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3 years ago

14

The gravitational force between two objects is 100 N. If the mass of each object is doubled, what is the gravitational force bet

ween the objects at the same distance? 25 N 50 N 200 N 400 N

2 answers:

aleksandrvk [35]3 years ago

5 0

The answer to the question is 400 N

Kryger [21]3 years ago

4 0

The answer is D) 400 N

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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago

Hai vận động viên chạy trên cùng 1 đoạn đường, vận động

Vladimir [108]

Answer:

Pemain A

Explanation:

Mengingat data berikut;

Kecepatan pemain A = 12 m/s

Kecepatan pemain B = 36 km/h

Untuk menentukan siapa pelari tercepat di antara dua pemain;

Pertama-tama, kita harus mengubah kecepatan menjadi satuan standar pengukuran yang sama.

Jadi, mari kita gunakan pengukuran umum meter per detik.

Konversi:

36 km/h = (36 * 1000)/(60 * 60)

36 km/h = 36000/3600

36 km/h = 10 m/s

Kecepatan pemain B = 10 m/s

Oleh karena itu, dibandingkan dengan kecepatan pemain A; pemain A lebih cepat.

6 0

3 years ago

A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a sp

Y_Kistochka [10]

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

0² = 345²+2 x a x 0.089

a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

4 0

3 years ago

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$

where we have:

$v=30,000 km/s = 3\cdot 10^7 m/s$ is the orbital speed

r is the orbital radius

$T=27 h \cdot 3600 =97,200 s$ is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

$r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m$

B. $6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s$

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

$M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg$

and considering the value of the solar mass

$M_s = 2\cdot 10^{30}kg$

the mass of the black hole as a multiple of our sun's mass is

$M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s$

C. $9.28\cdot 10^9 m$

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

$R=\frac{2MG}{c^2}$

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

$R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m$

8 0

3 years ago

Which of the following is NOT a property of long-period comets that suggests that they originate from the Oort cloud?

Lesechka [4]

Answer:

b. They orbit around the Sun in a counterclockwise direction, when viewed from above the ecliptic plane.

Explanation:

All the objects of the solar system revolve around the Sun in a counterclockwise direction. The comet coming from the Oort's cloud will also follow the same kind of orbit. That is why it can't be a property to distinguish an Oort's cloud comet.

All other properties are correct to identify an Oort's cloud comet as the Oort's cloud is a considered a spherical cloud just outside the Solar system.

7 0

3 years ago