Answer:
5 sq. root 3
Explanation:
theta= 60°
=> u sin theta = 10 × sin 60
= 10× sq. root 3/2
= 5 sq. root 3
The potential energy of the box when it gets to the top is
(mass) (gravity) (height)
= (7 kg) (9.8 m/s²) (5 m)
= 343 joules.
That's the work done against the force of gravity. Any
additional work is done against the force of friction.
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
Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B
Where 
=Mass of block B
Substitute the values
For block A
Where 
Mass of block A
Substitute the values
Hence, the friction force acting on block A=40 N
