The distance between the two cities is 513.24 km.
<h3>Time of motion when the two trains meet</h3>
The time spent on the journey when the two trains meet is calculated as follows;
(Va - Vb)t = d
where;
- d is the distance between the trains before meeting
(76 - 65)t = 40
11t = 40
t = 40/11
t = 3.64 hr
<h3>Distance traveled by the fast train</h3>
d1 = 76 km/h x 3.64 h
d1 = 276.64 km
<h3>Distance traveled by the slow train</h3>
d2 = 65 km/h x 3.64 h
d2 = 236.6 km
The distance between the two cities = 276.64 km + 236.6 km
= 513.24 km
Learn more about relative velocity here: brainly.com/question/17228388
Answer:
200 N
Explanation:
For a body moving in uniform circular motion, the force acting on it will be <em>centripetal force</em> and its direction is <em>radially inward</em> , pointing to the center.
The radially inward acceleration, or the centripetal acceleration is given by :
a = v² / r
where v is the speed at which the body is moving and r is the radius of the circle
Given-
m = 55kg
v = 14.1 m/s
r= 55m
We know that F = ma
⇒ F = m ( v²/ r )
⇒ F = 55 x 14.1 x 14.1 / 55
⇒ F =14.1 x 14.1 = 200 N
∴ <em>The force acting is 200 N</em>.
Answer:
Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
So, the new work is more than 130 J.
