This motion of all can be divided in to two, first the upward motion and second downward motion.
In case of projectiles time taken for upward motion = time taken for downward motion.
So we have time taken for upward motion = 8.14/2 = 4.07 seconds
We have equation of motion, v = u+at, where v is the final velocity , u is the initial velocity, a is the acceleration and t is the time taken.
In case of upward motion
v = 0 m/s
t = 4.07 seconds
a = -9.8 
Substituting
So initial velocity of ball = 39.886 m/s.
Given that,
Number of turns, N = 25
Magnetic field, B = 1.5 T
The coil has a crosssectional area of 0.80 m2
The coil exits the field in 1.0 s.
To find,
Induced emf in the coil.
Solution,
The induced emf in the coil is given by :
So, the induced emf in the coil is 30 V.
Answer:
θ = 47.75º East of North.
Explanation:
- Assuming no external forces acting during the collision, total momentum must be conserved.
- Since momentum is a vector, if we project it along E-W and N-S axes, the momentum components along these axes must be conserved too.
- So, for the N-S axis, we can write the following equation:
- Since the car moving due east has no speed component along the N-S axis, the initial momentum along this axis is simply:
where m₁ = 1570 kg, v₁₀ = 156 km/h
- In order to work with the same units, we need to convert the speed in km/h to m/s, as follows:
- Replacing by the values in the left side of (1), we get:
- Since the collision is inelastic, both cars stick together, so we can write the right side of (1) as follows:
- From (4) and (5) , we can solve for VfNorth:
- We can repeat exactly the same process for the E-W axis:
- Since the car moving due north has no speed component along the E-W axis, the initial momentum along this axis is simply:
- As we did with v₁₀, we need to convert v₂₀ from km/h to m/s, as follows:
- Replacing by the values in the left side of (7), we get:
- Since the collision is inelastic, both cars stick together, so we can write the right side of (7) as follows:
- From (10) and (11) , we can solve for VfEast:
- In order to find the angle East of North of the velocity vector, as we know the values of the horizontal and vertical components, we need just to apply a little bit of trigonometry, as follows:
- The angle θ, East of North, is simply tg⁻¹ (1.1):
θ = tg⁻¹ (1.1) = 47.75º E of N.
ANSWER:
How to Test Hypotheses
State the hypotheses. Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis. ...
Formulate an analysis plan. The analysis plan describes how to use sample data to accept or reject the null hypothesis. ...
Analyze sample data. ...
Interpret the results.
