Answer:
v_f = -25.9 m/s
Explanation:
- The complete question is as follows:
" Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction.
Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s to the baseball?"
Given:
- mass of baseball m = 0.145 kg
- Speed before impact v_i = 32 m/s
- Speed after impact v_f
- Impulse applied by the bat I = - 8.4Ns
Find:
What is the ball's velocity just after leaving the bat
Solution:
- Impulse is the change in linear momentum of the ball according to Newton's second law of motion:
I = m* ( v_f - v_i )
- Taking the + from pitcher to batsman and - from batsman to pitcher.
- Plug in the values:
-8.4 = 0.145* ( v_f - (32) )
v_f = -57.93103 + 32
v_f = -25.9 m/s
Answer:
Explanation:
Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:
Ratio of kilomoles oxygen to kilomole nitrogen is:
It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.
The empirical formula for the compound is:
Answer:
The magnitude of electric field is 
in downward.
Explanation:
Given that,
Force 
Charge q= -2.00 μC
We know that,
Charge is negative, then the electric field in the opposite direction of the exerted force.
We need to calculate the magnitude of electric field
Using formula of electric force
Negative sign shows the opposite direction of electric force.
Hence, The magnitude of electric field is in downward.
Answer:
Make sure you're getting paid. Unpaid interships are a complete and total waste.
Explanation:
The answer is both.
For kinetic energy:
KE = 1/2*m*v^2 = 0.5*20,000 grams*5 = 50,000 J
For gravitational potential energy:
Pe = mgh = 20,000 grams*9.81 m/s^2*2 m = 392.2 J
