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3 years ago

How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and

Physics

1 answer:

aev [14]3 years ago

4 0

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h ( g = 10 m/s²)

h= 46.66 m

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An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects he

Agata [3.3K]

Answer:

2.36 x 10^6 J

Explanation:

Tc = 0°C = 273 K

TH = 22.5°C = 295.5 K

Qc = heat used to melt the ice

mass of ice, m = 85.7 Kg

Latent heat of fusion, L = 3.34 x 10^5 J/kg

Let Energy supplied is E which is equal to the work done

Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J

Use the Carnot's equation

$\frac{Q_{H}}{Q_{c}}=\frac{T_{H}}{T_{c}}$

$Q_{H}=286.24\times 10^{5}\times \frac{295.5}{273}$

QH = 309.8 x 10^5 J

W = QH - Qc

W = (309.8 - 286.24) x 10^5

W = 23.56 x 10^5 J

W = 2.36 x 10^6 J

Thus, the energy supplied is 2.36 x 10^6 J.

8 0

3 years ago

In a given chemical reaction, the energy of the products is greater than the energy of the reactants. Which statement is true fo

Flura [38]

Energy is released in the reaction

Explanation:

In a given where the energy of the products is greater than that of the reactants, we can infer that energy is released in the reaction.

This indicates that the reaction is an exothermic or exergonic reaction.

These reaction types are accompanied by release of energy.

In an exothermic change energy is released to the surroundings.
The surrounding becomes hotter at the end of the change.
This applies in exergonic reaction which leaves a reaction having more energy than it originally started with.

Learn more:

Exothermic process brainly.com/question/10567109

#learnwithBrainly

3 0

3 years ago

Read 2 more answers

Based upon the information you have learned throughout this module on blood spatter analysis, do you feel that analyzing blood s

kompoz [17]

Answer:

I do not think that it is the most reliable way to gain information since it is very hard to do and can be easily messed up. No, I don't think you can charge someone on only evidence from blood spatter, but if there was additional evidence I think that this would definitely help with the case but not on its own, since it doesn’t give you physical evidence about the suspect.

Explanation:

6 0

2 years ago

A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju

stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

$-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90$

3 0

3 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$