The last one is correct (D)
Answer:
12N
Explanation:
when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied
Answer:
Explanation:
Given that,
Mass of star M(star) = 1.99×10^30kg
Gravitational constant G
G = 6.67×10^−11 N⋅m²/kg²
Diameter d = 25km
d = 25,000m
R = d/2 = 25,000/2
R = 12,500m
Weight w = 690N
Then, the person mass which is constant can be determined using
W =mg
m = W/g
m = 690/9.81
m = 70.34kg
The acceleration due to gravity on the surface of the neutron star is can be determined using
g(star) = GM(star)/R²
g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²
g (star) = 8.49 × 10¹¹ m/s²
Then, the person weight on neutron star is
W = mg
Mass is constant, m = 70.34kg
W = 70.34 × 8.49 × 10¹¹
W = 5.98 × 10¹³ N
The weight of the person on neutron star is 5.98 × 10¹³ N
The x- and y-coordinates are 9142.57 m and -304.425 m
<u>Explanation:</u>
As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell
in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of
and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by


For motion with constant acceleration, we know

Along the horizontal, x-axis, we might write this as

Measuring distances relative to the firing point means

we know that,

or,

By applying the values, we get,

The acceleration of gravity is vertically downward and is
, hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

we know,
and
, so,

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m