Answer:
06 Hours
Explanation:
As per the details given in the question it self, the neutron star X-1 is revolving around its companion star. The orbital period is 1.7 years which means it will complete the revolution in 1.7 years. During the movement in the orbit we will be able to detect the x-rays except for the time when it goes behind the companion star and eclipsed by it as seen from Earth.
Since the x-rays disappear completely for around 6 hours. This clearly means that eclipse period is 06 hours.
Answer:
Explanation:
Potential difference between two points in constant electric field is given by the formula
here we know that
also we know that
now we have
now change in potential energy is given as
You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final² = (V)initial²+2a(dx)
Given:
a=0.500m/s²
dx=6.32 m
(V)intial=0m
(V)final= UNKNOWN
(V)final= 2.51396m/s
Explanation:
Initial speed of the incident water stream, u = 16 m/s
Final speed of the exiting water stream, v = -16 m/s
The mass of water per second that strikes the blade is 48.0 kg/s.
We need to find the magnitude of the average force exerted on the water by the blade. The force acting on an object is given by :
Here,
So, the magnitude of the average force exerted on the water by the blade is 1536 N.
Answer: true
Explanation: a force can be anything that effects an object, as long as the object moves
b) What are the magnitude and direction of the velocity of the car at t= 8 sec?
c) What is the magnitude and direction of cars acceleration at t=8 sec
Given:
At max. flow density,
(a)
→
By substituting the value,
→
(b)
The speed will be:
→
(c)
The density be:
→
Thus the responses above are correct.
Find out more information about density here:
Answer:
a) capacity of the highway section = 4006.4 veh/h
b) The speed at capacity = 25 mph
c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi
Explanation:
q = 50k - 0.156k²
with q in veh/h and k in veh/mi
a) capacity of the highway section
To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.
q = 50k - 0.156k²
At maximum flow density, (dq/dk) = 0
(dq/dt) = 50 - 0.312k = 0
k = (50/0.312) = 160.3 ≈ 160 veh/mi
q = 50k - 0.156k²
q = 50(160.3) - 0.156(160.3)²
q = 4006.4 veh/h
b) The speed at the capacity
U = (q/k) = (4006.4/160.3) = 25 mph
c) the density when the highway is at one-quarter of its capacity?
Capacity = 4006.4
One-quarter of the capacity = 1001.6 veh/h
1001.6 = 50k - 0.156k²
0.156k² - 50k + 1001.6 = 0
Solving the quadratic equation
k = 21.5 veh/mi or 299 veh/mi
Hope this Helps!!!