I'm not sure about the distance to the nearest star, but it's probably about 4 light-years (L-y).
1 L-y = 1.86 * 10E5 mi/sec * 3600 sec/hr * 24 hr/day * 365 day/yr
1 L-y = 5.9 *10E12 mi and 4 L-y = 2.3 *10E13 mi distance to star
2.3 * 10E13 mi / 900 mi/hr = 2.6 * 10E10 hr hours to star
2.6 * 10E10 hr / (24 hr/day) = 1.1 * 10E9 day days to star
1.1 * 10E9 day / 365 day/yr = 3 * 10E6 yr = 3 million years to star
Answer:
Electromagnetic waves
Explanation:
Electromagnetic waves are waves that consist of oscillating electric and magnetic fields, that oscillate perpendicularly to each other and perpendicularly to the direction of propagation of the wave (for such a reason, these waves are also called transverse waves).
Electromagnetic waves always travel in a vacuum at the same speed, called speed of light:

and they are classified into 7 different types, according to their frequency. From lowest to highest frequency, we have:
Radio waves
Microwaves
Infrared
Visible light
Ultraviolet
X-rays
Gamma rays
Therefore, gamma rays, x-rays, visible light and radio waves are all types of electromagnetic waves with different frequencies.
The weight of an object when calculated by multiplying with the pull of the gravity is dependent on the mass of the object and the value of g. The value of g is constant however is still dependent on the distance of the object from the center of the Earth. Thus, the answers are <em>mass and distance. </em>
Answer:
14,700 N
Explanation:
The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:
(1)
the weight of the hyppo is

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
<u>For F₁</u>
<u />
<u />
<u>For F₂</u>
![F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D2%2Acos%2860%29%5C%5CF_%7Bx%7D%3D1%5BN%5D%5C%5CF_%7By%7D%3D-2%2Asin%2860%29%5C%5CF_%7By%7D%3D-1.73%5BN%5D)
<u>For F₃</u>
<u />
<u />
Now we can sum each one of the forces in the given axes:
![F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D1-0.866%3D0.134%5BN%5D%5C%5CF_%7By%7D%3D2-1.73%2B0.5%5C%5CF_%7By%7D%3D0.77%5BN%5D)
Now using the Pythagorean theorem we can find the total force.
![F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B%280.134%29%5E%7B2%7D%20%2B%280.77%29%5E%7B2%7D%7D%5C%5CF%3D%200.78%5BN%5D)