Answer:
We are given x= bt +ct²
So
A. bxt= m
Because m/s*s= m
So b= m/s and c= m/s²
B.
x= bt-ct²
So at x=0 t=0
x=0 t= 2
We have
bt = ct² so t = b/c at x= 0
So b-2ct= 0
B. To find velocity we use
dx / dt = b - 2 Ct
C. At rest wen V= 0
We have t= b/2c
D. To find acceleration we use
dv / dt = - 2C
Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is 
Here c is the peed of light with value 
The time taken to reach Retah from earth is 
The velocity of the spacecraft is mathematically evaluated as
substituting values
The time elapsed in the spacecraft’s frame is mathematically evaluated as
substituting value
![T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }](https://tex.z-dn.net/?f=T%20%20%3D%20%2090000%20%2A%20%20%5Csqrt%7B%201%20-%20%20%5Cfrac%7B%5B2.4%2A10%5E%7B8%7D%5D%5E2%7D%7B%5B3.0%2A10%5E%7B8%7D%5D%5E2%7D%20%7D)
=> 
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Answer:
Wn = 9.14 x 10¹⁷ N
Explanation:
First we need to find our mass. For this purpose we use the following formula:
W = mg
m = W/g
where,
W = Weight = 675 N
g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²
m = Mass = ?
Therefore,
m = (675 N)/(9.8 m/s²)
m = 68.88 kg
Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:
gn = (G)(Mn)/(Rn)²
where,
gn = acceleration due to gravity on surface of neutron star = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg
Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m
Therefore,
gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)
gn = 13.27 x 10¹⁵ m/s²
Now, my weight on neutron star will be:
Wn = m(gn)
Wn = (68.88)(13.27 x 10¹⁵ m/s²)
<u>Wn = 9.14 x 10¹⁷ N</u>
