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3 years ago

If the sun disappeared what would happen to the path the planets follow?

2 answers:

8 0

Regarding the Earth's motion (and the motion of every other planet),
there suddenly wouldn't be any centripetal force to curve the Earth's
path. So the Earth would sail off in a straight line, in the direction it
was moving at the instant when the sun's gravity disappeared.

Although everything would become very dark, you wouldn't feel
anything different. Even now, as we follow the curved orbit around
the sun, the direction of the Earth's motion only changes by about
one degree per day, so we don't feel it.

dmitriy555 [2]3 years ago

7 0

I would imagine that the planets would just go a drift in space or try to find another orbit like Jupiter sense it's the biggest planet in our solar system.

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1.) What causes ocean tides

BigorU [14]

Answer: Tides are periodic rises and falls of large bodies of water. Tides are caused by gravitational interaction between the earth and the moon. The gravitational attraction of the moon causes the oceans to bulge out in the direction of the moon.

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4 years ago

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To drive to the store from home, somebody first travels 5 km west, then they travel 5 km northwest. What is the straight line di

schepotkina [342]

Straight line distance between their home and the store can be solve using cosine law. first is solve the angle which is180 - 45 = 135 degree
C^2 = a^2 + b^2 - 2ab cos(135)c^2 = 5^2 + 5^2 - 2(5)(5) cos(135)c^2 = 85.35c = 9.24 km is the straight line distance between their home and the store

6 0

3 years ago

A rocket-driven sled Sonic Wind No. 2, is used for investigating the physiological effects of large accelerations on people. It

rewona [7]

Answer:

The best single equation in order to fin the acceleration is

$a=v_f/t=250/2=125m/s^2$

Explanation:

The best single equation in order to find the acceleration is the following kinematics equation:

$v_f=v_o+at$

As the rocket start from the rest, the initial velocity value is zero. Then the acceleration value is:

$a=v_f/t=250/2=125m/s^2$

7 0

3 years ago

Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric fiel

Delicious77 [7]

Answer:

q₁ = q₂ = Q = 14.8 pC

Explanation:

Given that

q₁ = q₂ = Q = ?

Distance between charges = r =7.3 cm = 0.073 m

Combined electric field = E₁ + E₂ = E = 50 N/C

Using formula

$E=2\frac{kQ}{r^2}$

Rearranging for Q

$Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}$

$Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC$

4 0

3 years ago

A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff

Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ( $E$ ) of the project is equal to the sum of gravitational potential energy ( $U_{g}$ ) and translational kinetic energy ( $K$ ), all measured in joules:

$E = U_{g} + K$ (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}$ (Eq. 1b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y$ - Initial height of the projectile above ground, measured in meters.

$v$ - Initial speed of the projectile, measured in meters per second.

If we know that $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y = 132\,m$ and $v = 126\,\frac{m}{s}$ , the initial mechanical energy of the earth-projectile system is:

$E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}$

$E = 498556.296\,J$

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

$W_{loss} = E_{o}-E_{1}$ (Eq. 2)

Where:

$E_{o}$ - Initial total mechanical energy, measured in joules.

$E_{1}$ - FInal total mechanical energy, measured in joules.

$W_{loss}$ - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$W_{loss} = E_{o}-K_{1}-U_{g,1}$

$W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}$ (Eq. 2b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ - Maximum height of the projectile above ground, measured in meters.

$v_{1}$ - Current speed of the projectile, measured in meters per second.

If we know that $E_{o} = 498556.296\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 297\,m$ and $v_{1} = 89.3\,\frac{m}{s}$ , the work losses due to air friction are:

$W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)$

$W_{loss} = 125960.4\,J$

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

$E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}$ (Eq. 3)

$K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}$

Where:

$E_{1}$ - Total mechanical energy of the projectile at maximum height, measured in joules.

$U_{g,2}$ - Potential gravitational energy of the projectile, measured in joules.

$K_{2}$ - Kinetic energy of the projectile, measured in joules.

$W_{loss}$ - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}$ (Eq. 3b)

$m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}$

$v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}$

$v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }$

If we know that $E_{1} = 372595.896\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{2} =0\,m$ and $W_{loss} = 125960.4\,J$ , the final speed of the projectile is:

$v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }$

$v_{2} \approx 82.475\,\frac{m}{s}$

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

7 0

3 years ago