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Rus_ich [418]
4 years ago
6

A car engine burns fuel and has moving parts. Sometimes, an engine can become too hot and overheat. What can prevent the engine

from overheating?
Physics
2 answers:
arsen [322]4 years ago
6 0

there many things you could do, one is try getting a cooling system, another thing is its extramly hot and you may need to tunr off your car to prevent the engine from over heating.


Ahat [919]4 years ago
4 0

If I'm not mistaken the only prevention I know of is coolant and the only thing I can think of is that the car ran out of coolant.  If your car is overheating it can either be because the coolant is not moving fast enough, the heat isn't getting transferred out of the engine to the coolant, or it is not getting transferred out of the coolant to the air through the radiator

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The physical and chemical properties of a substance are used to identify the substance and to see how it will behave in the pres
pentagon [3]
Im going to guess on this one, and say that it is chemical.
8 0
3 years ago
Read 2 more answers
A train of mass 320000 kg accelerate uniformly from rest. It takes 57 s to travel a distance of 810 m in a straight line. Find t
Nastasia [14]

Answer:

160,000 N

Explanation:

Given:

m = 320,000 kg

v₀ = 0 m/s

a = constant

t = 57 s

Δx = 810 m

Find: Fnet

Apply Newton's second law:

∑F = ma

Fnet = ma

To find Fnet, we must first find the acceleration.

x = x₀ + v₀ t + ½ at²

810 m = 0 m + (0 m/s) (57 s) + ½ a (57 s)²

a = 0.50 m/s²

Fnet = (320,000 kg) (0.50 m/s²)

Fnet = 160,000 N

8 0
4 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
4 years ago
As a horse finishes its trot around the corral, it slows from 4m/s to a stop in 3
german
As a horse finishes its trot around the corral, it slows from 4m/s to a stop in 3
seconds. Calculate the acceleration of the horse.
Can anybody help ?
8 0
3 years ago
Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
sergeinik [125]

Planet Y has rotated by 135.5° through during this time.

To find the answer, we need to know about the relation between angle and radius of orbit.

<h3>What's the expression of angle in terms of radius?</h3>
  • Angle= arc/radius
  • As arc = orbital velocity × time,

            angle= (orbital velocity × time)/radius

  • Orbital velocity= √(GM/radius), G= gravitational constant and M = mass of sun
  • So, angle = (√(GM)× time)/radius^3/2
<h3>What's is the angle rotated by planet Y after 5 years, if ratio of the radius of orbit of planet X and Y is 4:3 and planet X is rotated by 88°?</h3>
  • Let Ф₁= angle rotated by planet Y, Ф₂= angle rotated by planet X
  • As time = 5 years ( a constant)
  • Ф₁/Ф₂= (radius of planet X / radius of planet Y)^(3/2)
  • Ф₁= (radius of planet X / radius of planet Y)^(3/2) × Ф₂

   = (4/3)^(3/2) × 88°

   = 135.5°

Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.

Learn more about the orbital velocity here:

brainly.com/question/22247460

#SPJ1

8 0
2 years ago
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