Answer:
The new speed of the ball is 176.43 m/s
Explanation:
Given;
mass of the ball, m = 7 kg
initial speed of the ball, u = 5 m/s
applied force, F = 300 N
time of force action on the ball, t = 4 s
Apply Newton's second law of motion;
![F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v = \frac{Ft}{m} + u](https://tex.z-dn.net/?f=F%20%3D%20ma%20%3D%20%5Cfrac%7Bm%28v-u%29%7D%7Bt%7D%5C%5C%5C%5Cm%28v-u%29%20%3D%20Ft%5C%5C%5C%5Cv-u%20%3D%20%5Cfrac%7BFt%7D%7Bm%7D%5C%5C%5C%5Cv%20%3D%20%20%5Cfrac%7BFt%7D%7Bm%7D%20%2B%20u)
where;
v is new speed of the ball
![v = \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7BFt%7D%7Bm%7D%20%2B%20u%5C%5C%5C%5Cv%20%3D%5Cfrac%7B300%2A4%7D%7B7%7D%20%2B%205%5C%5C%5C%5Cv%20%3D%20176.43%20%5C%20m%2Fs)
Therefore, the new speed of the ball is 176.43 m/s
Answer:
a) 43.20V
b) 2.71W/s
c) 40.25s
d) 7.77Nm
Explanation:
(a) The emf of a rotating coil with N turns is given by:
![emf=NBA\omega sin(\omega t)](https://tex.z-dn.net/?f=emf%3DNBA%5Comega%20sin%28%5Comega%20t%29)
N: turns
B: magnitude of the magnetic field
A: area
w: angular velocity
the emf max is given by:
![emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V](https://tex.z-dn.net/?f=emf_%7Bmax%7D%3DNBA%5Comega%3D%2850%29%281.80T%29%280.200m%2A0.100m%29%2824.0rad%2Fs%29%5C%5C%5C%5Cemf_%7Bmax%7D%3D43.20V)
(b) the maximum rate of change of the magnetic flux is given by:
![\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D%3D%5Cfrac%7Bd%28A%5Ccdot%20B%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%28ABcos%5Comega%20t%29%3DAB%5Comega%20sin%28%5Comega%20t%29%5C%5C%5C%5C%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D_%7Bmax%7D%3D%28%5Cpi%280.200%2A0.100%29%29%281.80T%29%2824.0rad%2Fs%29%3D2.71%5Cfrac%7BW%7D%7Bs%7D)
(c) ![emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V](https://tex.z-dn.net/?f=emf%28t%3D0.050s%29%3D%2850%29%281.80T%29%280.200m%2A0.100m%29%2824rad%2Fs%29sin%2824.0rad%2Fs%280.050s%29%29%5C%5C%5C%5Cemf%28t%3D0.050s%29%3D40.26V)
(d) The torque is given by:
![\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm](https://tex.z-dn.net/?f=%5Ctau%3DNABIsin%5Ctheta%5C%5C%5C%5CNAB%5Comega%3Demf_%7Bmax%7D%5C%5C%5C%5C%5Ctau%3D%5Cfrac%7Bemf_%7Bmax%7D%7D%7B%5Comega%7D%5Cfrac%7Bemf_%7Bmax%7D%7D%7BR%7D%5C%5C%5C%5C%5Ctau%3D%5Cfrac%7B%2843.20V%29%5E2%7D%7B%2824.0rad%2Fs%29%2810.0%5COmega%29%7D%3D7.77Nm)
Answer:
When a bridge is used for long time It lises its elastic .Therefore ,the amount of strain in the bridge for a given stress will become large and ultimately, the bridge may collapse .This is why the bridge are declared unsafe after a long time.
Explanation:
Answer:
Pure Food and Drug Act of 1906
Explanation:
The Pure Food and Drug Act of 1906 prohibited the sale of misbranded or adulterated food and drugs in interstate commerce and laid a foundation for the nation's first consumer protection agency, the Food and Drug Administration
Answer:
Work done is 93.75J
Explanation:
Given mass, m = 30kg; Time = 5 seconds; force = 15N; u= 0
Acceleration, a = F/m ⇒ 15/30
⇒ 0.5ms
Displacement , s= ut+1/2at²
⇒ 0 +1/2 (0.5*5*5)
⇒ 0 + 1/2(0.5*25)
⇒ 0 + 1/2(12.5)
⇒ 0 + 6.25
⇒ 6.25m
work done = Fs=15*6.25= 93.75 J