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3 years ago

How would life on earth be different if earth’s axis were not tilted with respect to its orbit?

Physics

2 answers:

igor_vitrenko [27]3 years ago

5 0

Answer:

There would be no seasons.

Explanation:

Vladimir [108]3 years ago

4 0

If Earth's axis was "straight up and down" instead of tilted, then ...

<span>-- There would be no seasons.

-- The climate at any one place would be the same all year around.

-- The days would be the same length, everywhere,
    and all year around.

-- So would the nights.

-- The sun would be up a little more than 12 hours every day.
    It would be down a little less than 12 hours every day.

-- At the middle of the day, the sun would be at the same height
   in the sky all year around, not higher in some months and lower
   in others.

-- The equator would be the only place on Earth where the sun
    could ever be directly over your head.

-- If you were at the north pole or the south pole, the sun would be
   down on the horizon, and it would just go around and around you
   every day. It would never rise or set, and it would never get any
   higher or lower.

</span>

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5. Confirm dimensionally that the product (lvp/n) is

Sidana [21]

Answer:

How do you do it?

Explanation:

7 0

3 years ago

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb

Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 $in^{2}$

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = $\frac{F}{A}$

⇒ P = $\frac{6.4}{0.4}$

⇒ P = 16 $\frac{lb}{in^{2} }$

This is the pressure inside the cylinder.

Let force applied on the brake pad = $F_{1}$

Area of the brake pad ( $A_{1}$ )= 1.7 $in^{2}$

Thus the pressure on the brake pad ( $P_{1}$ ) = $\frac{F_{1} }{A_{1} }$

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = $P_{1}$

⇒ 16 = $\frac{F_{1} }{A_{1} }$

⇒ $F_{1}$ = 16 × $A_{1}$

Put the value of $A_{1}$ we get

⇒ $F_{1}$ = 16 × 1.7

⇒ $F_{1}$ = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = $\frac{F_{1} }{4}$ = $\frac{27.2}{4}$ = 6.8 lb

⇒ The force applied on one wheel during braking.

7 0

3 years ago

Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =

Yanka [14]

The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>

4 0

3 years ago

The scale on the horizontal axis is 7.5 s per

Julli [10]

The point on the graph is above or below the 3rd division on the x-axis. But that's all we know, since you've told us nothing about the motion.

4 0

3 years ago

You are given 2 hourglasses.one time is 4 mins another is 7.can u make exactly 9 mins?

SCORPION-xisa [38]

I don't think so it would be some where between 9 and 10

5 0

3 years ago