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.
Answer:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.
Generally, the relationship between temperature, density and rate of vertical pressure is given as;


where;
- <em>ρ is density</em>
- <em>T is temperature</em>
- <em>dP is rate of change of vertical pressure</em>
Thus, from the formula above, we can conclude the following relationship between temperature, density and the rate of vertical pressure change in spatial pattern of heights.
The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.
Learn more here:brainly.com/question/25395377
Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine
= 62 MPa = 62 x 10^6 Pa
we also know that
= ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine
= 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure
, we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>