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1 year ago

An object traveling at a constant

Physics

1 answer:

mel-nik [20]1 year ago

4 0

Answer:

pi / 2 radians / s

Explanation:

One revolution = 2 pi Radians in 4 seconds

2 pi / 4 = pi/2 radians / s

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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

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

2 years ago

A car begins at rest at a stoplight. After the light turn green, the car begins to accelerate 5 m/s2 over the span of 3.6 second

trasher [3.6K]

Answer:

d=9m

Explanation:

5 0

2 years ago

______ is most resistant to thermal energy transfers. a. copper wire b. aluminum foil c. water d. foam insulation

Reika [66]

Answer:

Water

Explanation:

Because it does not conduct much energy.

8 0

3 years ago

PS Final Exam

tatuchka [14]

Answer:

rolling friction

Explanation:

5 0

2 years ago