Answer:
d) Decrease, increase, remain constant
Explanation:
If one tire has a slow leak, the air pressure within the tire will_DECREASE____with time due to outflow of air , the surface area between the tire and the road will__INCREASE__in time,due to flattening of tire.
The net force the tire exerts on the road will_REMAIN CONSTANT____in time. It is so because force does not depend upon area. It is pressure which depends upon area. As there is no change in the weight of the car , force on the road will remain constant.
The frictional force is given by F = μmg
<span>where μ is the coeficient of friction. </span>
<span>Work done by frictional force = Fd = μmgd </span>
<span>Kinetic energy "lost" = 1/2 mv² </span>
<span>Fd = μmgd = 1/2 mv² </span>
<span>The m's cancel μgd = v² / 2 </span>
<span>d = v² / 2μg </span>
<span>d = 8² / 2(0.41)(9.8) </span>
<span>d = 32 / (0.41)(9.8) </span>
<span>d = 7.96 </span>
<span>Player slides 8 m . </span>
<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>
<span>d = v² / 2μg </span>
<span>= 4² / 2(0.46)(9.8) </span>
<span>= 8 / (0.46)(9.8) </span>
<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3