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Savatey [412]
2 years ago
14

The strength of the gravitational field of a source mass can be measured by the magnitude of the acceleration due to gravity at

a field point. Thus the gravitational field strength at a point depends on the distance from the source mass. For this problem assume that Earth’s mass is concentrated at its center and that Earth has a radius of RE = 6,378 km at sea level and a mass of ME = 5.972×1024 kg. Ignore any forces due to Earth’s rotation or due to other astronomical bodies.
Physics
1 answer:
Svetradugi [14.3K]2 years ago
5 0

Answer:

9.79211 m/s²

Explanation:

M = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6378000 m

g=G\frac{M}{r^2}\\\Rightarrow g=6.67\times 10^{-11}\frac{5.972\times 10^{24}}{(6378000)^2}\\\Rightarrow g=9.79211\ m/s^2

The acceleration due to gravity is 9.79211 m/s²

For any distance above the Earth's surface h

g=6.67\times 10^{-11}\frac{5.972\times 10^{24}}{(6378000+h)^2}\\\Rightarrow g=\frac{3.983324\times 10^{14}}{6378000+h}\ m/s^2

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A car travels 10 m/s east. Another car travels 10 m/s north. The relative speed of the first car with respect to the second is:
Thepotemich [5.8K]

Answer:

d. less than 20m/s

Explanation:

To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is

[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s

As 14.14m/s is less than 20m/s. d is the correct selection for this question.

3 0
2 years ago
A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w
slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0
2 years ago
Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sl
KonstantinChe [14]

Answer:

Acceleration, a=1.2\ m/s^2

Explanation:

Given that,

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg

Force exerted by the doges on the rope attached to the front sled, F = 240 N

To find,

The acceleration of the sleds.

Solution,

Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :

F = ma

a=\dfrac{F}{m}

a=\dfrac{240\ N}{2\times 100\ kg} (m = 2m)

a=1.2\ m/s^2

So, the acceleration of the sleds is 1.2\ m/s^2.

6 0
2 years ago
A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity
Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

\omega_i = 2\pi f_1

\omega_1 = 2\pi(\frac{610}{60})

\omega_1 = 63.88 rad/s

similarly final angular speed is given as

\omega_f = 2\pi f_2

\omega_2 = 2\pi(\frac{837}{60})

\omega_2 = 87.65 rad/s

angular acceleration of the turbine is given as

\alpha = 5.9 rad/s^2

now we have to find the angular displacement is given as

\theta = \omega t + \frac{1}{2}\alpha t^2

\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)

\theta = 234.62 radian

3 0
2 years ago
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