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svet-max [94.6K]
1 year ago
13

g) If two cars A and B are moving with velocity 60 km/hr and 80 km/hr respectively in the same direction. What will be the relat

ive velocity of B with respect to A? (20 kmhr)​
Physics
1 answer:
tester [92]1 year ago
4 0

Explanation:

Here,

Velocity of A=60km/hr

Velocity of B=80km/hr

Now,

Relative velocity of B with respect to A in the same direction=Velocity of B - Velocity of A

=80km/hr-60km/hr

=20km/hr

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. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T
Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

         I = F t = Δp

in this case the final velocity is zero

        F t = 0 -m v₀

        F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

        F₁ = -p₀ / 0.1

        F₁ = - 10 p₀

barrier collapses

         F₂ = -p₀ / 1

let's look for the relationship of the forces

        F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

5 0
2 years ago
What s physics?
Alex777 [14]

The answer is (B. The study of Matter and Energy) but technically you could consider physics all of these as engineering is based on physics and that would be the study of inventions, chemistry and biology were both discovered because of physics, and physics invokes more math than any other subject as it applies math to the entire Universe.

7 0
3 years ago
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Why is freshly distilled or deionized water used in this standardization?
sukhopar [10]

Answer:

The amount of carbon dioxide is little in deionized water.

Explanation:

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3 years ago
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A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s
Ede4ka [16]

When acceleration is constant, the average velocity is given by

\bar v=\dfrac{v+v_0}2

where v and v_0 are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}

where x,x_0 are the final/initial displacements, and t,t_0 are the final/initial times, respectively.

Take the car's starting position to be at t_0=0\,\mathrm s. Then

\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t

So we have

x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m

You also could have first found the acceleration using the equation

v=v_0+at

then solve for x via

x=x_0+v_0t+\dfrac12at^2

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of a anyway.

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