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
A. dependent variable
As the the independent variable (i.e. number of cats being sold) increases, the dependent variable (i.e. money made) will also increase. You can have 1,000,000 cats for sale and make no money, but you cannot make money without having sold some cats first.
Hope my weird example helped!
If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11
.
Calculation:
Step-1:
It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is
=5rev/min.
It is required to find the angular acceleration of a passenger at his lowest point during the ride.
The formula required to calculate the angular acceleration is,
.
Step-2:
Now substituting the given values into the equation to get the value of the angular acceleration.

The acceleration is towards upwards that means towards the center of the wheel.
Learn more about the angular acceleration:
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Answer:
Twice as fast
Explanation:
Solution:-
- The mass of less massive cart = m
- The mass of Massive cart = 2m
- The velocity of less massive cart = u
- The velocity of massive cart = v
- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.
- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.
- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.

- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:

- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:
m*u - 2*m*v = 0
Where,
( u ) and ( v ) are opposing velocity vectors in 1-dimension.
- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:
m*u = 2*m*v
u = 2*v
Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).
Answer:
can you please translate to english?
Explanation: