnification of the ornament is 0.25

8 0

2 years ago

Một oto đang chuyển động thẳng đều với vận tốc 36km/h thì hãm phanh sau 20s thì ô tô dừng lại quãng đường oto đi được là

saw5 [17]

36km/h=10m/s
v=s/t=>s=200m

3 0

3 years ago

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

const2013 [10]

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

$\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}$

we know that

force = pressure * area

therefore force is

$F =(\frac{T_{1}}{T_{2}}*P_{1})A$

$F =(\frac{378}{278}*101325)(17*10^{-2})^{2}$

F = 3.98 kN

5 0

3 years ago

What is the definition of physical change

jeka94

They are used to separate mixtures into their component compounds but can no usually be used to deprecate compounds into chemical elements or simpler compounds

7 0

3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago