(a) Let's convert the final speed of the car in m/s:
![v_f = 61 km/h = 16.9 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2061%20km%2Fh%20%3D%2016.9%20m%2Fs)
The kinetic energy of the car at t=19 s is
![K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J](https://tex.z-dn.net/?f=K%3D%20%5Cfrac%7B1%7D%7B2%7Dmv_f%5E2%3D%20%5Cfrac%7B1%7D%7B2%7D%281400%20kg%29%2816.9%20m%2Fs%29%5E2%3D2.00%20%5Ccdot%2010%5E5%20J%20%20)
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
![P= \frac{W}{\Delta t}](https://tex.z-dn.net/?f=P%3D%20%5Cfrac%7BW%7D%7B%5CDelta%20t%7D%20)
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
![P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W](https://tex.z-dn.net/?f=P%3D%20%5Cfrac%7BK%7D%7B%5CDelta%20t%7D%3D%20%5Cfrac%7B2.00%20%5Ccdot%2010%5E5%20J%7D%7B19%20s%7D%3D1.05%20%5Ccdot%2010%5E4%20W%20%20)
(c) The instantaneous power is given by
![P_i = Fv_f](https://tex.z-dn.net/?f=P_i%20%3D%20Fv_f)
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
![a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7Bv_f-v_i%7D%7B%5CDelta%20t%7D%3D%20%20%5Cfrac%7B16.9%20m%2Fs%7D%7B19%20s%7D%3D0.89%20m%2Fs%5E2%20)
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
Answer:
Friction
Explanation:
Friction is a force that slows down moving objects. If you roll a ball across a shaggy rug, you can see that there are lumps and bumps in the rug that make the ball slow down. The rubbing, or friction, between the ball and the rug is what makes the ball stop rolling. External Force is required.
Answer:
frequency of the sound = f = 1,030.3 Hz
phase difference = Φ = 229.09°
Explanation:
Step 1: Given data:
Xini = 0.540m
Xfin = 0.870m
v = 340m/s
Step 2: frequency of the sound (f)
f = v / λ
λ = Xfin - Xini = 0.870 - 0.540 = 0.33
f = 340 / 0.33
f = 1,030.3 Hz
Step 3: phase difference
phase difference = Φ
Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984
Φ = 3.9984 rad * (360°/2π rad)
Φ = 229.09°
Hope this helps!
Answer:
![\Delta v=5.77m/s](https://tex.z-dn.net/?f=%5CDelta%20v%3D5.77m%2Fs)
Explanation:
Newton's 2nd Law relates the net force <em>F</em> on an object of mass <em>m </em>with the acceleration <em>a</em> it experiments by <em>F=ma.</em> In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:
![a=\frac{F}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D)
Also, acceleration is defined by the change of velocity
in a given time t, so we have:
![a=\frac{\Delta v}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7Bt%7D)
Since we want the change in velocity, <em>mixing both equations</em> we conclude that:
![\Delta v=at=\frac{Ft}{m}](https://tex.z-dn.net/?f=%5CDelta%20v%3Dat%3D%5Cfrac%7BFt%7D%7Bm%7D)
Which for our values means:
![\Delta v=\frac{Ft}{m}=\frac{(25N)(15s)}{(65Kg)}=5.77m/s](https://tex.z-dn.net/?f=%5CDelta%20v%3D%5Cfrac%7BFt%7D%7Bm%7D%3D%5Cfrac%7B%2825N%29%2815s%29%7D%7B%2865Kg%29%7D%3D5.77m%2Fs)
A heat pump that uses work to move heat