Vector trigonometry can be used for this problem. Since the horizontal component is 12 meters per second, this is technically the hypotenuse (actual initial velocity) multiplied to cosine of 40 degrees. Therefore, to find the hypotenuse, we must divide 12 by cosine 40degrees. cos(40)= 0.766, and 12/0.766 = approximately 15.664, therefore our answer is (3) 15.7 m/s
Answer:
The impulse transferred to the nail is 0.01 kg*m/s.
Explanation:
The impulse (J) transferred to the nail can be found using the following equation:

Where:
: is the final momentum
: is the initial momentum
The initial momentum is given by:

Where 1 is for the hammer and 2 is for the nail.
Since the hammer is moving down (in the negative direction):
And because the nail is not moving:

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:
Since the hammer and the nail are moving in the negative direction:
=
= -9.7 m/s
Finally, the impulse is:

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.
I hope it helps you!
"The number of waves per second will increase" is the statement among the choices given in the question that <span>will be true if you increase the frequency of a periodic wave. The correct option among all the options that are given in the question is the first option or option "A". I hope that the answer has helped you.</span>
Answer:Angular width of the electromagnetic wave after it emerges from between the buildings is 2.761 degree
Explanation:
The wavelength of the electromagnetic wave is

Consider two buildings as a single slit, for single slit diffraction dark fringes are located an angle of

where
is frequency,
is space between buildings.
For the first dark fringes

Angular width of the electromagnetic wave after it emerges from between the buildings is

Explanation:
Given that,
Mass of the ball, m = 1.2 kg
Initial speed of the ball, u = 10 m/s
Height of the floor from ground, h = 32 m
(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :



v = -25.04 m/s (negative as it rebounds)
The impulse acting on the ball is equal to the change in momentum. It can be calculated as :


J = -42.048 kg-m/s
(b) Time of contact, t = 0.02 s
Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :



F = 0.8409 N
Hence, this is the required solution.