Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:
where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,
Now, we will calculate the value of g on the neutron star:
where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,
Therefore, the weight on the surface of the neutron star will be:
<u>W' = 1.66 x 10¹⁴ N</u>
Answer: are fluffy
form 1-1.5 miles above Earth
can extend up to 10 miles into the sky
Explanation:
Cumulus clouds are referred to as the clouds that have have flat bases and its appearance resembles that of floating cotton.
Cumulus clouds simply shows a fair weather. Cumulus cloud also forms 1-1.5 miles above Earth and can also extend up to 10 miles into the sky.
Answer:
look for the "W" in the North. Remember that, the "W" may be on its side or inverted to form an "M." If you can recognize the Big Dipper (Ursa Major), the two stars at the edge of the Dipper point toward the North Star (Polaris)
Explanation:
Momentum is conserved throughout this scenario.
Before the man does anything, the total momentum of him and his book is zero. So we know that it'll be zero after he throws the book.
Momentum = (mass) x (velocity)
The man gives the book (1.2 kg)x(10 m/s north) = 12 kg-m/s north
of momentum.
Since the total momentum must be zero, the man himself picks up 120 kg-m/s of momentum south.
(his mass)x(his v) = 120 kg-m/s south = (770 kg-m/s^2/9.8 m/s^2)x(V).
His velocity southward = (120 x 9.8) / (770) m/s .
He needs to reach the shore 10m away.
Time = distance/speed
= (10 x 770) / (120 x 9.8) seconds
= 6.55 seconds
Answer:
a) Angular width of the central maximum = 0.5294°
b) The distance of the third dark band from the central bright band = 0.72 cm
Explanation:
a) Width of the slit, d = 0.1 mm = 0.0001 m
Wavelength of the light,
The distance of the screen, R = 1 m
where is the angle at which first minima is visible
The angular width of the central maximum =
Angular width of the central maximum = 2 * 0.2647
Angular width of the central maximum = 0.5294°
b) d = 0.1 mm = 0.0001 m
R = 40 cm = 0.4 m
The distance from central bright band to third dark band, is given by the formula:
y = 0.72 cm