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Anettt [7]
3 years ago
11

Technician A says that the radiator should always be inspected for leaks and proper flow before installing a rebuilt engine. Tec

hnician B says that overheating during slow city driving can be caused by a defective electric cooling fan. Who is right?
A. Technician A
B. Technician B
C. Both Technicians
D. Neither Technician A nor Technician B
Physics
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

Both the technicians are right so option C is correct

Explanation:

Both the technicians are right so option C is correct

while installing a rebuilt engine, it is necessary to inspect the leaks and proper flow because if there is leak then cooling oil will flow out of the engine which results in engine heating. Secondly flow should be proper otherwise the oil will not reach to all parts and may cause heating of some parts of engine.

If the cooling fan is defective it won't be suck enough heat from engine and it will also cause heating.

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Why can concave mirror is used in cosmetic mirror​
tatuchka [14]

Answer:

"When face is placed between the concave mirror and its focus, it produces a magnified image. This enlarged image of face is helpful in makeup as even pores of skin are clearly visible."

4 0
3 years ago
Discuss the differences between Seismic Reflection and Seismic Refraction
Maru [420]

Explanation:

Seismic refraction is the bending of the seismic waves as they pass geological layers of the earth due to different densities. This is especially true for Primary waves because they can pass through all the layers of the earth both liquid and solid.

Seismic reflectivity is the bouncing back of seismic waves at a boundary of geological layers due to different densities or subsurface formation. This is especially true for secondary waves that are unable to pass through liquid layers of the earth like the outer core and mantle. When they reach the boundary of these layers they bounce back towards the earth's surface.

These properties of P and S waves are used to ‘auscultate’ the epicenter of an earthquake by triangulation.

Learn More:

For more on Seismic Reflection and Seismic Refraction check out;

brainly.com/question/13502364

#LearnWithBrainly

6 0
3 years ago
How can inclined planes change the amount of force to move an object? A. Increase height of the inclined plane B. Increase dista
mash [69]

Answer:

Option ( B ) is correct .

Explanation:

To lift a heavy weight , inclined plane is used . Use of inclined plane , makes the task easier because instead of force mg , force mg sinθ is to be used which is less than mg . Here θ is inclination of inclined plane.

If h be the height by which weight is to be lifted

potential energy acquired by weight = mgh

work done by force mg sinθ = mgsinθ x d where d is displacement required .

mg sinθ x d = mgh  ( work done by force = potential energy stored in luggage )

d = h / sinθ

d will be more than h

Hence inclined plane increases the distance to be covered by force applied though it decreases the force itself.

Hence option ( B ) is correct .

4 0
3 years ago
Why is visual constancy important?
weqwewe [10]

Answer:

C.

Explanation:

Visual constancy is a key mechanism that allows the perception to remain.the same even as images change

#<em><u>CARRYONLEARNING</u></em>

4 0
2 years ago
The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

h_{Hg-TOP}=675mmHg=0.675m the barometric reading at the top of the building

h_{Hg-BOT}=695mmHg=0.695m the barometric reading at the bottom of the building

\rho _{air}=1.18 kg/m^{3} density of air

\rho _{Hg}=13600 kg/m^{3} density of mercury

g=9.8/m^{2} gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building P_{TOP} and the perssure at the bottom P_{BOT}:

P_{TOP}=\rho _{Hg} g h_{Hg-TOP} (1)

P_{BOT}=\rho _{Hg} g h_{Hg-BOT} (2)

From (1): P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa (3)

From (2): P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure \Delta P:

\Delta P=P_{BOT} - P_{TOP} (5)

\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa (6)

On the other hand, we have a column of air with a cross-section area A and the same height of the building, lets name it h_{air}.

As pressure is defined as the force F exerted on a specific area A, we can write:

\Delta P=\frac{F}{A} (7)

If we isolate F we have:

F= A \Delta P (8)

Also, the force gravity exerts on this column of air (its weight) is:

F=m_{air} g (9)

Knowing the density of air is: \rho_{air}=\frac{m_{air}}{V_{air}} (10)

where the volume of air can be written as: V_{air}=(A)(h_{air}) (11)

Substituting (1) in (10):

\rho_{air}=\frac{m_{air}}{(A)(h_{air}} (12)

Isolating m_{air}:

m_{air}=(\rho_{air}) (A) (h_{air}) (13)

Substituting (13) in (9):

F=(\rho_{air}) (A) (h_{air}) (g) (14)

Matching (8) and (14)

A \Delta P=(\rho_{air}) (A) (h_{air}) (g) (15)

Isolating h_{air}:

h_{air}=\frac{\Delta P}{g \rho_{air}} (16)

Substituting the known and calculated values:

h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})} (17)

Finally:

h_{air}=230.50 m This is the height of the building

8 0
3 years ago
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