Answer:
B. Our technology has not advanced enough to make faster-than-light travel possible
Explanation:
So far, we don't know of anything that can go faster then the speed of light.
Given Information:
Mass = m = 500 kg
Acceleration = a = 10 cm/s²
Required Information:
Magnitude of rightward net force = F = ?
Answer:
Magnitude of rightward net force = 50 N
Explanation:
From the Newton's second law of motion
F = ma
Where m is the mass and a is the acceleration
To get force in Newtons first convert 10 cm/s² into m/s²
10/100 = 0.1 m/s²
F = 500*0.1
F = 50 N
Therefore, the magnitude of rightward net force acting on it is 50 Newtons.
Answer:
Elliptical galaxies
Explanation:
Edwin Hubble classified galaxies into three categories
Elliptical
Spiral
Lenticular
The elliptical galaxies have an elipsoidal shape roughly. They have stars which are old and the primary light source of the galaxy. The formation of new stars is very limited. This increases the brightness of the galaxy. The mass of the stars are low. So, far the percentage elliptical galaxies is low compared to other galaxies.
Answer:
Moving them farther apart
Explanation:
The electric field between the two charges Q and q separated by a distance r is given by
![E = \frac{K\times Q\times q}{r^{2}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BK%5Ctimes%20Q%5Ctimes%20q%7D%7Br%5E%7B2%7D%7D)
It shows that the electric field is inversely proportional to the square of the distance between two charges.
So, as the distance between two charges increases, the electric filed between the two charges decreases.
Answer:
1.43 s
Explanation:
The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.
The vertical distance covered by an object in free fall is given by
![S=ut + \frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3Dut%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
u = 0 is the initial vertical speed
t is the time
a= g = 9.8 m/s^2 is the acceleration
since u=0, it can be rewritten as
![S=\frac{1}{2}gt^2](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
And substituting S=10.0 m, we can solve for t, to find the duration of the fall:
![t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2S%7D%7Bg%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%2810.0%20m%29%7D%7B9.8%20m%2Fs%5E2%7D%7D%3D1.43%20s)