Convert 32 cm/s to km/s

The main difference between internal combustion engine and the diesel engine is?

Sav [38]

<span>The difference between a internal combustion engine and a diesel engine is the ignition, But a Diesel engine is an internal combustion engine. The both burn internal one uses compression to fire the other uses ignition system.</span>

8 0

3 years ago

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A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Pig. P7.41). (a) How fast

Leona [35]

Answer:

a)Velocity of car =v=16 m/s

b)Force against the track at point B=1.15* $10^{4}$ N

Explanation:

Given mass of roller coaster=m=350 kg

Position of A=Ha=25 m

Position of B=Hb=12 m

Net potential energy=mg(ha-hb)

Net potential energy=(350)(9.80)(25-12)

Net potential energy=44590 J

Using energy conservation

net kinetic energy=net potential energy

(1/2)mv^2=mg(ha-hb)

m=350

velocity=v=16 m/s

b)There two force acting,centripetal force upward and gravity downward.

Thus net force acting will be

Net force=(mv^2/r)-mg

Net force=14933.33-3430

Net force=1.15* $10^{4}$ N

3 0

3 years ago

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If the distance between two objects increased, which of the following things could keep the force of gravity between the two obj

AURORKA [14]

The answer A hope this helped

7 0

3 years ago

a wall switch is connected to a wall outlet. A lamp is plugged into the same outlet. You turn the switch on, but the lamp stays

jolli1 [7]

1: only half the outlet is switched and the lamp is in the other half
2: the lamp is turned off.
3: The light bulb is burned out
4: the switch might be broken
5: the fuse might be blown
6: the electricity might be off

5 0

3 years ago

Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.

alina1380 [7]

Answer:

The time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

Explanation:

Given the intensity of the energy is $1.36\ kW/m^2$

And the arriving energy is $3.55\times 10^9\ J$

Also, the area in which energy is being arriving is $4.25\ m^2$

Now, we will use relation between energy $(E)$ , intensity of energy $(p)$ , area $(A)$ and time $(T)$ .

Where energy is in Joule, intensity is in $kW/m^2$ , area is in $m^2$ and time is in seconds.

The equation is

$E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds$

So, the time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

4 0

4 years ago