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A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

5 0

3 years ago

A virtual image formed by convex lens is always

natita [175]

Answer:

2. erect and magnified

4 0

3 years ago

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

Svetradugi [14.3K]

Answer:

Total contraction on the Bar = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

Hole diameter = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180 × 10³ N

Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

5 0

3 years ago

Per person, the energy consumption in the United States is about 3.3 × 10^8 Btu/yr. This is equivalent to a power of:________. A

zhuklara [117]

Answer:

$3.3\times 10^8\ btu/yr=11\ kW$

Explanation:

Btu of British thermal unit is a unit of heat. The relation between btu/year and watts is given by :

$1\ btu/yr=3.343\times 10^{-5}\ W\\\\3.3\times 10^8\ btu/yr=3.343\times 10^{-5}\times 3.3\times 10^8\ W\\\\3.3\times 10^8\ btu/yr=11031.9\ W\\\\3.3\times 10^8\ btu/yr=11\ kW$

So, $3.3\times 10^8\ btu/yr$ is equal to 11 kW. Hence, the correct option is (d).

4 0

3 years ago

Which group of measurements is the most precise? A) 2 g, 3 g, 4 g B) 2 g, 2.5 g, 3 g C) 2.0 g, 3.0 g, 4.0 g, 5.0 g D) 2.0 g, 3.0

siniylev [52]

The answer to your question is A

3 0

3 years ago