<h3>a. The impulse</h3>
The impulse is 100.0 Ns
The impulse I = Ft where
- F =average force = 50.0 N and
- t = time = 2.0 s
Substituting these values into the equation, we have
I = Ft
I = 50.0 N × 2.0 s
I = 100.0 Ns
The impulse is 100.0 Ns
<h3>b. Change in momentum</h3>
The change in momentum is 100 kgm/s
Since change in momentum Δp = I where I = impulse.
Since I = 100.0 Ns,
Substituting this into the equation, we have
Δp = I
= 100.0 Ns
= 100 kgm/s
The change in momentum is 100 kgm/s
<h3>c. Mass's change in velocity</h3>
The change in velocity is 25.0 m/s
Since change in momentum Δp = mΔv where
- m = mass = 4.0 kg and
- Δv = change in velocity.
Making Δv subject of the formula, we have
Δv = Δp/m
Substituting the values of the variables into the equation, we have
Δv = Δp/m
Δv = 100.0 kgm/s/4.0 kg
Δv = 25.0 m/s
The change in velocity is 25.0 m/s
Learn more about impulse here:
brainly.com/question/25700778
Explanation:
Initial speed of blood, u = 0
Final speed of the blood, v = 26 cm/s
(a) Displacement of the blood, d = 2 cm
Let a be its acceleration. It can be calculated using third equation of motion as :
![a=\dfrac{v^2-u^2}{2s}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2s%7D)
![a=\dfrac{(26)^2}{2\times 2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B%2826%29%5E2%7D%7B2%5Ctimes%202%7D)
![a=169\ cm/s^2](https://tex.z-dn.net/?f=a%3D169%5C%20cm%2Fs%5E2)
(b) Let t is the time taken by the blood to reach its final speed. It can be calculated as :
<u>
</u>
![t=\dfrac{26\ cm/s}{169\ cm/s^2}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B26%5C%20cm%2Fs%7D%7B169%5C%20cm%2Fs%5E2%7D)
t = 0.15 s
Hence, this is the required solution.
Heya!!
For calculate final velocity, lets applicate formula
![\boxed{V=V_o+a*t}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3DV_o%2Ba%2At%7D)
<u>Δ Being Δ</u>
V = Final Velocity = ?
Vo = Initial velocity = 0 m/s
a = Aceleration = 5 m/s²
t = Time = 12 s
⇒ Let's replace according the formula:
![\boxed{V=0\ m/s +5\ m/s*12\ s}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3D0%5C%20m%2Fs%20%2B5%5C%20m%2Fs%2A12%5C%20s%7D)
⇒ Resolving
![\boxed{V=60\ m/s}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3D60%5C%20m%2Fs%7D)
Result:
The velocity after 10 sec is <u>60 meters per second (m/s)</u>
Good Luck!!
Complete Question
The question image is in the first uploaded image
Answer:
![E=\frac{KQ*4xa}{(x^2-a^2)^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BKQ%2A4xa%7D%7B%28x%5E2-a%5E2%29%5E2%7D)
Explanation:
From the question we are told that
Distance b/w Q mid point and P is given as x
Generally the equation for magnitude of the electric field at the point P is given as
![E=\frac{kQ}{d^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BkQ%7D%7Bd%5E2%7D)
where
![k=\frac{1}{4\pi e_0}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7B4%5Cpi%20e_0%7D)
![d=x^2-a^2](https://tex.z-dn.net/?f=d%3Dx%5E2-a%5E2)
Therefore
![E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20e_0%7D%20%5Cfrac%7BQ%7D%7B%28x%5E2-a%5E2%29%5E2%7D-%20%5Cfrac%7B1%7D%7B4%5Cpi%20e_0%7D%20%5Cfrac%7BQ%7D%7B%28x%5E2%2Ba%5E2%29%5E2%7D)
![E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7BQ%7D%7B4%5Cpi%20e_0%7D%20%28%5Cfrac%7B1%7D%7B%28x%5E2-a%5E2%29%5E2%7D-%20%20%5Cfrac%7B1%7D%7B%28x%5E2%2Ba%5E2%29%5E2%7D%29)
Therefore equation for magnitude of the electric field at the point P is
![E=\frac{KQ*4xa}{(x^2-a^2)^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BKQ%2A4xa%7D%7B%28x%5E2-a%5E2%29%5E2%7D)
Answer:
v_f = 10.85 m/s
Explanation:
We will apply the law of conservation of momentum here:
![m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\](https://tex.z-dn.net/?f=m_%7B1%7Dv_%7B1i%7D%20%2B%20m_%7B2%7Dv_%7B2i%7D%20%3D%20m_%7B1%7Dv_%7B1f%7D%2Bm_%7B2%7Dv_%7B2f%7D%5C%5C)
where,
m₁ = mass of roller skater = 47 kg
m₂ = mass of bag = 6 kg
v_1i = initial speed of roller skater = 12 m/s
v_2i = initial speed of the bag = 0 m/s
v_1f = final speed of the roller skater = ?
v_2f = final speed of the bag = ?
Both the bag and the skater will have same speed at the end because kater is carrying the bag:
v_1f = v_2f = v_f
Therefore, the equation will become:
![(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\](https://tex.z-dn.net/?f=%2847%5C%20kg%29%2812%5C%20m%2Fs%29%2B%286%5C%20kg%29%280%5C%20m%2Fs%29%3D%2847%5C%20kg%29%28v_%7Bf%7D%29%2B%285%5C%20kg%29%28v_%7Bf%7D%29%5C%5C564%5C%20N.s%20%3D%20%2847%5C%20kg%2B5%5C%20kg%29%28v_%7Bf%7D%29%5C%5Cv_%7Bf%7D%20%3D%20%5Cfrac%7B564%5C%20N.s%7D%7B52%5C%20kg%7D%5C%5C)
<u>v_f = 10.85 m/s</u>