<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
Answer:
-3,89
Explanation:
First to consider is that the reaction is raising the temperature of the solution, so it's an exothermic reaction.
The enthalpy can be calculated by the following equation:
![H=m*C_{p} *ΔT](https://tex.z-dn.net/?f=H%3Dm%2AC_%7Bp%7D%20%2A%CE%94T)
First we need to calculate the mass using the density and the volume:
⇒ ![m=d*V](https://tex.z-dn.net/?f=m%3Dd%2AV)
![m=250[mL]*1,25[\frac{g}{mL}]=312,5 [g]](https://tex.z-dn.net/?f=m%3D250%5BmL%5D%2A1%2C25%5B%5Cfrac%7Bg%7D%7BmL%7D%5D%3D312%2C5%20%5Bg%5D)
Then we have Cp = 3,74 Joules/gram°K ΔT = -3,33 °C (cause of exothermic reaction)
Replacing in the formula:
![H=312,5*3,74*(-3,33)=-3891 [J]=-3,81[kJ]](https://tex.z-dn.net/?f=H%3D312%2C5%2A3%2C74%2A%28-3%2C33%29%3D-3891%20%5BJ%5D%3D-3%2C81%5BkJ%5D)
Answer:
seasonl changes at the poles only
Explanation:
A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s