1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Aleksandr [31]
3 years ago
14

Please help!!!!!!!!!

Physics
1 answer:
IgorC [24]3 years ago
5 0
In an Internal Combustion Engine, the fuel is singed in the chamber or vessel. Example: Diesel or Petrol motor utilized as a part of Cars. 
The internal engine has its vitality touched off in the barrel, as 99.9% of motors today. In an External Combustion Engine, the inner working fuel is not consumed. Here the liquid is being warmed from an outer source. The fuel is warmed and extended through the interior instrument of the motor bringing about work. Eg. Steam Turbine, Steam motor Trains. An outer burning case is a steam motor where the warming procedure is done in a kettle outside the motor.
You might be interested in
A subatomic particle that has a positive charge and that is located in the nucleus of an atom
wolverine [178]

Answer:

c

Explanation:

3 0
3 years ago
Read 2 more answers
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
3 years ago
When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine
JulsSmile [24]
Choice - B is the correct one.

At the top of the arc, at one end of the swing:
-- it's not going to get any higher, so the potential energy is maximum
-- it stops moving for an instant, so the kinetic energy is zero

At the bottom of the arc, in the center of the swing:
-- it's not going to get any lower, so the potential energy is minimum
-- it's not going to move any faster, so the kinetic energy is maximum
7 0
3 years ago
Can somone pls help me??!! i’m very stuck
Alchen [17]

Answer:

its the third one

Explanation:

3 0
3 years ago
A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

3 0
3 years ago
Other questions:
  • The angle turned through the flywheel of a generator during a time interval t is given by theta = at + bt^3 -ct^4, where a, b, a
    12·1 answer
  • When baking soda and vinegar react, the surface bubbles. What does this most likely indicate?
    12·3 answers
  • The small particles that produce a streak of light upon entering earth’s atmosphere are called
    6·2 answers
  • So what would the answer for D be?
    15·1 answer
  • Enter the expression 2gΔym−−−−−√, where Δ is the uppercase Greek letter Delta.
    9·1 answer
  • A box is being pushed but two stellar science students, one on each side of the box. Peter is pushing box with a force of 10 N t
    15·1 answer
  • Which of the following is a phase change?
    11·1 answer
  • A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t
    13·1 answer
  • A car's speed changes from 10m/s to 35m/s in 15.0s along a straight road while moving in one way. Find the magnitude of the acce
    6·1 answer
  • give an example of one living and one non-living thing that uses the force of buoyancy to function. explain how they work.
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!