Answer:
F= 600 N
Explanation:
Given that
Initial velocity ,u= 0 m/s
Final velocity ,v= 30 m/s
mass ,m = 0.5 kg
time ,t= 0.025 s
The change in the linear momentum is given as
ΔP= m (v - u)
ΔP= 0.5 ( 30 - 0 ) kg.m/s
ΔP= 15 kg.m/s
We know that from second law of Newtons
Now by putting the values
F= 600 N
Answer:
It is C on edge.
Explanation:
Because I just figured it out and got it right and because it says so in the link provided from the question.
The time taken for the tiny saliva to travel is 0.55 second.
The horizontal distance traveled at speed of 4 m/s is 2.2 m.
The horizontal distance traveled at speed of 20 m/s is 11 m.
<h3>
Time of motion of the tiny saliva</h3>
The time taken for the tiny saliva to travel is calculated as follows;
h = vt + ¹/₂gt²
where;
- v is initial vertical velocity = 0
- g is the acceleration due to gravity
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
t = √(2h/g)
Substitute the given parameters and solve for time of motion;
t = √(2 x 1.5 / 10)
t = 0.55 second
<h3>Horizontal distance traveled at speed of 4 m/s</h3>
X = Vx(t)
X = (4 m/s)(0.55)
X = 2.2 m
<h3>Horizontal distance traveled at speed of 20 m/s</h3>
X = (20)(0.55)
X = 11 m
Learn more about time of motion here: brainly.com/question/2364404
#SPJ1
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Hi!
The answers would be <u>chunking</u> & <u>short-term</u>
1. <u>Chunking </u>involves organizing and breaking down information into easier groups to expand capacity.
<h3>Explanation:</h3>
Chunking is a mental process that is observed to increase short-term memory by taking the information and categorizing it into small groups. For instance, a longer number taken as a single unit is harder to recall then when it is divided into smaller units. 235469350 is harder to instantly recall as compared to when it is chunked into 3 groups: 235 469 350.
This allows more information to be stored in, thereby increasing the capacity of the mind to store information.
2. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of <u>short-term</u> memory.
<h3>Explanation: </h3>
Short-term memory is lost after a couple of seconds or minutes, for instance even if you chunk the information, you might not recall it after 30 seconds. Rehearsing or repetition of information, either loudly or mentally, extends the time a particular information is retained.
So you depending on the number of times you repeat the number 235 469 350, the more your short term memory improves .
Hope this helps!
