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sladkih [1.3K]
2 years ago
6

The increase of kinetic energy at the square of the speed of your vehicle has a major influence on all motor vehicles in three p

articular situations. What are they
Physics
1 answer:
Viktor [21]2 years ago
5 0

This implies that stopping distance and impact force grow as a function of speed. The best ways to improve manoeuvrability and lessen crash severity are to drive at an appropriate pace and to slow down as soon as you spot dangers in front of you.

Keep in mind that stopping distance increases with speed; at 50 mph, it is four times longer than at 25 mph, and at 75 mph, the force of impact is nine times greater.

<h3>What is the impact of speed on kinetic energy ?</h3>

When your car expends or absorbs energy to speed up or slow down, you may feel a pull or a jolt, called impulse. Impulse increases as the energy or force increases, and increases as the duration of the force decreases. You'll feel a harder jolt if you speed up or slow down suddenly.

  • Consider: coming to a stop from 60 mph in ten seconds doesn't hurt you or your vehicle because the force of this event is spread out over a long time. But if you hit a wall and come to a stop in just half a second, you'll feel twenty times the impulse, causing severe damage.

Learn more about Kinetic energy here:

brainly.com/question/25959744

#SPJ4

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A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.
aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c} Where m_{c} is the calorimeter mass and n_{s} is the number of moles of the samples; H_{c} is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product m_{c} *Cp is the heat capacity, so the balance equation is:

6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}

So, the enthalpy of combustion can be calculated:

H_{c}=-5899.5\frac{kJ}{mol}

I will be happy to solve any doubt you have.

4 0
3 years ago
If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

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