I need help with 1 -7 please .

melamori03 [73]

<span>B. Compression causes upward vertical movement.</span>

5 0

3 years ago

Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Eart

Temka [501]

The distance from the Earth's center to the point outside the Earth is 55800 Km

<h3>How to determine the distance from the surface of the Earth</h3>

Acceleration due to gravity of Earth = 9.8 m/s²
Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Mass of the Earth (M) = 5.97×10²⁴ Kg
Distance from the surface of the Earth (r) =?

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by g

r² = GM / g

Take the square root of both sides

r = √(GM / g)

r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]

r = 4.94×10⁷ m

Divide by 1000 to express in Km

r = 4.94×10⁷ / 1000

r = 4.94×10⁴ Km

<h3>How to determine the distance from the center of the Earth</h3>

Distance from the surface of the Earth (r) = 4.94×10⁴ Km
Radius of the Earth (R) = 6400 Km
Distance from the centre of the Earth =?

Distance from the centre of the Earth = R + r

Distance from the centre of the Earth = 6400 + 4.94×10⁴

Distance from the centre of the Earth = 55800 Km

Learn more about gravitational force:

brainly.com/question/21500344

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5 0

2 years ago

The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of

Fed [463]

Answer:

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

Explanation:

From Physics we get that expansion of the rod portion is found by this formula:

$\Delta l = \alpha\cdot l_{o}\cdot (T_{f}-T_{o})$ (Eq. 1)

Where:

$\Delta l$ - Expansion of the rod portion, measured in meters.

$\alpha$ - Linear coefficient of expansion for titanium, measured in $\frac{1}{^{\circ}C}$ .

$l_{o}$ - Initial length of the rod portion, measured in meters.

$T_{o}$ - Initial temperature of the rod portion, measured in Celsius.

$T_{f}$ - Final temperature of the rod portion, measured in Celsius.

If we know that $\alpha = 5\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $l_{o} = 10\,m$ , $T_{o} = 0\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , the expansion experimented by the rod portion is:

$\Delta l = \left(5\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (10\,m)\cdot (400\,^{\circ}C-0\,^{\circ}C)$

$\Delta l = 0.02\,m$

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

4 0

3 years ago

If two bodies weighing 10 kg and 20 kg respectively are standing at the

kvv77 [185]

Answer:

potential energy = mgh , where m is mass of body , g is acceleration due to gravity and h is given height

since both of the body are at same height above the ground , so the body having greater mass will have more potential energy

that is 20 kg body will have higher P.E

Explanation:

7 0

3 years ago

When the instantaneous voltage and current in an ac circuit are in phase, we know that?

Vaselesa [24]

We know that the centripetal acceleration is zero, since the summation of forces is the mass times the acceleration in the x- component.

7 0

4 years ago